FE CBT Mechanical Problems Solutions

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NCEES Fundamentals of Engineering (FE) Examination

FE-CBT –Mechanical

Dr. Sri Susarla

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Mathematics Total Questions 6-9

A. B. C. D. E. F.

Analytic geometry Calculus Linear algebra Vector analysis Differential equations Numerical methods

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PART-I QUESTION 1. The center of the ellipse (𝑥 + 𝑦 − 2)2 (𝑥 − 𝑦)2 + = 1 𝑖𝑠 9 16 A. B. C. D.

(0,0) (1,1) (1,0) (0,1)

QUESTION 2. The length of the tangent from (0, 0) to the circle 2(x2 +y2) + x –y + 5 = 0 is A. B. C. D.

√5 √5/2 √2 √5/2

QUESTION 3. The area enclose by the parabola y2 = 8x and the line y =2x is: 4

A. 3 B. ¾ C. ¼ D. ½

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QUESTION 4. The eccentricity of the ellipse 9x2 + 5y2 -30y = 0 is A. B.

1 3 2 3

C. ¾ D. ½

QUESTION 5. If the length of the major axis of an ellipse is three times the length of minor axis, its eccentricity is: A. 1/3 1 B. C. D.

√3 1

√2 2√2 3

QUESTION 6. The length of the latus rectum of the hyperbola x2 – 4y2 = 4 is A. B. C. D.

1 2 3 4

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QUESTION 7. The area of the triangle formed by the coordinate axes and the line 4x + 5y = 20 is: A. 5 B. 10 C. 15 D. 20

QUESTION 8. The angle between the lines formed by joining (2, -3), (-5, 1) and (7,-1) and (0, 3) is A. B. C. D.

π/2 π/4 0 π/6

QUESTION 9. Axis of the parabola x2 – 3y – 6x + 6 = 0 is: A. B. C. D.

x = -3 x=3 y = -1 y=1

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QUESTION 10. The eccentricity of the hyperbola 9x2 - 16 y2 – 72x -32y – 16 = 0 is: A. B. C. D.

5 4 4 5 9 16 16 9

QUESTION 11. Area bounded by the curves y = x and y = x3 is: A. ¼ 1 B. 6 C. ½ 1 D. 12

QUESTION 12. The area bounded by the curve 𝑦 = 1 + A. B. C. D.

8 𝑥2

and x =2 and x = 4 is:

2 3 4 5

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QUESTION 13. A polygon has 35 diagonals. The number of sides of the polygon is: A. B. C. D.

10 15 20 25

QUESTION 14. The eccentricity of the ellipse 9x2 + 16y2 = 144 is: A. B. C. D.

4 √7 2 √7 √7 4 √7 3

QUESTION 15. An ellipse has the coordinate axes as its axes on its foci at (4,0) and its eccentricity is 4/5. The equation of the ellipse is: A. B. C. D.

𝑥2

+ 25 𝑥2 9 𝑥2 5 𝑥2 4

+ + +

𝑦2 9 𝑦2 25 𝑦2 4 𝑦2 5

=1 =1 = 25 = 25

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QUESTION 16. The equation of a circle with center (4,1) and touching the tangent 3x + 4y -1 = 0 is: A. B. C. D.

x2 + y2 -8x -2y -8 = 0 x2 + y2 -8x -2y +8 = 0 x2 + y2 -8x +2y +-8 = 0 x2 + y2 -8x -2y +4 = 0

QUESTION 17. The equation of the directrix to the parabola y2 – 2x - 6y – 5 = 0 is: A. B. C. D.

2x + 15 = 0 x+5=0 2x + 3 = 0 x+2=0

QUESTION 18. The equation r = 5cos θ + 12 sin θ represents: A. B. C. D.

A circle An ellipse A parabola A Straight line

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QUESTION 19. The equation of the straight line making an intercept of 3 units on the y-axis, and inclined at 45o to the x-axis is: A. B. C. D.

y=x–1 y=x+3 y= 45x + 3 y = x + 45

QUESTION 20. The vortex of the parabola x2 + 12x – 9y = 0 is A. B. C. D.

(6,-4) (-6,4) (6,4) (-6,-4)

QUESTION 21. The focus of the parabola y2 – 4y - 8x - 4 = 0 is: A. B. C. D.

(1,1) (1,2) (2,0) (2,2)

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QUESTION 21 A. 1

1

3

The equation 𝑟 = 8 + 8 𝑐𝑜𝑠𝜃 represents: A. B. C. D.

Circle Parabola Ellipse Hyperbola

QUESTION 22. The equation of the tangent to the curve y = x3 – 2x + 7 at point (1,6) is: A. B. C. D.

y=x+5 x+y=7 2x + y = 8 X + 2y = 13

QUESTION 23. The angle between the curves y2 = 4x and x2 = 4y at (4,4) is: 1

A. 𝑡𝑎𝑛−1 (2) 3

B. 𝑡𝑎𝑛−1 (4) 1

C. 𝑡𝑎𝑛−1 (3) D.

𝜋 2

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QUESTION 24. The equation of the tangent to the circle x2 + y2 + 6x + 4y -3 = 0 at (1,-2) is: A. B. C. D.

y+1=0 y+2=0 y+3=0 y–2=0

QUESTION 25. The equation 16x2 + y2 + 8xy – 74x – 78y + 212 = 0 represents: A. B. C. D.

Circle Parabola Ellipse Hyperbola

QUESTION 26. The equation of the curve represents: 1 𝜃 = 2𝑠𝑖𝑛2 𝑟 𝑟 A. B. C. D.

Straight line Parabola Circle Ellipse

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QUESTION 27. The equation of the tangent to the curve 6y = 7 – x3 at (1,1) is: A. B. C. D.

2x + y = 3 x + 2y = 3 x + y = -1 x+y+2=0

QUESTION 28. The equation of the parabola with the focus (3,0) and the directrix x + 3 is: A. B. C. D.

y2= 3x y2= 6x y2= 12x y2= 2x

QUESTION 29. If e and e’ are the eccentricities of the ellipse 5x2 + 9y2 = 45 and the hyperbola 5x2 – 4y2 = 45, respectively, then ee’ = A. B. C. D.

1 4 5 9

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QUESTION 30. The angle between the curves y = sin x and y = cos x is: A. B. C. D.

𝑡𝑎𝑛−1 (2√2) 𝑡𝑎𝑛−1 (3√2) 𝑡𝑎𝑛−1 (3√3) 𝑡𝑎𝑛−1 (5√2)

QUESTION 31. The eccentricity of the conic 36x2 + 144y2 - 36x – 96y -119 = 0 is: A. B. C. D.

√3 2 1 2 √3 4 1 √3

QUESTION 32. 1

The polar equation cos 𝜃 + 7 sin 𝜃 = 𝑟 represents: A. B. C. D.

Circle Parabola Straight line Hyperbola

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QUESTION 33. The center of the circle r2 – 4r (cos θ + sinθ) -4 = 0 is: A. B. C. D.

(1,1) (-1,-1) (2,2) (-2,-2)

QUESTION 34. The Cartesian form of the polar equation 𝜃 = 𝑡𝑎𝑛−1 2 is: A. B. C. D.

x = 2y y = 2x x = 4y y = 4x

QUESTION 35. The equation of the units circle concentric with x2 + y2 -8x +4y -8 = 0 is: A. B. C. D.

x2 + y2 -8x +4y - 8 = 0 x2 + y2 -8x +4y + 8 = 0 x2 + y2 -8x +4y - 28 = 0 x2 + y2 -8x +4y + 19 = 0

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QUESTION 36. The smallest possible value of x satisfying the equation: logcos x (sin x) + logsin x (cos x) = 2 is A. B. C. D.

0 1 π/4 π/2

QUESTION 37. The value of the expression: log4 (x3 + x2) – log4 (x + 1) = 2 A. B. C. D.

1 2 3 4

QUESTION 38. log 3√2 324 = A. B. C. D.

3 4 6 8

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QUESTION 39. If x = 27 and y = log3 4, then xy = A. 16 B. 64 C. 128 D. 256

QUESTION 40. log8128 = 1

A. 16 B. 16 3 C. 7 D.

7 3

QUESTION 41. lim

𝑥 →1

A. B. C. D.

√𝑥 − 1 + √𝑥 − 1 √𝑥 2 − 1

=

½ √2 1 1/√2

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QUESTION 42. lim

𝑥→0 √𝑥

A. B. C. D.

𝑥 +4−2

=

½ √2 4 8

QUESTION 43. [3𝑥 2 + 1] lim = 𝑥→∞ [2𝑥 2 + 1] 2

A. − 3 3

B. − 2 2

C.

3 3

D.

2

QUESTION 44. 2𝑥 + 7 sin 𝑥 = 𝑥→∝ 4𝑥 + 3 cos 𝑥 lim

A. 1 B. -1 C. ½ 1

D. − 2

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QUESTION 45. 3𝑥+4

𝐴

𝐵

If 𝑥 2 − 3𝑥+2 = 𝑥−2 − 𝑥−1 , then values of A and B are: A. B. C. D.

7,10 10, 7 10, -7 -10, 7

QUESTION 46. If

1 (1−2𝑥)(1+3𝑥)

A. B. C. D.

=

𝐴

+ 1−2𝑥

𝐵 1+3𝑥

, then 2B =

A 2A 3A 4A

QUESTION 47. 𝑥3 𝐵 𝑐 𝐷 =𝐴+ + + (2𝑥 − 1)(𝑥 + 2)(𝑥 − 3) 2𝑥 − 1 𝑥 + 2 𝑥 − 3 A= A. ½ 1 B. − 50 8

C. − 25 D.

25 27

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QUESTION 48. The roots of the equation x3 – 3x – 2 = 0 are: A. B. C. D.

-1,1-2 -1,-1,2 -1,2,-3 -1,-1,-2

QUESTION 49. 𝑑2 𝑦

If x = t2, y = t3, then 𝑑𝑥 2 =

A. 3/2 B. C. D.

3 4𝑡 3 2𝑡 3𝑡 2

QUESTION 50. 𝑑𝑦

If 3 sin (xy) + 4 cos (xy), then 𝑑𝑥 = A. B. C. D.

–y/x tan (x+y) x y

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QUESTION 51. The solution of the differential equation is: 𝑑𝑦 𝑑𝑥

A. B. C. D.

=

(1+ 𝑦 2 ) 1+𝑥 2

y –x = c(1 + xy) y + x = c (1 + xy) y + x = c ( 1- xy) y –x = c( (1- xy)

QUESTION 52. The solution of the differential equation: 𝑑𝑦

− 𝑑𝑥

2𝑥𝑦 1+𝑥 2

= 0 is

A. y = A (1 + x2) 𝐴 B. y =1+𝑥 2 C. y = A√1 + 𝑥 2 D. y = A/√1 + 𝑥 2

QUESTION 53. The order of the differential equation is: 𝑑𝑦 𝑑𝑦 ( )4 + ( )2 + 𝑦 4 = 0 𝑖𝑠 𝑑𝑥 𝑑𝑥 A. B. C. D.

1 2 3 4

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QUESTION 54. 𝑑𝑦

The solution of 2𝑥𝑦 𝑑𝑥 = 1 + 𝑦 2 is A. B. C. D.

1 - y2 = cx 1 + y2 =cx 1 – x2 = cy 1 + x2 = cy

QUESTION 55. The solution of x dx + y dy = x2ydy – xy2 dx is: A. B. C. D.

x2 -1 = c(1+ y2) x2 + 1 = c(1- y2) x2 - 1 = c(1+ y3) x2 +1 = c(1+ y3)

QUESTION 56. 𝑑𝑦

The solution of 𝑥 2 + 𝑦 2 𝑑𝑥 = 4 is: A. B. C. D.

x2 + y2 = 12x + c x2 + y2 = 3x + c x3 + y3 = 3x + c x3 + y3 = 12x + c

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QUESTION 57. 𝑑𝑦

𝑦

The solution of 𝑑𝑥 + 3 = 1 is A. B. C. D.

y = 3 + cex/3 y = 3 + ce-x/3 3y = c + ex/3 3y = c + e-x/3

QUESTION 58. 𝑑𝑦

The solution of 𝑦 + 𝑥 2 = 𝑑𝑥 is: A. B. C. D.

y + x2 + 2x + 2 = cex y + 2x2 + 2x + 2 = cex y + x2 + x + 2 = ce2x y2 + x2 + 2x + 2 = cex

QUESTION 59. The maximum value of x3 -3x in the interval [0,2] is: A. B. C. D.

-2 0 1 2

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QUESTION 60. The maximum value of f(x) = 2x3 – 21x2 + 36x + 20 in the interval 0 ≤ x ≤ 2 is: A. B. C. D.

30 32 37 44

QUESTION 61. A particle is projected vertically upward follows the relation S = 60 t – 16 t2. The velocity (m/s) of the particle when hits the ground is: A. B. C. D.

30 45 60 90

QUESTION 62. A particle moving along a straight line as S = 6t – ½ t3, the maximum velocity (m/s) during the motion is: A. 3 B. 6 C. 9 D. 12

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QUESTION 63. The distance travelled by a particle is given by x (m) = t3 – 12t2 + 6t + 8. The velocity (m/s) of the particle, when acceleration is zero equals to: A. B. C. D.

-48 -42 42 48

QUESTION 64. In a triangle ABC, the maximum value of cos A + cos B + cos C is equal to: A. B. C. D.

½ 1 3/2 2

QUESTION 65. The values of θ (0 < θ < 360o) that satisfies the equation are: cosec θ + 2 = 0 A. B. C. D.

210, 300 340, 300 210, 240 210, 330

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QUESTION 66. If x = y cos (2π/3) = z cos (4π/3), then xy + yz + ax = A. B. C. D.

-1 0 1 2

QUESTION 67. If sinθ1 + sinθ2 + sinθ3 = 3, then cosθ1 + cosθ2 + cosθ3 = A. B. C. D.

0 1 2 3

QUESTION 68. If A lies in the third quadrant and 3tan A – 4 = 0, then 5 sin 2A + 3 Sin A + 4 cos A

A. 0 B. − C. D.

24 5

24 5 48 5

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QUESTION 69. If sin x + sin2x = 1, then Cos8x + 2 cos6x + cos4x =

A. B. C. D.

-1 0 1 2

QUESTION 70. 1

2

tan-1 (4) + tan-1 (9) = 3

A. ½ cos-15 3

B. ½ tan-1 5 13

C. ½ tan ( − 5 ) D. tan-1 ½

QUESTION 71. The domain of sin-1 x is: A. B. C. D.

(0, 2π) (-1, 1) (-α, α) (1,1)

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QUESTION 72. 2𝜋

4𝜋

8𝜋

14𝜋

The value of cos (15 ) cos ( 15 ) cos ( 15 ) cos ( 15 ) = A.

1 16 1

B. 8 C. ¾ 1 D. 12

QUESTION 73. sin 𝜃 + sin 2𝜃 = 1 + 𝑐𝑜𝑠𝜃 + 𝑐𝑜𝑠 2𝜃 A. B. C. D.

sin θ cos θ tan θ cot θ

QUESTION 74. If tan θ + cot θ = 2, then sin θ = A. ± B. ± C. ± D. ±

1 √2 1 √3 1 √4 1 √5

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QUESTION 75. In a ∆ABC, if b = 20, c = 21 and sin A = 3/5, then a = A. B. C. D.

12 13 14 15

(No solutions for problems 76 through 104) QUESTION 76. The distance between the foci of the hyperbola x2 – 3y2 - 4x – 6y – 11 = 0 is: A. 4 B. 6 C. 8 D. 12

QUESTION 77. If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8, then the eccentricity is: A. ¼ B. ½ 3 C. 5 D.

4 5

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QUESTION 78. A polygon has 54 diagonals. The number of sides is: A. 7 B. 9 C. 10 D. 12

QUESTION 79. The equation of the line passing through (-5,2) and (4,-3) is: A. B. C. D.

5x + 9y + 7 = 0 2x + 3y -7 = 0 x+3–y=0 2x + 2y + 7 = 0

QUESTION 80. The characteristic equation 4x2 – 11xy + 26y2 + 3x + 4y – 12 = 0 is: A. B. C. D.

Straight line Circle Ellipse Hyperbola

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QUESTION 81. The angle (degrees) between the lines 2x – y + 1 = 0 and x – y + 3 = 0 is: A. 9 B. 12 C. 18 D. 24

QUESTION 82. A circle of radius 5 is drawn about the origin as center. The equation of the tangent to the circle at the point (-3,-4) is: A. B. C. D.

2x + 4y = 16 3x – 4y = 16 3x – 4y = 25 3x + 4y = 25

QUESTION 83. The distance from the line 4x – 3y + 15 = 0 to the point (-3,6) is: A. B. C. D.

1 2 3 4

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QUESTION 84. The equation of the circle (2,-3) and tangent to the line 12x - 5y + 13 = 0 is: A. B. C. D.

x2 + y2 – 4x + 6y – 3 = 0 x2 + y2 + 4x + 6y + 3 = 0 x2 + y2 – 6x + 8y – 13 = 0 x2 + y2 – 14x + 16y –23 = 0

QUESTION 85. The center of the equation is 9x2 + 9y2 – 12x + 54y + 49 = 0 is: 1

A. ( ,2) 2 2

B. (3 , 4) C. (4,6) 2

D. (3 , −3)

QUESTION 86. The length of the tangent from the point (3,-2) to the circle 2x2 + 2y2 – x + 3y – 1 = 0 is : A. B. C. D.

1.8 2.8 3.8 4.8

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QUESTION 87. The slope of the curve 3x2 – 8xy + y2 – 6x + 3y – 21 = 0 at (1,-3) is: A. B. C. D.

11 24 23 24 24 11 22 17

QUESTION 88. The tangent to the ellipse x2 + 4y2 = 100, parallel to the line 3x + 8y – 7 = 0 is: A. B. C. D.

3x + 8y -50 = 0 2x + 2y = 11 3x – 8y + 50 = 0 2x + 8y – 45 = 0

QUESTION 89. The equation of the hyperbola whose directrix is the line 3x – 4y + 14 = 0 is: A. B. C. D.

8x2 -24xy + 15y2 + 90x – 108y + 183 = 0 6x2 -24xy + 5y2 + 90x – 108y - 183 = 0 8x2 -12xy - 15y2 + 90x + 108y + 183 = 0 x2 -24xy + y2 + 90x – 108y + 183 = 0

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QUESTION 90. The intercept of the line 3x – 6y – 8z + 24 = 0 is: A. B. C. D.

3, -4,-8 1,1,1 2,3,4 3,4,-8

QUESTION 91. The distance from the plane whose intercepts are 3,-2,1 to the point (2,3-4) is: A. B. C. D.

-2 -3 -5 -9

QUESTION 92. 2x + 3y = 8 3x + 2y = 2 The values of x and y: A. B. C. D.

0, 2 2, 2 -2, 4 3,5

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QUESTION 93. The roots of the equation x3 = x2 + 6x are: A. B. C. D.

0,3,-2 1,0,0 2,3,1 1,1,1

QUESTION 94. The roots of the equation 12x2 – 8x – 15 = 0 are: A. ½, ½ B. C.

2 3 3

5

,−6 5

,−6 2 1

D. ¼, − 4

QUESTION 95. The roots of the equation x4 – 13x2 + 36 = 0 are: A. B. C. D.

2,-2,-3 2,2,3 1,2,-3 1,-4,5

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QUESTION 96. √4𝑥 + 8 + 1 = 𝑥 The value of x = A. B. C. D.

1 -4 5 7

QUESTION 97. 2x + 4y – 3z = -9 3x + y -2z = 4 5x + 2y + 4z = 28 The values of x, y, z are: A. B. C. D.

1,1,1 3,-2,4 3,4,5 4,-2,3

QUESTION 98. (√3 + 𝑖) 7 + (√3 − 𝑖) 7 = A. B. C. D.

−256√3 256√3 −128√3 128√3

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QUESTION 99. The value of (−2√3 + 2𝑖). (3 + 3√3𝑖) is: A. 8√5 − 12𝑖 B. 12√3 − 12𝑖 C. 12√5 + 12𝑖 D. 12√5 − 10𝑖

QUESTION 100. The sum of the series: 1 + 2 +3 +4 …..n A. B.

𝑛 2 𝑛

is:

(𝑛 + 1)

2

C. (n + 1) D. N

QUESTION 101. The sum of the progression 32, -16, 8, ……..1/8 is close to: A. B. C. D.

11 21 33 44

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QUESTION 102. The sum of first fifteen numbers in the series is: 10 20 40 5+ + + +⋯ 3 9 27 A. 5 B. 10 C. 15 D. 20

QUESTION 103. The sum of the series: 2.1 + 0.021 + 0.00021 + ….. is: A. B. C. D.

1 2 3 4

QUESTION 104. The sum of the series: 1 1 1 1 1 + + + + ….. 𝑛 2 4 8 2 −1 A. B. C. D.

1 2 3 4

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PART-II QUESTION 1. The function y for the following second order linear homogeneous differential equation is: y'' + 6y' + 9y = 0, when y(0) = 0, y' (0) = 3 A. 3x + e3x B. 3xe–3x C. 2x2e1/x D. x + e3x–1

QUESTION 2. The function y for the following first order linear homogeneous differential equation is: y' + 5y = 0, when y(0) = 1 A. e5t B. e-5t C. 5et D. -5et

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QUESTION 3. The solution for (xy + x)2y' + (xy + y)2 = 0 is:

a) 2 x  2 y 

1  ln x 2 y  C xy

 1 1  b) xy   2  2   ln xy 2  C y  x  1 1 c) x 2  y 2      ln xy  C  xy y  1 1 d) x  y      ln x 2 y 2  C x y

QUESTION 4. Given the following information, the function y equals: y' = 3(xy)2; y(1) = 1 A. y = (2 – x3)-1 B. y = (3 – 2x2)-2 C. y = (2 + 2x – x2)–1 D. y = 3x(2 – x)–3

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QUESTION 5. A general solution for the following differential equation is: (2 – x)y' = y2 A. y2 + 1 / (ln | x – 2 |) = C B. y / (ln | x – 2 |) = C C. y – ln | x – 2 | = C D. –y-1 + ln | x – 2 | = C

QUESTION 6. The unit vector perpendicular to the following vectors is: V1 = (1, 2, 1) and V2= (2, 1, 2) A. (

1 √2

, 0, − 1

1

)

√2

B. (√2, 2 , 0)

1

C. (0,√2, − 2) 1

D. −1, − 2 , 1/√2)

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QUESTION 7. The determinant of A is:

1 2  1 A  3 0 2  2  2  1

A. 4 B. 16 C. 24 D. -16

QUESTION 8. The inverse of the following matrix is: 1 4    3 11

−11 4 ] 3 −1 −11 3 B. [ ] 4 1 11 −4 C. [ ] 3 −1 11 −4 D. [ ] −3 1 A. [

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QUESTION 9. For the differential equation 2y’ = 3xy +1 The integrating factor is close to: A. 3x 3 B. 2 𝑥

3 2

C. 𝑒 −2𝑥

3 2

D. 𝑒 −4𝑥

QUESTION 10. Given 2y’ = 3xy +1 The solution is: 3

A. 𝑦 = ln (2 𝑥 2 ) + 𝐶 3

B. 𝑦 = 2 𝑥 + 𝐶 3 2

1

C. 𝑦 = 𝐶𝑒 4𝑥 − 3 3 2

1

−3 2 𝑑𝑥

D. 𝑦 = [𝑒 4𝑥 ] [2 ∫ 𝑒 4 𝑥

+ 𝐶

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QUESTION 11. The solution for the differential equation is: 8𝑦 = 𝑒 −2𝑥 − 10𝑦 ′ − 2𝑦′′ y(0) =1 y’(0) = -3/2 9

A. 𝑦 = 4 𝑒 𝑥 − ln(2𝑥) 9

B. 𝑦 = 4 𝑒 𝑥 − 2𝑒 4𝑥 41

11

1

C. 𝑦 = 108 𝑒 −𝑥 − 108 𝑒 −4𝑥 + 36 𝑒 −2𝑥 1

1

D. 𝑦 = 𝑒 −𝑥 + 4 𝑒 −4𝑥 − 4 𝑒 −2𝑥

QUESTION 12. The general solution of the following differential equation: 𝑑𝑦 (𝑥 2 + 9) = 𝑥𝑦 𝑑𝑥 A. B. C. D.

𝑦 = √(𝑥 2 + 9) + 𝐶 𝑦 = 𝑥2 + 9 𝑦 = 𝐶√(𝑥 2 + 9) 𝑦 = 𝑥2 − 9

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QUESTION 13. The approximate value for the following integral for n =5 is: 1

∫ √𝑥 2 + 1 𝑑𝑥 0

A. B. C. D.

1.00 1.25 1.50 2.00

QUESTION 14. 3 𝑑𝑥

Approximate ∫2 A. B. C. D.

𝑥=1

using Simpson’s Rule with n=4

0.187 0.287 0.555 0.875

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QUESTION 15. A unit vector parallel to the resultant of vectors r1 = 2i +4j -5k, r2 = i+2j+3k is: A. B. C. D.

3 7 1 7 3 5 1

6

2

2

3

2

3

𝑖 + 7𝑗 − 7 𝑘 𝑖 + 7𝑗 − 7𝑘 𝑖 + 7𝑗 + 4𝑘

4𝑖

6

6

+ 7𝑗 + 7𝑘

QUESTION 16. The angle between A = 2i +2j –k and B = 6i -3j +2k is: A. B. C. D.

29 49 79 90

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QUESTION 17. The projection of the vector A = I -2j +k on the vector B =4i -4j +7k is: A. B. C. D.

1 19 2 19 19 9 3 19

QUESTION 18. Determine a unit vector perpendicular to the plane of A = 2i -6j -3k and B = 4i +3j –k is: A.

3 7 1

2

6

𝑖 − 7𝑗 + 7𝑘 3

B. 𝑖 + 𝑗 − 𝑘 7 7 C. 𝑖 − 𝑗 + 𝑘 D. 2𝑘 − 3𝑗 + 4𝑘

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QUESTION 19. The work done in moving an object along a vector r = 3i +2j -5k, if the applied force is F = 2i –j –k is: A. 3 B. 6 C. 9 D. 12

QUESTION 20. The distance from the origin to the plane of vectors A = 2i +3j +6k and B = i+5j+3k is: A. B. C. D.

2 5 7 9

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QUESTION 21. If A x B =0 and if A and B are not zero, then the angle between A and B is: A. B. C. D.

0 25 45 90

QUESTION 22. |𝐴𝑥𝐵|2 + |𝐴. 𝐵|2 = A. B. C. D.

A.B |𝐴|2 |𝐵|2 𝐴2 𝐵 2 𝐴. 𝐵 2

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QUESTION 23. If A = 2i -3j-k and B = i+4j -2k then, (A+B) x (A-B) is: A. B. C. D.

-10i +10j -12k -20i – 6 j- 22k -5i +2j +k 10i +10k

QUESTION 24. If A = 3i –j +2k and B = 2i +j –k, C = i -2j +2k, then A x (B x C) is: A. B. C. D.

24i +7j -5k 12i –j –k 24i -5k +7k 7i -24j -5k

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QUESTION 25. The area of the triangle with vertices at P(1,3,2), Q(2,-1,1) and R(-1,2,3) is:

A. B. C. D.

1 2 1 3 1 2 5 6

√54 √107 √107 √10

QUESTION 26. A unit vector perpendicular to the plane of A = 2i -6j -3k and B = 4i +3j –k is:

A.

3 7

2

6

𝑖 − 7 𝑗 + 7𝑘 3

2

6

B. − 7 𝑖 − 7 𝑗 + 7 𝑘 2

2

6

C. 7 𝑖 − 7 𝑗 − 7 𝑘 D. 15𝑖 − 10𝑗 + 30𝑘

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QUESTION 27. The value of (2i-3j).[(i+j-k) x(3i-k)] is:

A. B. C. D.

1 2 3 4

QUESTION 28. 𝑑𝑟

R = Sint i + Cost j + tk, the magnitude of |𝑑𝑡 | = A. B. C. D.

√2 1 √12 5

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QUESTION 29. A particle moves along a curve whose parametric equations are x = e-t, y =2cos 3t and z = 2sin 3t, where t is the time. The magnitude of its velocity at time t = 0 is:

A. B. C. D.

√37 9 √101 21

QUESTION 30. A particle moves along the curve x = 2t2, y =t2 -4t, z = 3t-5, where t is the time. The magnitude of the direction at time t =1 in the direction i-3j +2k is:

16

A. √14 B. 14 C. 12 D. 16

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SOLUTIONS

PART-I QUESTION 1 Center: (𝑥 − 1)2 (𝑦 − 1)2 + =1 9 16 The center is (1,1) QUESTION 2 𝑥2 + 𝑦2 + Therefore:

𝑥 𝑦 5 − + =0 2 2 2

√0 + 0 + 0 − 0 +

5 5 = √ 2 2

QUESTION 3 Given: 𝑦 2 = 8𝑥 𝑎𝑛𝑑 𝑦 = 2𝑥 𝑦 2 = 4𝑥 2 Equating the two ys 4𝑥 2 = 8𝑥, 𝑜𝑟 4𝑥 2 − 8𝑥 = 0 𝑥 = 0, 𝑜𝑟 2 The area under the curve: 2

𝐴 = ∫ 2𝑥 − √8√𝑥 𝑑𝑥 0

𝐴 = [𝑥 2 −

4√2 3 𝑥 2] 3

Substituting the limits: 𝐴 = 4−

16 4 = 3 3

QUESTION 4 9𝑥 2 + 5𝑦 2 − 30𝑦 = 0 9𝑥 2 + 5(𝑦 2 − 6𝑦) = 0 9𝑥 2 + 5(𝑦 − 3)2 = 45 Or 𝑥 2 (𝑦 − 3)2 + =1 5 9 The eccentricity can be found using: 𝑏 2 = 𝑎2 (1 − 𝑒 2 ) 5 = 9(1 − 𝑒 2 )𝑜𝑟 𝑒 = 2/3

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QUESTION 5 𝑏 2 = 𝑎2 (1 − 𝑒 2 ) Given, 2a =3; and a = 3b 𝑏 2 = 9𝑏 2 (1 − 𝑒 2 ) 𝑜𝑟 𝑒 =

2√2 3

QUESTION 6 Length of latus rectum of 𝑥 2 − 4𝑦 2 = 4 𝑥2 𝑦2 − =1 4 1 2𝑏 2 2 𝐿𝑒𝑛𝑡ℎ = = =1 𝑎 2 QUESTION 7 4𝑥 + 5𝑦 = 20 𝑥 𝑦 + =1 5 4 Area: 1 𝑥 20 = 10 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠 2 QUESTION 8 4 Slope of (2,-3) and (-5,1) is −7 = 𝑚1 4

Slope of (7,-1) and (0,3) is −7 = 𝑚2 Since the slopes are equal, the lines are parallel to each other and the angle is zero QUESTION 9 Given: 𝑥 2 − 3𝑦 − 6𝑥 + 6 = 0 (𝑥 − 3)2 = 3𝑦 − 6 − 9 (𝑥 − 3)2 = 3(𝑦 − 1) Axis of the parabola x =3 QUESTION 10 9𝑥 2 − 16𝑦 2 + 72𝑥 − 32𝑦 − 16 = 0 9(𝑥 2 + 8𝑥) − 16(𝑦 2 + 2𝑦) = 16 9(𝑥 + 4)2 − 16(𝑦 + 1)2 = 16 − 16 + 144 (𝑥 + 4)2 (𝑦 + 1)2 − =1 16 9 𝑏 2 = 𝑎2 (1 − 𝑒 2 ) 5 9 = 16 (𝑒 2 − 1)𝑜𝑟 𝑒 = 4

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QUESTION 11 Given 𝑦 = 𝑥 𝑎𝑛𝑑 𝑦 = 𝑥 3 Finding the roots of equation for x; x = 0 or 1 1

𝐴 = ∫ (𝑥 − 𝑥 3 )𝑑𝑥 0

[𝑥 − 𝑥 4 /4 ] Substituting the limits for x 0 and 1, we have the area under the curve as ¼

QUESTION 12 Given 𝑦 = 1 + 8/𝑥 2 The area under the curve is: 4

∫ [1 + 8/𝑥 2 ]𝑑𝑥 2

Or [𝑥 − 8/𝑥]4 Substituting the limits, the area under the curve is: 4 units QUESTION 13 The number of diagonals of a polygon is: 𝑛(𝑛 − 3) = 35 𝑜𝑟 𝑛 = 10 2 QUESTION 14 Eccentricity is calculated as: 𝑎2 − 𝑏 2 16 − 9 √7 √ 𝑒= √ = = 𝑎2 16 4 QUESTION 15 4 Given (𝑎𝑒, 0) = (4,0); 𝑒 = 5 a = 5; 𝑏 2 = 𝑎2 (1 − 𝑒 2 ) = 9 𝑥2 𝑦2 + =1 25 9 QUESTION 16 Given 3x +4y -1 = 0, is a tangent to the circle with a center (4,1) Radius of the circle is: (3)(4) + 4 − 1 =3 √32 + 42 The equation of the circle is: (𝑥 − 4)2 + (𝑦 − 1)2 = 9 Or 𝑥 2 + 𝑦 2 − 8𝑥 − 2𝑦 + 8 = 0 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 17 Equation of the parabola is: 𝑦 2 − 2𝑥 − 6𝑦 − 5 = 0 𝑦 2 − 6𝑦 + 9 = 2𝑥 + 14 (𝑦 − 3)2 = 2(𝑥 + 7) Equation of the directrix is: 𝑥+7+ QUESTION 18

1 = 0 𝑜𝑟 2𝑥 + 15 = 0 2

𝑟 = 5𝑐𝑜𝑠𝜃 + 12 𝑠𝑖𝑛𝜃 𝑟 2 = 5𝑟𝑐𝑜𝑠𝜃 + 12𝑟 𝑠𝑖𝑛𝜃 𝑥 2 + 𝑦 2 = 5𝑥 + 12𝑦 2 𝑥 + 𝑦 2 − 5𝑥 − 12𝑦 = 0 Therefore the equation is a circle. QUESTION 19 Equation of the line: 𝑦 = 𝑚𝑥 + 𝑐 Slope m = 1 y intercept c =3 𝑦=𝑥+3 QUESTION 20 Given: 𝑥 2 + 12𝑥 − 19𝑦 = 0 (𝑥 + 6)2 = 9(𝑦 + 4) Vertex therefore (-6,-4) QUESTION 21 Given 𝑦 2 − 4𝑦 − 8𝑥 − 4 = 0 (𝑦 − 2)2 = 8(𝑥 + 1) Focus: 𝑆(ℎ + 𝑎, 𝑘) = (−1 + 2, 2) = (1,2) QUESTION 22 Given: 1 1 3 = + 𝐶𝑜𝑠𝜃 𝑟 8 8 8 = 1 + 3𝑐𝑜𝑠𝜃 𝑟 e=3, therefore the equation is a hyperbola

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QUESTION 23 Given: 𝑦 = 𝑥 3 − 2𝑥 + 7 𝑑𝑦 = 3𝑥 2 − 2 𝑑𝑥 dy/dx at (1,6) 3(1)2 − 2 = 1 Therefore, the equation of the tangent is: 𝑦 − 6 = 1(𝑥 − 1) 𝑦=𝑥+5 QUESTION 24 Given y2 = 4x 𝑑𝑦 1 𝑎𝑡 (4,4) = (𝑚1 ) 𝑑𝑥 2

For x2 = 4y

𝑑𝑦 𝑎𝑡 (4,4) = 2 (𝑚2 ) 𝑑𝑥 𝑚1 − 𝑚2 𝑡𝑎𝑛𝜃 = 1 + 𝑚1 𝑚2 Substituting the values, 3 𝜃 = 𝑡𝑎𝑛−1 ( ) 4 QUESTION 25 Given 𝑥 2 + 𝑦 2 + 6𝑥 + 4𝑦 − 3 = 0 𝑇ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑖𝑠 (−3, −2) The tangent that passes through the center at (1,-2) is: 𝑦+2=0 QUESTION 26 Given: 16x2 + y2 + 8xy – 74x – 78y + 212 = 0 ℎ2 − 𝑎𝑏 = 16 − 16 = 0 Therefore, the equation is a parabola. QUESTION 27. Given: 1 𝜃 = 2𝑠𝑖𝑛2 𝑟 𝑟 1 = 1 − 𝑐𝑜𝑠𝜃 𝑟 Or e= 1, the equation is a circle

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QUESTION 28 Given: 6y = 7 – x3 𝑑𝑦 1 𝑎𝑡 (1,1) = − 𝑑𝑥 2 Equation of the tangent is: 1 𝑦 − 1 = − (𝑥 − 1) 2 𝑥 + 2𝑦 − 3 = 0 QUESTION 29 Focus is (a,0) or (3,0), therefore a =3 Directrix: 𝑥 + 𝑎 = 0 𝑜𝑟 𝑥 + 3 = 0 Equation of the parabola is: 𝑦 2 = 4𝑎𝑥 𝑦 2 = 12𝑥 QUESTION 30 Ellipse: 𝑥2 𝑦2 + =1 9 5 𝑒= √

𝑎2 − 𝑏 2 9−5 2 √ = = 𝑎2 9 3

Hyperbola: 𝑦2 𝑥 − 45 = 1 9 4 2

45 (9 + 4 ) 3 𝑎2 + 𝑏 2 √ 𝑒 =√ = = 𝑎2 9 2 ′

Therefore e.e’: 2 3 = 𝑥 =1 3 2 QUESTION 31 Given: 𝑦 = 𝑠𝑖𝑛𝑥, 𝑦 = 𝑐𝑜𝑠𝑥 𝑑𝑦 𝜋 𝑚1 = = 𝑐𝑜𝑠𝑥 = cos ( ) = 1/√2 𝑑𝑥 4 𝑑𝑦 𝜋 𝑚2 = = −𝑠𝑖𝑛𝑥 = − sin ( ) = −1/√2 𝑑𝑥 4 𝑚1 − 𝑚2 −1 𝜃 = 𝑡𝑎𝑛 [ ] 1 + 𝑚1 𝑚2 𝑡𝑎𝑛−1 (2√2)

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QUESTION 32 Given: 36x2 + 144y2 - 36x – 96y -119 = 0 𝑒= √

144 − 36 √3 = 144 2

QUESTION 33 Given: r2 – 4r (cos θ + sinθ) -4 = 0 The Cartesian equation of the circle is: 𝑥 2 + 𝑦 2 − 4𝑥 − 4𝑦 − 4 = 0 (𝑥 − 2)2 + (𝑦 − 2)2 = −4 Center is: (2,2) QUESTION 34 Given: 𝜃 = 𝑡𝑎𝑛−1 2 𝑡𝑎𝑛𝜃 = 2 𝑦 − 2 = 0 𝑜𝑟 𝑦 = 2𝑥 𝑥 QUESTION 35 Given: x2 – 3y2 - 4x – 6y – 11 = 0 (𝑥 − 2)2 (𝑦 + 1)2 + =1 12 4 Distance between the foci can be calculated as: 2ae = 2√𝑎2 + 𝑏 2 = 8 QUESTION 36 𝑥 2 + 𝑦 2 − 8𝑥 + 4𝑦 + 𝐾 = 0 𝐶𝑒𝑛𝑡𝑒𝑟 ∶ (4, −2); 𝑟𝑎𝑑𝑖𝑢𝑠: √16 + 4 − 𝐾 √20 − 𝐾 = 1 𝑜𝑟 𝐾 = 19 Therefore, 𝑥 2 + 𝑦 2 − 8𝑥 + 4𝑦 + 19 = 0 QUESTION 37 𝑎 = log cos 𝑥 sin 𝑥 1 𝑎 + = 2 𝑜𝑟 𝑎2 − 2𝑎 + 1 = 0 𝑎 (𝑎 − 1)2 = 0 𝑜𝑟 𝑎 = 1 𝜋 log cos 𝑥 sin 𝑥 = 1 𝑜𝑟 𝑥 = 4 QUESTION 38 log 4 [𝑥 3 + 𝑥 2 ] − log 4 (𝑥 + 1) = 2 𝑥3 + 𝑥2 log 4 =2 (𝑥 + 1) 2 log 4 𝑥 = 2 𝑜𝑟 𝑥 = 4 QUESTION 39 log 3√2 324 = log 3√2 [3√2]4 = 4

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QUESTION 40 𝑥 = 27, 𝑦 = log 3 4 𝑥 𝑦 = (27)log3 4 = 33 log3 4 = 3log3 64 = 64 QUESTION 41 lim

√𝑥 − 1 + √𝑥 + 1

𝑥→1

√𝑥 2 − 1

=

1 √𝑥 + 1

=

1 √2

QUESTION 42 𝑥(√𝑥 + 4 + 2 =4 𝑥→0 (𝑥 + 4) − 4 lim

QUESTION 43 1 3+ 2 3 3𝑥 2 + 1 𝑥 lim = lim = 1 𝑥→∞ 2𝑥 2 + 1 𝑥→∞ 2+ 2 2 𝑥 QUESTION 44 7𝑠𝑖𝑛𝑥 2+ 𝑥 2𝑥 + 7 sin 𝑥 2 1 lim = lim = 𝑜𝑟 3𝑐𝑜𝑠𝑥 4 2 𝑥→∞ 4𝑥 + 3𝐶𝑜𝑠 𝑥 𝑥→∞ 4 + 𝑥 QUESTION 45 3𝑥 + 4 𝐴 𝐵 = + − 3𝑥 + 2 (𝑥 − 2) 𝑥 − 1 3𝑥 + 4 𝐴 𝐵 = + (𝑥 − 2)(𝑥 − 1) (𝑥 − 2) 𝑥 − 1 6+4 3+4 𝐴= = 10 𝑎𝑛𝑑 𝐵 = = −7 1 −1 𝑥2

QUESTION 46 1 𝐴 𝐵 = + (1 − 2𝑥)(1 + 3𝑥) 1 − 2𝑥 1 + 3𝑥 1 = 𝐴(1 + 3𝑥) + 𝐵(1 − 2𝑥) 1 3 𝑥 = − ;𝐵 = 3 5 1 𝑥 = ; 𝐴 = 2/5 2 6 2𝐵 = = 3𝐴 5 QUESTION 47 𝑥3 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 3 𝑖𝑛 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝐵 𝐶 𝐷 = + + + 3 (2𝑥 − 1)(𝑥 + 2)(𝑥 − 3) 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 𝑖𝑛 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 2𝑥 − 1 𝑥 − 2 𝑥 − 3 1 1 𝑇ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑒𝑛𝑡 𝑜𝑓 𝑥 3 = 𝑜𝑟 𝐴 = 2 2

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QUESTION 48 By observing the equation, the roots are -1,1, and -2 (one can work backwards) QUESTION 49 𝑥 = 𝑡2; 𝑦 = 𝑡3 3𝑡 2 3 𝑦′ = = 𝑡 2𝑡 2 3 𝑑𝑡 3 3 𝑦 ′′ = 𝑜𝑟 = 𝑡 2 𝑑𝑥 2.2𝑡 4 QUESTION 50 3 sin(𝑥𝑦) + 4 cos(𝑥𝑦) = 5 𝐴𝑠𝑠𝑢𝑚𝑒 𝑡 = 𝑥𝑦 3 sin 𝑡 + 4 cos 𝑡 = 5 𝑑𝑡 [3 cos 𝑡 − 4 sin 𝑡] = 0 𝑑𝑥 𝑑𝑦 [3 cos 𝑡 − 4 sin 𝑡] [𝑥 + 𝑦] = 0 𝑑𝑥 𝑑𝑦 𝑦 = − 𝑑𝑥 𝑥 QUESTION 51 𝑑𝑦 1 + 𝑦2 =∫ 𝑑𝑥 1 + 𝑥2 𝑑𝑦 𝑑𝑥 ∫ = ∫ 2 1+𝑦 1 + 𝑥2 tan−1 𝑦 − tan−1 𝑥 = tan−1 𝑥 𝑦−𝑥 = 𝑐 𝑜𝑟 𝑦 − 𝑥 = 𝑐(1 + 𝑥𝑦) 1 + 𝑥𝑦 ∫

QUESTION 52 𝑑𝑦 2𝑥𝑦 = 𝑑𝑥 1 + 𝑥 2 1 2𝑥 𝑑𝑦 − 𝑑𝑥 = 0 𝑦 1 + 𝑥2 log 𝑦 − log(1 + 𝑥 2 ) = log 𝐴 𝑦 = 𝐴 (1 + 𝑥 2 ) QUESTION 53 The order of the differential equation is 1 QUESTION 54 𝑑𝑦 = 1 + 𝑦2 𝑑𝑥 2𝑦 𝑑𝑥 ∫ 𝑑𝑦 = ∫ 2 1+𝑦 𝑥 2) log(1 + 𝑦 = log 𝑥 + log 𝐶 1 + 𝑦 2 = 𝑐𝑥 2𝑥𝑦

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QUESTION 55 𝑥𝑑𝑥 + 𝑥𝑑𝑦 = 𝑥 2 𝑦𝑑𝑦 − 𝑥𝑦 2 𝑑𝑥 𝑥(1 + 𝑦 2 )𝑑𝑥 = −𝑦(1 − 𝑥 2 )𝑑𝑦 𝑥 𝑦 − 𝑑𝑥 = 𝑑𝑦 2 1−𝑥 1 + 𝑦2 log(𝑥 2 − 1) = log(1 + 𝑦 2 ) + 𝑙𝑜𝑔𝑐 𝑥 2 − 1 = 𝑐(1 + 𝑦 2 ) QUESTION 56 𝑑𝑦 =4 𝑑𝑥 (4 − 𝑥 2 )𝑑𝑥 = 𝑦 2 𝑑𝑦 𝑥2 + 𝑦2

Integrating both sides: 𝑥 3 + 𝑦 3 = 12𝑥 + 𝑐 QUESTION 57 𝑑𝑦 𝑦 + =1 𝑑𝑥 3 Integrating factor: 1

𝑥

𝑒 ∫3𝑑𝑥 = 𝑒 3 𝑥

𝑥

𝑦𝑒 3 = ∫ 𝑒 3 𝑑𝑥 𝑥

𝑥

𝑦𝑒 3 = 3𝑒 3 + 𝐶 𝑥

𝑦 = 3 + 𝐶𝑒 −3

QUESTION 58 𝑦 + 𝑥2 =

𝑑𝑦 𝑑𝑥

𝑑𝑦 − 𝑦 = 𝑥2 𝑑𝑥 Integrating factor:

= 𝑒 ∫ −1𝑑𝑥 = 𝑒 −𝑥 Solution: 𝑦𝑒 −𝑥 = ∫ 𝑥 2 𝑒 −𝑥 𝑑𝑥 𝑦𝑒 −𝑥 = 𝑒 −𝑥 [𝑥 2 + 2𝑥 + 2] + 𝑐 𝑦 + 𝑥 2 + 2𝑥 + 2 = 𝑐𝑒 −𝑥 QUESTION 59 𝑓(𝑥) = 𝑥 3 − 3𝑥 𝑓 ′ (𝑥) = 3𝑥 2 − 3 = 0; 𝑥 = ±1 𝑓 ′′ (𝑥) = 6𝑥 < 0; 6𝑥 = −1 𝑀𝑎𝑥𝑖𝑚𝑢𝑚𝑣𝑎𝑙𝑢𝑒 = −1 + 3 = 2

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QUESTION 60 𝑓(𝑥) = 2𝑥 3 − 21𝑥 2 + 36𝑥 + 20 𝑓 = 6𝑥 2 − 42𝑥 + 36 0𝑟 𝑥 = 1 𝑜𝑟 2 𝑓 ′′ (𝑥) = 12𝑥 − 42 𝑓 ′ (1) = −30 < 0 𝑓 ′ (2) = 24 − 42 = −18 < 0 𝑓(1) = 37 𝑎𝑛𝑑 𝑓(2) = 24 Therefore, the maximum value is 37 ′ (𝑥)

QUESTION 61 𝑑ℎ = 60 − 32𝑡 𝑑𝑡 For initial velocity t=0, h = 60 QUESTION 62

𝑑2 𝑠

𝑑𝑠 3 = 6 − 𝑡2 𝑑𝑡 2 𝑑2𝑠 = −3𝑡 𝑑𝑡 2

For maximum velocity, 𝑑𝑡 2 = 0 At t= 0, the maximum velocity is 6 m/s QUESTION 63 𝑆 = 𝑡 3 − 12𝑡 2 + 6𝑡 + 8 𝑑𝑠 𝑣= = 3𝑡 2 − 24𝑡 + 6 𝑑𝑡 𝑑𝑣 𝑎= = 6𝑡 − 24 = 0 𝑑𝑡 At t= 4, V = -42 m/s QUESTION 64 𝐴 = 𝐵 + 𝐶 = 60𝑜 1 1 1 3 𝐶𝑜𝑠𝐴 + 𝐶𝑜𝑠 𝐵 + 𝐶𝑜𝑠 𝐶 = + + = 2 2 2 2 QUESTION 65 𝑐𝑜𝑠𝑒𝑐 𝜃 + 2 = 0 𝑐𝑜𝑠𝑒𝑐 𝜃 = −2 𝑜𝑟 sin 𝜃 = − 𝜃 = 120 𝑎𝑛𝑑 330 𝑑𝑒𝑔𝑟𝑒𝑒𝑠

1 2

QUESTION 66 2𝜋 4𝜋 𝑥 = 𝑦 cos = 𝑧 cos 3 3 𝑦 𝑧 𝑥 = − = − 𝑜𝑟 𝑦 = 𝑧 2 2 𝑦2 𝑦2 𝑥𝑦 + 𝑦𝑧 + 𝑥𝑧 = + 𝑦2 − =0 2 2 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 67 sin 𝜃1 + sin 𝜃2 + sin 𝜃3 = 3 𝜋 𝜃1 +𝜃2 + 𝜃3 = 2 cos 𝜃1 + cos 𝜃2 + cos 𝜃3 = 0 QUESTION 68 3 𝑡𝑎𝑛𝐴 − 4 = 0 4 tan 𝐴 = 3 24 4 3 5 sin 2𝐴 + 3 sin 𝐴 + 4 𝐶𝑜𝑠 𝐴 = 5𝑥 − 3 𝑥 − 4𝑥 = 0 25 5 5 QUESTION 69 𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛2 𝑥 = 1 sin 𝑥 = 𝑐𝑜𝑠 2 𝑥 𝑐𝑜𝑠 8 𝑥 + 2𝑐𝑜𝑠 6𝑥 + 𝑐𝑜𝑠 4 𝑥 = (𝑐𝑜𝑠 4 𝑥 + 𝑐𝑜𝑠 2 𝑥)2 (𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥)2 = 1 QUESTION 70 1 2 + 1 2 1 tan−1 + tan−1 = tan−1 4 9 = tan−1 2 4 9 2 1 − 36 QUESTION 71 Domain of sin−1 𝑥 𝑖𝑠 [0. 2𝜋] QUESTION 72 cos

2𝜋 4𝜋 8𝜋 14𝜋 1 𝑥 cos 𝑥 cos 𝑥 cos = 15 15 15 15 16

QUESTION 73 sin 𝜃 + sin 2𝜃 sin 𝜃 + 2 sin 𝜃 cos 𝜃 = 1 + cos 𝜃 + 𝑐𝑜𝑠2𝜃 cos 𝜃 + 2𝑐𝑜𝑠 2 𝜃 sin 𝜃(1 + 2𝑐𝑜𝑠𝜃) = = tan 𝜃 cos 𝜃(1 + 2𝑐𝑜𝑠 𝜃) QUESTION 74 1 =2 𝑡𝑎𝑛𝜃 𝜃 = 45𝑜 𝑜𝑟 225 1 sin 𝜃 = ± √2

tan 𝜃 +

QUESTION 75 3 4 , 𝐶𝑜𝑠 𝐴 = 5 5 𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 𝐶𝑜𝑠 𝐴 sin 𝐴 =

Substituting the values, a = 13

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PART-II QUESTION 1. Characteristic equation: r2 + 6r + 9 = 0 a2 = 4b General homogeneous solution: y = c1e-3x + c2xe-3x Initial Condition

y(0) = 0 0 = c1e0 + 0 c1 = 0 y = c2xe-3x and y' = c2e-3x – 3c2xe-3x

Initial Condition

y'(0) = 3 3 = c2 – 0 c2 = 3

Initial value problem solution: y = 3xe-3x QUESTION 2. General first order homogeneous solution: y’ + ay = 0

with solution of y = Ce-at

so, y’ + 5y = 0 y = Ce-5t Initial Condition y(0) = 1: 1 = Ce-5(0) C=1 Initial value problem solution: y = e-5t

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QUESTION 3. This is a separable first order differential equation and we write it as: (xy + y)2dx + (xy + x)2dy = 0 y2(x + 1)2dx + x2(y + 1)2dy = 0

x  12 dx   y  12 dy  0 x2

y2

 x2  2x 1   y2  2 y 1    dy  0 dx   x2 y2     Therefore, the solution is given by: 

2

 2 1  1 dx   1   2 dy  c 2    y y 

 1  x  x

Where, c is an arbitrary constant. By integration we obtain: 1 1 x  y      lnx 2 y 2  c x y

QUESTION 4. y1  3(xy)2 dy Rewrite:  3x 2 y 2 dx dy  3x 2 dx y2 dy 2  yz   3x dx  c 1 –  x3  c y

Initial values: -1 = 1+c c = -2 −

1 = 𝑥3 − 2 𝑦

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1 = 2 − 𝑥3 𝑦 1 𝑦= 2 − 𝑥3 QUESTION 5. dy y2  dx 2  x y 2 dy  

y

2

dy  



dx x–2

dx C x2

1  ln x  2  C y

QUESTION 6. First, find a vector orthogonal to v1 and v2 by taking the cross-product:

v 3  v1  v 2

    i j k  1 2 1  (22 – 12 , – 2  2, 12 – 22 )    2 1 2  

= (3, 0, –3) Now, make it a unit vector by dividing it by its own length: v unit 

v3  v3

 3,0, 3 2 32  02   3



 3,0, 3 3 2

1  1  ,0,    2 2

QUESTION 7.

1 2  1 A  3 0 2  2  2  1

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A = (1)(0)(-1) + (2)(2)(2) + (-1)(3)(-2) – (-1)(0)(2) – (2)(3)(-1) – (1)(2)(-2) A = (0) + (8) + (6) – (0) – (-6) – (-4) A = 8 + 6 + 6 + 4 = 24

QUESTION 8. We write down an augmented matrix consisting of the given one followed by the identity matrix. Then we perform row operations so that the identity matrix appears on the left. 1 4 3 11  1 4 0 –1 

1 0 0 1 0 –3 1 1

1 4 0 1 

1 0 3 –1

1 0 0 1 

–11 3

The inverse is

Multiply first row by (–3) and add to second row

Multiply second row by –1 Multiply second row by (–4) and add to first row

4 –1  –11 4   3 –1 

QUESTION 9. The differential equation can be arranged as: 𝑦’ =

3 𝑥𝑦 + 1 2

The integrating factor is: 𝑥

3

3 2

𝑒 ∫0 −2𝑥 𝑑𝑥 = 𝑒 −4𝑥 QUESTION 10. 𝑦=

1 (∫ 𝑢(𝑥)𝑔(𝑥)𝑑𝑥 + 𝐶) 𝑢(𝑥)

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=

1

3 − 𝑥2 1 4 ( ) 𝑑𝑥 + 3 2 (∫ 𝑒 2 − 𝑥 𝑒 4 3 2 1 3 2 𝑒 4𝑥 ( ∫ 𝑒 −4𝑥 𝑑𝑥 + 𝐶)

𝐶)

2

QUESTION 11. The homogeneous equation: 2𝑦 ′′ + 10𝑦 ′ + 8𝑦 = 0 The characteristic equation is: 𝑟2 + (

10 8 )𝑟 + = 0 2 2

The roots of the equation are: (𝑟 − 1)(𝑟 − 4) = 0; 𝑟 = −1; −4 Therefore, 5 25 𝑎 2 = ( )2 = >𝑏=4 2 4 𝑦ℎ = 𝐶1 𝑒 𝑟1 𝑥 + 𝐶2 𝑒 𝑟2 𝑥 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −4𝑥 Assume the particular solution is of the form e-2x since that is the form of the non homogeneous forcing function. 𝑦𝑝 = 𝐶3 𝑒 −2𝑥 The first and second derivatives are: 𝑦𝑝′ = −2𝐶3 𝑒 −2𝑥 𝑦𝑝′′ = 4𝐶3 𝑒 −2𝑥 2𝑦 ′′ + 10𝑦 ′ + 8𝑦 = 𝑒 −2𝑥 2(4𝐶3 𝑒 −2𝑥 ) + 10(−2𝐶3 𝑒 −2𝑥 ) + 8(𝐶3 𝑒 −2𝑥 ) = 𝑒 −2𝑥 1 8𝐶3 − 20𝐶3 + 8𝐶3 = 1; 𝐶3 = − 4 Complete solution: 𝑦 = 𝑦ℎ + 𝑦𝑝 1 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −4𝑥 − 𝑒 −2𝑥 4 Evaluate the unknown coefficients: 1 𝑦(0) = 1 = 𝐶1 𝑒 −(0) + 𝐶2 𝑒 −4(0) − 𝑒 −2(0) 4 1 1 = 𝐶1 + 𝐶2 − 4 𝐶1 + 𝐶2 = 5/4 3 𝑦 ′ (0) = − 2 1 = 𝐶1 𝑒 −(0) − 4𝐶2 𝑒 −4(0) + 𝑒 −2(0) 2 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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3 1 − = −𝐶1 − 4𝐶2 + 2 2 Solving for C1 and C2; 𝐶1 = 1; 𝐶2 =

1 4

The complete solution is: 𝟏 𝟏 𝒚 = 𝒆−𝒙 + 𝒆−𝟒𝒙 − 𝒆−𝟐𝒙 𝟒 𝟒 QUESTION 12. Separate the variables: (𝑥 2 + 9)𝑑𝑦 = (𝑥𝑦)𝑑𝑥 𝑑𝑦 𝑥 =( 2 )𝑑𝑥 𝑦 𝑥 +9 Integrate 𝑑𝑦 𝑥 = ∫( 2 )𝑑𝑥 𝑦 𝑥 +9 1 𝑙𝑛(𝑦) = ln( 𝑥 2 + 9) + 𝐶 2 𝑦 = ±𝑒 𝐶 √𝑥 2 + 9 ∫

The general solution is: 𝑦 = 𝐶 √𝑥 2 + 9 QUESTION 13. Here a = 0 and b =1 𝑏−𝑎 1−0 = = 0.2 𝑛 5 𝑦(0) = 𝑓(𝑎) = 𝑓(0) = √02 + 1 = 1 ∆𝑥 =

𝑦1 = 𝑓(𝑎 + ∆𝑥) = 𝑓(0.2) = √0.22 + 1 = 1.0198039 𝑦2 = 𝑓(𝑎 + 2∆𝑥) = 𝑓(0.4) = √0.42 + 1 = 1.0770330 𝑦3 = 𝑓(𝑎 + 3∆𝑥) = 𝑓(0.6) = √0.62 + 1 = 1.166904 𝑦4 = 𝑓(𝑎 + ∆4𝑥) = 𝑓(0.8) = √0.82 + 1 = 1.2806248 𝑦1 = 𝑓(𝑎 + 5∆𝑥) = 𝑓(1.0) = √12 + 1 = 1.4142136 So the area: 1 1 = 0.2 ( 𝑥 1 + 1.01198039 + 1.0770330 + 1.1661904 + 1.2806248 + 𝑥1.4142136) = 1.150 2 2

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∆𝑥 =

3−2 = 0.25 4

1 = 0.3333 2+1 1 𝑦1 = 𝑓(𝑎 + ∆𝑥) = 𝑓(2.25) = = 0.3076923 2.25 + 1 1 𝑦2 = 𝑓(𝑎 + 2∆𝑥) = 𝑓(2.5) = = 0.0.2857142 2.5 + 1 1 𝑦3 = 𝑓(𝑎 + 3∆𝑥) = 𝑓(2.75) = = 0.266667 2.75 + 1 1 𝑦4 = 𝑓(𝑎 + ∆4𝑥) = 𝑓(3.0) = = 0.25 3+1 𝑦(0) = 𝑓(𝑎) = 𝑓(2) =

𝑏

𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥 = 𝑎

0.25 [0.333333 + 4(0.3076923) + 2(0.2857142) + 4(0.2666667) + 0.25 = 0.2876831 3

QUESTION 15. Resultant R = r1 + r2 (2𝑖 + 4𝑗 − 5𝑘) + (𝑖 + 2𝑗 + 3𝑘) = 3𝑖 + 6𝑗 − 2𝑘 𝑅 = |𝑅| = |3𝑖 + 6𝑗 − 2𝑘| = √32 + 62 + −22 = 7 A unit vector parallel to R is R/7. 𝑅 3𝑖 + 6𝑗 − 2𝑘 3 6 2 = = 𝑖+ − 7 7 7 7𝑗 7𝑘 QUESTION 16. 𝐴. 𝐵 = 𝐴𝐵𝐶𝑜𝑠𝜃 𝐴 = √22 + 22 + (−1)2 = 3; 𝐵 = √62 + (−3)2 + 22 = 7 𝐴. 𝐵 4 𝐶𝑜𝑠𝜃 = = = 0.1905; 𝜃 = 79 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 (3)(7) 𝐴𝐵 QUESTION 17. 𝐵

A unit vector in the direction B is b = |𝐵| =

4𝑖 − 4𝑗 + 7𝑘

=

4 4 7 𝑖− 𝑗+ 𝑘 9 9 9

√42 + (−4)2 + 72 The projection of A on the vector B = A.b 4 4 7 4 4 7 19 (𝑖 − 2𝑗 + 𝑘) ( 𝑖 − 𝑗 + 𝑘) = (1) ( ) + (−2) (− ) + (1) ( ) = 9 9 9 9 9 9 9

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QUESTION 18. Let vector C =c1i +c2j +c3k be perpendicular to the plane of A and B. Then C is perpendicular to A and also to B. Hence: 𝐶. 𝐴 = 2𝑐1 − 6𝑐2 − 3𝑐3 = 0 𝐶. 𝐵 = 4𝑐1 + 3𝑐2 − 𝑐3 = 0 Solving the above equations, we have: 1 𝑐1 = 𝑐3 2 −1 𝑐2 = 𝑐 3 3 1 1 𝐶 = 𝑐3 ( 𝑖 − 𝑗 + 𝑘) 2 3 Then a unit vector in the direction of C is C/[C] 1 1 𝑐3 (2 𝑖 − 3 𝑗 + 𝑘) 3 2 6 = = 𝑖− + 𝑘 2 7 7𝑗 7 √𝑐32 [1 + (− 1)2 + 12 ] 2 3 QUESTION 19. Work done = (magnitude of force in the direction of motion)(distance moved) (𝐹𝐶𝑜𝑠𝜃)(𝑟) = 𝐹. 𝑟 (2𝑖 − 𝑗 − 𝑘)(3𝑖 + 2𝑗 − 5𝑘) = 6 − 2 + 5 = 9 QUESTION 20. The distance from the origin to the plane is the projection of B on A A unit vector in the direction of A is a 𝐴 2𝑖 + 3𝑗 + 6𝑘 2 3 6 𝑎= = = 𝑖+ 𝑗+ 𝑘 |𝐴| 7 7 √22 + (3)2 + 62 7 Then projection of B on A is B.a 2 3 6 (𝑖 + 5𝑗 + 3𝑘). ( 𝑖 + 𝑗 + 𝑘) = 5 7 7 7 QUESTION 21. 𝐼𝑓 𝐴𝑥𝐵 = 𝐴𝐵𝑆𝑖𝑛𝜃𝑢 = 0, 𝑡ℎ𝑒𝑛 𝑆𝑖𝑛𝜃 = 0, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝜃 = 0 𝑜𝑟 180 QUESTION 22. 𝐴𝑥𝐵 = 𝐴𝐵 𝑆𝑖𝑛𝜃 = |𝐴𝐵𝑠𝑖𝑛𝜃𝑢|2 + |𝐴𝐵𝐶𝑜𝑠𝜃|2 𝐴2 𝐵 2 𝑆𝑖𝑛2 𝜃 + 𝐴2 𝐵 2 𝐶𝑜𝑠 2 𝜃 = 𝐴2 𝐵 2 = |𝐴|2 |𝐵|2

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QUESTION 23. 𝐴 + 𝐵 = (2𝑖 − 3𝑗 − 𝑘) + (𝑖 + 4𝑗 − 2𝑘) = 3𝑖 + 𝑗 − 3𝑘 𝐴 − 𝐵 = (2𝑖 − 3𝑗 − 𝑘) − (𝑖 + 4𝑗 − 2𝑘) = 𝑖 − 7𝑗 + 𝑘 Then (𝐴 + 𝐵)𝑥 (𝐴 − 𝐵) = (3𝑖 + 𝑗 − 3𝑘) 𝑥 ( 𝑖 − 7𝑗 + 𝑘) 𝑖 [3 1

𝑗 𝑘 1 −3 3 1 3 −3 ]−𝑗[ ]+𝑘[ ] = −20𝑖 − 6𝑗 − 22𝑘 1 −3] = 𝑖 [ −7 1 1 1 1 −7 −7 1

QUESTION 24. 𝑖 𝑗 𝑘 𝐵𝑥𝐶 = [2 1 −1] = −0𝑖 − 5𝑗 − 5𝑘 1 −2 2 𝑖 𝑗 𝑘 𝐴𝑥(𝐵𝑥𝐶) = (3𝑖 − 𝑗 + 2𝑘)𝑥(−5𝑖 − 5𝑘) = [3 −1 2] = 15𝑖 + 15𝑗 − 15𝑘 0 −5 5 QUESTIOIN 25. 𝑃𝑄 = (2 − 1)𝑖 + (−1 − 3)𝑗 + (1 − 2)𝑘 = 𝑖 − 4𝑗 − 𝑘 𝑃𝑅 = (−1 − 1)𝑖 + (−2 − 3)𝑗 + (3 − 2)𝑘 = −2𝑖 − 𝑗 + 𝑘 The area of triangle: 1 1 |𝑃𝑄 𝑋𝑃𝑅| = |(𝑖 − 4𝑗 − 𝑘)𝑥(−2𝑖 − 𝑗 + 𝑘)| 2 2 𝑖 𝑗 𝑘 1 1 1 1 [ 1 −4 −1] = |−5𝑖 + 𝑗 − 9𝑘| = √−52 + 12 + −92 = √107 2 2 2 2 −21 −1 1 QUESTION 26. Ax B is a vector perpendicular to the plane of A and B 𝑖 𝑗 𝑘 𝐴𝑥𝐵 = [2 −6 −3] = 15𝑖 − 10𝑗 + 30𝑘 4 3 −1 QUESTION 27. 2 −3 0 [1 1 −1] = 4 3 0 −1

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QUESTION 28. 𝑑𝑟 𝑑 𝑑 𝑑 = (𝑠𝑖𝑛𝑡)𝑖 + (cos 𝑡)𝑗 + (𝑡)𝑘 = 𝑐𝑜𝑠𝑡 𝑖 − 𝑠𝑖𝑛𝑡 𝑗 + 𝑘 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑟 | | = √(𝑐𝑜𝑠𝑡)2 + (−𝑠𝑖𝑛𝑡)2 + (1)2 = √2 𝑑𝑡 QUESTION 29. The position vector of the particle: 𝑟 = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 Then the velocity is: 𝑑𝑟 𝑣= = −𝑒 −𝑡 𝑖 − 6 sin 3𝑡 𝑗 + 6 cos 3𝑡 𝑘 𝑑𝑡 At t = 0, 𝑑𝑟 = −1 + 6𝑘 𝑑𝑡 The magnitude of the velocity at t =0 is: √(−1)2 + (6)2 = √37 QUESTION 30. Velocity: 𝑑𝑟 𝑑 = [2𝑡 2 𝑖 + (𝑡 2 − 4𝑡)𝑗 + (3𝑡 − 5)𝑘] 𝑑𝑡 𝑑𝑡 = 4𝑡𝑖 + (2𝑡 − 4)𝑗 + 3𝑘 = 4𝑖 − 2𝑗 + 3𝑘 𝑎𝑡 𝑡 = 1 Unit vector in the direction i-3j +2k is: 𝑖 − 3𝑗 + 2𝑘

=

𝑖 − 3𝑗 + 2𝑘

√14 √12 + (−3)2 + (2)2 Then the component of the velocity in the given direction is: (4𝑖 − 2𝑗 + 3𝑘). (𝑖 − 3𝑗 + 2𝑘) 16 = √14 √14

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Probability and Statistics Total Questions 4-6 A. Probability distributions B. Regression and curve fitting

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PART-I QUESTION 1. The probability of drawing a pair of aces in two cards, when an ace has been drawn on the first card is: A. B. C. D.

1/13 1/26 3/51 4/51

QUESTION 2. There are ten defective parts per 1000 parts of a product. The probability that there is one and only one defective part in a random lot of 100 is: A. B. C. D.

99 x 0.0199 0.01 0.5 0.9999

QUESTION 3. The probability that both stages of a two-stage missile will function correctly is 0.95. The probability that the first stage will function is 0.98. The probability that the second stage will function correctly given that the first one does is: A. B. C. D.

0.95 0.97 0.98 0.99

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QUESTION 4. A standard deck of 52 playing cards is thoroughly shuffled. The probability that the first four cards dealt from the deck will be four aces is closet to: A. B. C. D.

4 x 10-6 4 x10-4 8 x 10-2 2 x 10-1

QUESTION 5. The number of teams of four can be formed from 35 people is: A. 25,000 B. 50,000 C. 75,000 D. 100,000

QUESTION 6. The number of three-letter codes may be formed from the English alphabet if no repetitions are allowed is: A. 8 B. 5900 C. 15,600 D. 22,100

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QUESTION 7. A tool has three parts, A, B and C with probabilities of 0.1, 0.2 and 0.25, respectively of being defective. The probability that exactly one of these parts is defective is: A. B. C. D.

0.005 0.375 0.55 0.95

QUESTION 7. If three students work on a certain math question, student A has a probability of success of 0.5, student B, 0.4 and student C, 0.3. If they work independently, the probability that no one works the question successfully is: A. B. C. D.

0.12 0.21 0.25 0.32

QUESTION 9. The probability of drawing an ace or a spade or both from a deck of cards is: A. B. C. D.

3/13 4/17 4/13 5/13

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QUESTION 10. Six dice are thrown simultaneously. The probability that all will show different faces is: A. B. C.

5! 65 5! 63 5! 64

D. 36

QUESTION 11. If standard deviation for two variable X and Y is 3 and 4, respectively and their covariance is 8, the correlation coefficient between them is: A. B. C. D.

2/3 2/9 1/6 1/3

QUESTION 12. Two cards drawn in succession from a pack of 52 cards. First card should be a king and the second a queen. The probability when the first card is replaced is: A. B. C. D.

1/169 2/663 4/169 3/169

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QUESTION 13. From a pack of regular playing cards, two cards are drawn at random. The probability that both cards will be kings, if the first card is not replaced is: A. B. C. D.

1.26 1/52 1/169 1/221

QUESTION 14. Standard deviation for 7, 9,11,13,15 is: A. B. C. D.

2.2 2.4 2.6 2.8

QUESTION 15. A bin contains 12 good parts and 6 minor defective parts and 2 major defective parts. A part is chosen at random, the probability the part is good is: A. B. C. D.

3/5 3/10 2/5 ½

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QUESTION 16. When two balls are drawn from a bag containing 2 white, 4 red and 6 black balls, the probability that both of them would be red is: A. B. C. D.

4/11 3/11 2/11 1/11

QUESTION 17. A box contains 10 parts out of which 4 are defective. Two parts are taken out together, one of them is found to be good, and the probability that the other part is also good is: A. B. C. D.

1/3 8/15 5/13 2/3

QUESTION 18. A question is given to three students. The chance of solving it individually is 1/3, ¼, 1/5. The probability that the question will be solved is: A. B. C. D.

1/5 2/5 3/5 4/5

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QUESTION 19. The probability of getting a total of 10 in a single throw of two dice is: A. B. C. D.

1/9 1/12 1/6 5/36

QUESTION 20. The probability of getting exactly 2 tails from 6 tosses of a fair coin is: A. B. C. D.

3/8 ¼ 15/64 49/44

QUESTION 21. How many different committees of 5 can be formed from 6 men and 4 women on which exactly 3 men and 2 women serve? A. 6 B. 20 C. 60 D. 120

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QUESTION 22. The probability of success that A can solve a question is 2/3 and B can solve is ¾. What is the probability that the question can get solved is: A. B. C. D.

11/12 7/12 5/12 9/12

QUESTION 23. Six coins are tossed simultaneously. The probability of getting at least 4 heads is: A. B. C. D.

11/64 11/32 15/44 21/32

QUESTION 24. If P(A) = ¼, P(B) = ½, P(A u B) = 5/8, then P(A∩B) is: A. B. C. D.

3/8 1/8 7/8 5/8

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QUESTION 25. The probability that a number selected at random from the set of numbers (1, 2, 3…100) is a perfect cube is: A. B. C. D.

1/25 2/25 3/25 4/25

QUESTION 26. When two dice are thrown, the probability of getting a total of 10 or 11 is: A. B. C. D.

7/36 5/36 5/18 7/18

QUESTION 27. If A and B are two events such that P(A u B) = 5/6 P (A∩B) = 1/3, P(A) = 2/3 then A and B are: A. B. C. D.

Dependent events Independent events Mutually exclusive events Mutually exclusive and independent

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QUESTION 28. Two dice are thrown at a time and the sum of the numbers on them is 6. The probability of getting the number 4 on any of the dice is: A. B. C. D.

2/5 1/5 2/3 1/3

QUESTION 29 A coin is tossed 3 times. The probability of getting head once and tail two times is: A. B. C. D.

1/3 ¼ 3/8 ½

QUESTION 30. The probabilities of two events A and B are 0.25 and 0.40, respectively. The probability of that both A and B occur is 0.15. The probability that neither A nor B occurs is: A. B. C. D.

0.35 0.65 0.50 0.75

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QUESTION 31. A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random, the probability that it is a black or red ball is: A. B. C. D.

1/3 ¼ 5/12 2/3

QUESTION 32. In a binominal distribution, the probability of getting a success is ¼ and the standard deviation is 3. The mean is: A. 6 B. 8 C. 10 D. 12

QUESTION 33. A bag contains 6 white and 4 black balls. Two balls are drawn at random. The probability that they are of the same color is: A. B. C. D.

1 15 2 5 4 15 7 15

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QUESTION 34. Given that the events A and B are such that P(A) = ¼, P(A│B) = ½ and P(B│A) = 2/3, then P(B) is: A. B. C. D.

½ 1/6 1/3 2/3

QUESTION 35. The mean of the numbers a, b, 8, 5 and 10 is 6 and the variance is 6.80. The possible values of a and b: A. B. C. D.

3,4 0,7 5,2 1,6

QUESTION 36. Suppose A and B are two events that P(A∩B) = 3/25, and P(A-B) = 8/28. The P(B) is: A. B. C. D.

11/25 3/11 1/11 9/11

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90

QUESTION 37. In a normal distribution the mean plus two standard distributions estimates the _____ percentile of the distribution.

A. B. C. D.

97.5 95.0 Lower 90.0 Middle 84.0

QUESTION 38. Given the following set of numbers 10, 12, 14, and 18 what is the Standard Deviation (SD)?

A. B. C. D.

13.5 11.67 5.6 3.4

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PART-II QUESTION 1. A firm rents cars from three rental agencies: 60% from agency D, 20% from agency E, and the rest from agency F. If 12% of the cars from D have bad tires, 4% from E have bad tires, and 10% from F have bad tires, what is the probability that a car that is rented will have bad tires? A. B. C. D.

0.02 0.10 0.20 0.24

QUESTION 2. Suppose a set of ten parts is known to contain 20% defects. What is the best answer for the probability of selecting two good parts when sampling without replacement?

A. 0.80 B. 0.64 C. 0.62 D. 0.16

QUESTION 3. A machine is producing metal pieces that are cylindrical in shape. A sample of the pieces is taken and the diameters (in cm) are: 1.01, 0.97, 1.03, 1.04, 0.99, 0.98, 1.01, and 1.03. The variance for the sample is close to: A. B. C. D.

0.000057 0.00057 0.0057 0.057

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QUESTION 4. A production process outputs plastics with strengths that follow a normal distribution function. The mean strength of plastics produced is 32.0kN/m2 and the standard deviation is 2.5 kN/m2. What is the probability that a sample of plastic chosen from this process with have a strength less than 36.0 kN/m2?

A. 0.8159 B. 0.9032 C. 0.9192 D. 0.9452

QUESTION 5. The traffic light at Main Street and Broadway is either green, red, or yellow for Main Street traffic with the following probabilities: P(green) = 0.7 P(red) = 0.25 P(yellow) = 0.05 What is the probability that 3 out of 5 cars on Main Street get a green light at the intersection? A. 0.1852 B. 0.3087 C. 0.4242 D. 0.6030

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QUESTION 6. The standard deviation of the daily production at a steel mill is s = 10 tons. After sampling for the past 30 days, the average output was 125 tons. What is the best choice for a 90% confidence interval for the true population mean of daily production?

A. [122, 128] B. [115, 135] C. [118, 132] D. [110, 140]

QUESTION 7. From a large group of people, 4 persons were weighed as: 139, 152, 160, and 173 lbs. What is a good estimate for a 95% confidence interval for the whole group mean? A. [153, 159] B. [148, 164] C. [145, 167] D. [133, 179]

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QUESTION 8. What is most nearly the sample standard deviation using the following data?

A. B. C. D.

0.091 0.098 0.198 0.320

QUESTION 9. Testing has shown that, on the average, 3% of the bearings produced at a factory are defective. 12 bearings are chosen at random. The probability that exactly two of them are defective is most nearly A. 0.036 B. 0.044 C. 0.059 D. 0.066

QUESTION 10. The final scores of students in a graduate course are distributed normally with a mean of 72 and a standard deviation of 10. What is the probability that a student’s score will be between 65 and 78?

A. B. C. D.

0.28 0.38 0.48 0.58

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95

QUESTION 11. The least squares method is used to plot a straight line through the data points (5,−5), (3,−2), (2,3), and (−1, 7). The correlation coefficient is most nearly A. −0.80 B. −0.88 C. −0.92 D. −0.97

QUESTION 12. The best curve of the form y  ab x

is fitted to the (x, y) points (2.0, 8.8), (3.1, 9.5), (5.5, 10.8), and (6.8, 11.3). The value of b is most nearly

A. −0.015 B. 0.48 C. 0.52 D. 2.1

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QUESTION 13. When operating properly, a plant has a daily production rate that is normally distributed with a mean of 144 kg/d and a standard deviation of 25 kg/d. During an analysis period, the output is measured on 30 consecutive days, and the mean output is found to be 135 kg/d. The probability (%) that the plant is not operating properly is most nearly A. 1 B. 5 C. 95 D. 99

QUESTION 14. Suppose that the life of light bulbs forms a normal distribution with a mean life of 5000 hours and a standard deviation of 1000 hours. The probability that the life of a randomly selected light bulb will last more than 6500 hours most nearly is:

A. B. C. D.

0.0500 0.0668 0.1023 0.1732

QUESTION 15. Four data points have been observed as follows: i 1 2 3 4

xi yi 2.0 5.1 1.5 4.2 3.6 7.5 5.7 10.4

Using linear least-square regression, the equation that best fits this data is:

A. B. C. D.

y = 2.3 +1.5x y = 2.3 +2.1x y = 1.5 +2.1x y = 1.5 +1.5x FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

97

QUESTION 16. The yield of a chemical process is being studied. The past 5 days of plant operation have resulted in the following yields: 91.5, 88.7, 90.8, 89.9 and 92.1. Test hypotheses are H0 = mean yield, µ=90% versus H1; µ≠90%. The P-value of this statistical test most nearly is:

A. B. C. D.

0.0500 0.2515 0.3125 0.4975

QUESTION 17. The life in hours of batteries is known to be approximately normally distributed with a standard deviation of 25 hours. A random sample of 10 batteries resulted in the following data: 535, 541, 562, 551, 573, 528, 565, 548, 543, 567 hours. The 95% two sided confidence interval on the mean battery life is:

A. B. C. D.

[525.1, 560.4] [535.8, 566.8] [528.0, 573.0] [545.3, 557.3]

QUESTION 19. The sample standard deviation of 5 data points 1,3,4,6 and 6 is:

2

A. 3√5 B. 3/√2 C. 3 D. 5

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QUESTION 20. Two cards are randomly selected from a deck of 52 playing cards (excluding the two jokers). The probability that the both selected cards are diamonds most nearly is:

A. B. C. D.

1 52 1 26 1 17 1 13

QUESTION 21. The number of messages sent per hour over a computer network has the following distribution: x= number of messages 10 11 12 13 14 15 f(x) 0.08 0.15 0.30 0.20 0.20 0.07 The expected number of messages sent per hour over the computer network is:

A. B. C. D.

11.0 11.5 12.0 12.5

QUESTION 22. The number of messages sent per hour over a computer network has the following distribution: x= number of messages 10 11 12 13 14 15 f(x) 0.08 0.15 0.30 0.20 0.20 0.07 The standard deviation of the number of messages sent per hour over the computer network is:

A. B. C. D.

1.00 1.25 1.36 1.50 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

99

QUESTION 23. The amount of a particular impurity in a batch of a certain chemical product is a random variable with mean value 4.0g and a standard deviation 1.5g. If 50 batches are independently prepared, what is probability that the sample average amount of impurity 𝑋̅ is between 3.5 and 3.8 g?

A. B. C. D.

0.1345 0.1445 0.1545 0.1645

QUESTION 24. A mutual fund company offers its customers several different funds: a money market fund, three different bond funds (moderate and high-risk) and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: Money market:20% Short bond: 15% Intermediate bond: 10% Long bond: 5% High risk stock: 18% Moderate risk stock: 25% Balanced: 7%

A customer who owns shares in just one fund is randomly selected. The probability that the selected individual does not won shares in a stock fund is nearly:

A. B. C. D.

0.27 0.37 0.47 0.57

QUESTION 25. Consider the type of clothes dryer (gas or electric) purchased by each of five different customers at a certain store. If the P (all five purchase gas) is 0.116 and P(all five purchase electric) = 0.005, the probability that at least one of each type is purchased is:

A. B. C. D.

0.478 0.599 0.687 0.879 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

100

QUESTION 26. A little league team has 15 players on its roster. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right handed players?

A. B. C. D.

1,700 2,100 2,300 2,700

QUESTION 27. A box in a certain supply room contains four 40-W light bulbs, five 60-W bulbs, and six 75bulbs. If two bulbs are randomly selected from the box, and at least one of them is found to be rated 75W, the probability that both of them are 75-W bulbs is:

A. B. C. D.

0.1872 0.2022 0.2174 0.3234

QUESTION 28. A chemical engineer is interested in determining whether a certain impurity is present in a product. An experiment has a probability of 0.80 of detecting impurity if it is present. The probability of not detecting the impurity if it is absent is 0.90. The prior probabilities of the impurity being present and being absent are 0.40 and 0.60, respectively. Three separate experiments result in only two detections. The probability that the impurity is present:

A. B. C. D.

0.715 0.825 0.905 0.995 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 29. A system consists of two components. The probability that the second compound functions in a satisfactory manner during its design life is 0.9, the probability that at least one of the two components does so is 0.96 and the probability that both components do so is 0.75. Given that the first component functions in a satisfactory manner throughout its design life, the probability that the second one does also is nearly:

A. B. C. D.

0.526 0.666 0.816 0.926

QUESTION 30. The number of major defects on a randomly selected appliance of a certain type is: x P(x)

0 0.08

1 0.15

2 0.45

3 0.27

4 0.05

The expected value is close to: A. B. C. D.

1.06 2.06 3.06 4.06

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QUESTION 31. The number of major defects on a randomly selected appliance of a certain type is: x P(x)

0 0.08

1 0.15

2 0.45

3 0.27

4 0.05

The variance is: A. B. C. D.

0.7342 0.8123 0.9364 0.9786

QUESTION 32. A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb lots. The distribution of the lots is given below: x P(x)

1 0.20

2 0.40

3 0.30

4 0.10

The expected number of pounds left after the next customer’s order is shipped is nearly: A. B. C. D.

75.5 88.5 95.5 98.5

QUESTION 33. A company that produces fine crystal knows from experience that 10% of its goblets have cosmetic flaws and must be classified as seconds. Among six randomly selected goblets, how likely is it that only one is second?

A. B. C. D.

0.1435 0.3543 0.4167 0.5000

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QUESTION 34. Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

A. B. C. D.

0.1478 0.2466 0.3476 0.4989

QUESTION 35. Suppose that 90% of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? A. 0.207 B. 0.317 C. 0.407 D. 0.555

QUESTION 36. If the temperature at which a certain compound melts is a random variable with mean value 120C and standard deviation 2C, the standard deviation is oF is: A. 1.6 B. 2.6 C. 3.6 D. 4.6

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QUESTION 37. Suppose that force acting on a column that helps to support a building is normally distributed with mean 15.0 kips and standard deviation 1.25 kips. The probability that the force is almost 18 kips is close to:

A. B. C. D.

0.1452 0.5674 0.8765 0.9452

QUESTION 38. If a normal distribution has μ=30 and σ = 5, what is the 91st percentile of the distribution? A. 30.1 B. 33.6 C. 36.7 D. 39.9

QUESTION 39. The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.256 ohms and 5% having a resistance smaller than 9.671 ohms. The standard deviation of the resistance distribution is:

A. B. C. D.

0.10 0.20 0.30 0.40

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QUESTION 40. Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ = 1 (which is identical to a standard gamma distribution with α=1). The probability that 2 ≤ X ≤5 occurring is nearly:

A. B. C. D.

0.101 0.115 0.129 0.155

QUESTION 41. The inside diameter of randomly selected piston ring is a random variable with mean 12 cm and standard deviation 0.04 cm. The distribution of diameter is normal. The P (11.99 ≤ X ≤12.01) when n =16 is close to:

A. B. C. D.

0.6826 0.7224 0.8114 0.9867

QUESTION 42. Rockwell hardness of a certain type is known to have a mean value of 50 and a standard deviation of 1.2. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51?

A. B. C. D.

0.0040 0.0051 0.0062 0.0075

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QUESTION 43. A gas station sells three grades of gasoline; regular, extra and super. These are priced at $21.20, $21.35 and $21.50 per gallon, respectively. Let X1, X2, and X3 denote the amounts of these grades purchased (gallons) on a particular day. The Xi are independent with μ1 = 1000, μ2 = 500 and μ3 = 300; σ1 =100, σ2 = 80 and σ3 = 50. The standard deviation is nearly:

A. B. C. D.

222 322 422 505

QUESTION 44. Let X1, X2, and X3 represent the times necessary to perform three repair tasks at a certain service facility. If μ1 = μ2=μ3 = 60 and σ12 = σ22 = σ23 = 12, the value P(X1 + X2 + X3 ≤ 200) is:

A. B. C. D.

0.1198 0.2298 0.7899 0.9986

QUESTION 45. For a normal distribution, σ is known. What is the confidence level for the interval𝑥̅ ± 2.81𝜎/√𝑛: A. 89.5% B. 91.5% C. 95.5% D. 99.5%

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QUESTION 46. Assume that the helium porosity (in percentage) of coal samples taken away from any particular seam is normally distributed with true standard deviation 0.75. The 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85 is:

A. B. C. D.

(4.52, 5.18) (4.66; 5.22) (4.87; 5.22) (5.22, 6.18)

QUESTION 47. A sample of 50 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16 and the sample standard deviation was 164.43. The 95% CI for true average CO2 level in the population of all homes from which the sample was selected is close to:

A. B. C. D.

(608.58; 699.74) (609.23; 701.24) (607.99; 700.25) (608.58; 688.74)

QUESTION 48. The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with σ = 0.200 kg. Let μ denote the true average weight reading on the scale. Suppose the scale is to be recalibrated if either 𝑥̅ ≥ 10.1032 𝑜𝑟 𝑥̅ ≤ 9.8968. What is the probability that recalibration is carried out when it is actually unnecessary?

A. B. C. D.

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QUESTION 49. For which of the given P-values would the null hypothesis not rejected when performing a level 0.05 test?

A. B. C. D.

0.021 0.078 0.047 0.039

QUESTION 50. Let μ1 denote the true average tread life for a premium brand of P205/65R15 radial tire and let μ 2 denote the true average tread life for an economy brand of the same size. Test H 0: μ1-μ2 = 5000 versus, Ha: μ1-μ2 >5000 at level 0.01 using the following data: m = 45, 𝑥̅ = 42,500, s1 =2200, n =45, 𝑦̅ = 36,800, and s2 = 1500.

A. B. C. D.

Life of radial tires is better than economy brand tires Life of economy tires is better than radial tires Both have similar life span None of the above

QUESTION 51. The data on corn yield x and peanut yield y (mT/ha) for eight different types of soil is given below: X Y

2.4 1.33

3.4 2.12

4.6 1.80

3.7 1.65

2.2 2.00

3.3 1.76

4.0 2.11

2.1 1.63

The sample correlation coefficient is: A. B. C. D.

0.147 0.247 0.347 0.447

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SOLUTIONS

PART-I QUESTION 1 This is a conditional probability problem. Let B be “draw an ace” and let A be “draw a second ace”. 4 3 𝑃(𝐵) = ; 𝑃(𝐴) = 52 51 Therefore, the probability of getting an ace when an ace is already drawn is 3/51 QUESTION 2 𝑃𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 =

10 = 0.01 1000

The probability that one item is good is: 𝑃𝑔𝑜𝑜𝑑 = 1 − 0.01 = 0.99 The probability that exactly one defective will be found in random sample of 100 items is: = 100 𝑥 (0.01)(0.99)99 = (0.99)99 QUESTION 3 Given the probability of success of first stage is 0.98 and the overall probability is 0.98 The probability of success of second stage is: 0.95 𝑃(𝑠𝑒𝑐𝑜𝑛𝑑 𝑠𝑡𝑎𝑔𝑒) = = 0.97 0.98 QUESTION 4 The probability of drawing an ace on the first card: 4/52 The probability of drawing a second ace: 3/51 The probability of drawing a third ace: 2/50 The probability of drawing a fourth ace: 1/49 Overall probability: 4 3 2 1 𝑥 𝑥 𝑥 = 3.7𝑥10−6 52 51 50𝑥 49 QUESTION 5 The number of teams that can be formed is: 35.34.33.32 = 52,360 4.3.2.1 QUESTION 6 There are 26 choices for the first letter, 25 remain for the second and 24 for the third. The number of words can be formed is: 26.25.24 QUESTION 7 The probability that only A is defective is: = 0.1𝑥(1 − 0.2)𝑥(1 − 0.25) = 0.06 The probability that only B is defective: = (1 − 0.1)𝑥 0.2 𝑥 (1 − 0.25) = 0.135 The probability that only C is defective: (1 − 0.1)𝑥(1 − 0.2)𝑥0.25 = 0.18 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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The final probability is: = 0.06 + 0.135 + 0.18 = 0.375 QUESTION 8 Multiplying the complimentary probabilities: (1 − 0.5)𝑥(1 − 0.4)𝑥(1 − 0.3) = 0.21 QUESTION 9 4 The probability of drawing an ace from a deck of cards: 52 The probability of drawing a card of spades:

13 52 1

The probability of drawing an ace of spades: 52 Since the two events, i.e, a card being an ace and a spade are not mutually exclusive, the probability of drawing an ace or a spade is: 4 13 1 4 = + − = 52 52 52 13 QUESTION 10 Number of ways different numbers can occur on six dice = 6! The total number of ways if five dice are thrown 65 6! 5! The probability: = 64 65 QUESTION 11 𝑐𝑜𝑣(𝑥, 𝑦) 8 2 𝑃(𝑥, 𝑦) = = = 𝜎𝑥 𝜎𝑦 3𝑥4 3 QUESTION 12 4 1 The probability of drawing a king: = 13 52 If the card is replaced, the deck will have 52 cards. 1 The probability of drawing a queen: 13 The two events being independent, the probability of drawing both cards in succession is: 1 1 𝑥 = 1/169 13 13 QUESTION 13 The probability of both cards being kings: 4 3 1 𝑃= 𝑥 = 52 51 221 QUESTION 14 7 + 9 + 11 + 13 + 15 𝑚𝑒𝑎𝑛 = = 11 5 𝜎= √

(11 − 7)2 + (11 − 9)2 + (11 − 11)2 + (11 − 13)2 +(11 − 15)3 = 2.2 5

QUESTION 15 The total number of parts: The probability of good part: QUESTION 16 Favorable events: Total number of events:

12+6+2 = 20 12 3 =5 20 4𝐶2 = 6 12𝐶2 = 66 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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Probability:

6

1

= 66 = 11

QUESTION 17 6𝐶2 6𝑥5 1 = = 10𝐶2 10𝑥9 3 QUESTION 18 1 1 2 ; 𝑃(𝐴̅) = 1 − = 3 3 3 1 1 3 𝑃(𝐵) = ; 𝑃(𝐵̅ ) = 1 − = 4 4 4 1 1 4 𝑃(𝐶) = ; 𝑃(𝐶̅ ) = 1 − = 5 5 5 Probability of solving the problem: 2 3 4 3 1− 𝑥 𝑥 = 3 4 5 5 QUESTION 19 Favorable events are: (5,5), (4,6) and (6,4) 3 1 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦: = 36 12 QUESTION 20 1 4 1 6𝐶2 = ( ) 𝑥 ( )4 = 15/64 2 2 QUESTION 21 6𝐶3 𝑥 4𝐶2 = 20 𝑥 6 = 120 QUESTION 22 2 2 1 𝑃(𝐴) = ; 𝑃(𝐴̅) = 1 − = 3 3 3 3 3 1 𝑃(𝐵) = ; 𝑃(𝐵̅ ) = 1 − = 4 4 4 1 𝑃(𝐴 ⋂ 𝐵̅ ) = 12 1 11 𝑃(𝐴 𝑈 𝐵) = 1 − = 12 12 QUESTION 23 1 1 𝑃(4) = 6𝐶4 ( )4 ( )2 2 2 1 5 11 𝑃(5) = 6𝐶5 ( ) ( ) 2 2 1 0 1 6 𝑃(6) = 6𝐶6 ( ) ( ) 2 2 15 + 6 + 1 22 11 = = 64 64 32 𝑃(𝐴) =

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QUESTION 24 1 𝑃(𝐴) = 𝑃(𝐵) 4 1 5 = 𝑃(𝐴 𝑈 𝐵) = 2 8 5 1 1 1 = + − 𝑃 (𝐴 ⋂ 𝐵) = 8 4 2 8 QUESTION 25 Between 1 and 100, there are 4 cubes, 1, 8, 27, 64 4 1 Probability: = 25 100 QUESTION 26 Getting a total of 10: Getting a total of 11: Probability:

(5,5), (6,4) and (4,6) (three possibilities) (5,6) and (6,5) (two possibilities) 5 5 = 6𝑥6 = 36

QUESTION 27 𝑃(𝐴 𝑈 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ⋂ 𝐵) 5 2 1 1 = + 𝑃(𝐵) − 𝑃(𝐵) = 6 3 3 2 2 1 1 𝑃(𝐴). 𝑃(𝐵) = 𝑥 = = 𝑃(𝐴 ⋂ 𝐵) 3 2 3 QUESTION 28 Let E1 be the event of getting the sum of 6 and E2 be the event of getting the number 4 on any of the dice. 𝑛(𝐸1) = (1,5), (2,4), (3,3), (4,2)𝑎𝑛𝑑 (5,1) 𝑛(𝐸2) = (2,4)𝑎𝑛𝑑 (4,2) 2 𝑃= 5 QUESTION 29 11 1 3 3 3𝐶1 = ( ) 𝑥 ( ) = 2 2 8 QUESTION 30 𝑃(𝐴) = 0.25; 𝑃(𝐵) = 0.4 𝑃(𝐴 ⋂ 𝐵) = 0.15 𝑃(𝐴 𝑈 𝐵) = 1 − [𝑃(𝐴) + 𝑃(𝐵) − 𝑃 (𝐴 ⋂ 𝐵)] = 0.5 QUESTION 31 𝑛(𝑠) = 12𝐶1 = 12 𝑛(𝐸) = 8𝐶1 = 8 8 2 𝑃= = 12 3

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QUESTION 32 1 ; √𝑛𝑝𝑞 = 3 4 3 𝑞= 4 1 3 𝑛 ( ) ( ) = 9 𝑜𝑟 𝑛 = 48 4 4 1 𝑀𝑒𝑎𝑛 = 48 𝑥 = 12 4 𝑃=

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PART-II QUESTION 1. Let B = event of having bad tires P(D) = 0.6 P(E) = 0.2 P(F) = 0.2 (probabilities of renting from agencies) P(B|D) = 0.12

P(B|E) = 0.04

P(B|F) = 0.10

P(B) = P(D) P(B|D) + P(E) P(B|E) + P(F) P(B|F) = (0.6)(0.12) + (0.2)(0.04) + (0.2)(0.10) = 0.10 QUESTION 2. P{ A | B} 

P{ A  B} P{B}

P(drawing two good parts) = P(good 1st draw) x P(good 2nd draw½good 1st draw) = (

8 7 ) 𝑥 ( ) = 0.62 10 9

QUESTION 3. Sample Mean: n

X  1 / n  X i i 1

 (1/ 8)0.97  0.98  0.99  1.01  1.01  1.03  1.03  1.03  1.00625 Sample Variance: n

s 2  1 /( n  1) ( X i  X ) 2 i 1





 (1 / 7) (0.97  1.01) 2  (0.98  1.01) 2  ...  0.00057

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QUESTION 4. To find a probability for “less than 36.0”, we want to find a z value in order to locate F(z) in the standard probability table. z

x





36.0  32.0  1.6 2. 5

F (1.6)  0.9452 QUESTION 5. Use the binomial distribution because there are 2 outcomes (green or not green).

p = probability of green = 0.7 q = probability of not green = 0.25 + 0.05 = 0.3 n= 5

and

x=3

P5 (3)  C (5,3)(0.7) 3 (0.3) 53 

5! (0.7) 3 (0.3) 53 3!(5  3)!

The correct answer is (b) 0.3087 QUESTION 6. Given: σ = 10 n = 30 90% CI:   0.10

X  125

Z a / 2  1.6449   X  Za / 2

 n

 125  1.6449

10  125  3.003 30

CI: [(125 – 3), (125 + 3)] QUESTION 7. Given sample: 139, 152, 160, 173

n=4

= (1/4)(139 + 152 + 160 + 173) = 156 σ is not known, therefore we calculate s:

s

1 /( n  1) X i  X 2  n

i 1





(1 / 3) (139  156) 2  (152  156) 2  ...  14.26 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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  0.05 df = 4 – 1 = 3, and from t-table:

ta / 2  3.182

  X  ta / 2

s 14.26  156  (3.182)  156  22.69 n 4 CI: [(156 – 23), (156 + 23)]

QUESTION 8.

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QUESTION 9. Use the binomial distribution. p = 0.03 q = 1− p = 0.97

QUESTION 10. Calculate standard normal values for the points of interest, 65 and 78. Z

x



78  72 10  0.60

Z 78 

65  72 10  0.70

Z 65 

The probability of a score falling between 65 and 78 is equal to the area under the unit normal curve between these two standard normal values. Determine this area by subtracting F(Z65) from F(Z78). Although the F(Z) statistic is not tabulated for negative x values, the curve’s symmetry allows the R(x) statistic to be used instead. F(x) = R(-x) P(65 < x < 78) = F(Z78) – F(Z65) = F(Z78) – R(-Z65) = F(0.60) – R(0.70) = 0.7257– 0.2420 = 0.4837

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QUESTION 11. First calculate the following values.

The correlation coefficient is: Sxy =

QUESTION 12. Obtain a linear relationship by letting z equal √𝑥 Then plot a straight line of the form y = a+bz through the points (1.414, 8.8), (1.761, 9.5), (2.345, 10.8), and (2.608, 11.3). Calculate the following values. ∑ 𝑧𝑖 = 1.414 + 1.761 + 2.345 + 2.608 = 8.128

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∑ 𝑦𝑖 = 8.8 + 9.5 + 10.8 + 11.3 = 40.4

The slope of the line y = z + bz is:

QUESTION 13. Because a specific direction in the variation is not given, a two-tail hypothesis test is used. The zvalue corresponding to the confidence level is

Comparing the calculated value of z = 1.97 with a table of z-values for various confidence levels in the Table for the Values of Z/2 in the NCEES FE Handbook p. 44 shows that the confidence interval is most nearly 95%.

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QUESTION 14. If X is a normal random variable with a mean µ and a standard deviation σ, then 𝑍=

𝑋−𝜇 𝜎

is a standard normal random variable. Let X be the life of a light bulb selected at random. Then X is normal with µ=5000 and σ =1000. Hence P(the life of a randomly selected light bulb will last more than 6500 hours) = 𝑃(𝑋 > 6500) = 𝑃(𝑍 >

6500 − 5000 ) 1000

= 𝑃(𝑍 > 1.5) = 0.0668 QUESTION 15. The data averages are: 2 + 1.5 + 3.6 + 5.7 = 3.2 4 5.1 + 4.2 + 7.5 + 10.4 𝑦̅ = = 6.8 4 𝑥̅ =

The numerator: 4

∑(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅) = 15.74 𝑖=1

The denominator: 4

∑(𝑥𝑖 − 𝑥̅ )2 = 10.74 𝑖=1

Therefore, 𝑏̇ =

∑(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅) 15.74 = = 1.47 ∑(𝑥𝑖 − 𝑥̅ )2 10.74

And 𝑎̇ = 𝑦̅ − 𝑏̇𝑥̅ = 6.8 − 1.47𝑥3.2 = 2.1 Therefore, the equation that best fits this data is: 𝑦 = 1.5 + 2.1𝑥

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QUESTION 16. HINT: Since the variance of the yield is unknown, t-distribution must be used. The P-value for a two-sided test is 2P(Tn-1 > t0), where, n-1 are the degrees of freedom. Reject the null hypothesis H0 at (1-α) significance level if α 0.745) = 0.4975 QUESTION 17. HINT: when the variance is known, the standard normal distribution is used, and the 100(1-α)% confidence interval on mean µ is 𝑥̅ −

𝑧𝛼/2 𝜎, √𝑛

, 𝑥̅ +

𝑧𝛼/2 𝜎, √𝑛

The sample size n =10. A 95% confidence interval means, 𝛼 = 0.05 𝑎𝑛𝑑 𝑧0.025 = 1.96 Compute the sample average as: 𝑥̅ =

535 + 541 + 562 + 551 + 573 + 528 + 565 + 548 + 543 + 567 = 551.3 10

Calculate µ: 551.3 ±

1.96 𝑥25 √10

= [535.8, 566.8

QUESTION 18. HINT: Two events A and B are independent if P(A∩B)=P(A)P(B) Events A and B are mutually exclusive, that is: 𝐴∩𝐵 =0 Therefore, 𝑃(𝐴 ∩ 𝐵) = 𝑃(0) = 0 On the other hand, 𝑃(𝐴)𝑃(𝐵) = 0.5 𝑥 0.4 = 0.2 ≠ 𝑃(𝐴 ∩ 𝐵)

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Hence events A and B are independent. QUESTION 19. HINT: Sample standard deviation is given by: ∑(𝑥𝑖 − 𝑥̅ )2 𝑠= √ 𝑛−1 n=5 1+3+4+6+6 =4 5 [(1 − 4)2 +(3 − 4)2 +(4 − 4)2 +(6 − 4)2 +(6 − 4)2 ] 𝑠2 = = 9/2 (5 − 1) 𝑥̅ =

Therefore, 𝑠=

9 2

QUESTION 20. N = 52, M =13 and n = x =2 𝑀 𝑁− 𝑀 13 52 − 13 ) ( )( ) ( )( 2 𝑥 𝑛 − 𝑥 2 − 2 𝑃(𝑥 = 2) = = 𝑁 52 ( ) ( ) 𝑛 2 13! 39! 𝑥 1 2! (13 − 2)! 0! (39 − 0)! = 52! 17 2! (52 − 2)! QUESTION 21. HINT: Given the probability mass function f(xj) of a discrete random variable X with possible values xj (1,2…n), the expected value of X, E(X) is computed by: 𝑛

𝐸(𝑋) = ∑ 𝑥𝑗 𝑓(𝑥𝑗 ) 𝑗=1

n =6 6

𝐸(𝑋) = ∑ 10𝑥0.08 + 11𝑥0.15 + 13𝑥0.2 + 14𝑥0.2 + 15𝑥0.07 = 12.5 𝑗=1

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QUESTION 22. HINT: Given the probability mass function f(xj) of a discrete random variable X with possible values xj(1,2,3…n), the standard deviation of X denoted as σ is computed by: 𝑛

𝜎 = √𝑉𝑋 = √∑ 𝑥𝑗2 𝑓(𝑥𝑗 ) − (𝐸(𝑋))2 𝑗=1 6

𝐸(𝑋) = ∑ 10𝑥0.08 + 11𝑥0.15 + 12𝑥0.3 + 13𝑥0.2 + 14𝑥0.2 + 15𝑥0.07 = 12.5 𝑗=1 6

∑ 𝑥𝑗2 𝑓(𝑥𝑗 ) = 102 𝑥0.08 + 112 𝑥0.15 + 122 𝑥0.3 + 132 𝑥0.2 + 142 𝑥0.2 + 152 𝑥0.07 = 158.1 𝑗=1

𝜎 = √158.1 − 12.52 = 1.36 QUESTION 23. Given n =50, is large enough for the Central Limit Theorem to be applicable. 𝑋̅ then has approximately a normal distribution with mean 𝜇𝑋̅ = 4.0 and 1.5 𝜎𝑥̅ = = 0.2121 √50 So, 3.5 − 4.0 3.8 − 4.0 𝑃(3.5 ≤ 𝑋̅ ≤ 3.8) = 𝑃 [ ≤𝑍 ≤ ] 0.2121 0.2121 = Φ(−0.94) − Φ(−2.36) = 0.1645 QUESTION 24. Let event A = selected customer owns stocks. Then the probability that a selected customer does not own a stock can be represented by: 𝑃(𝐴′ ) = 1 − 𝑃(𝐴) = 1 − (0.18 + 0.25) = 0.57 QUESTION 25. Let event A be the event that all purchase gas. Let event B be the event that all purchase electric. All other possible outcomes are those in which at least one of each type is purchased. Thus the desired probability: 1 − 𝑃(𝐴) − 𝑃(𝐵) = 1 − 0.116 − 0.005 = 0.879

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QUESTION 26. 10 5 [ ]𝑥 [ ] = 2,100 5 3 QUESTION 27. The first desired possibility is P(both bulbs are 75 watts│at least one is 75 watt). P(at least one is 75 watt) = 1- P(none are 75 watt) 9 ( ) 36 69 1− 2 =1− = 15 105 105 ( ) 2 6 ( ) 15 = 𝑃(𝑏𝑜𝑡ℎ 𝑎𝑒 75𝑤𝑎𝑡𝑡) = 2 = 15 105 ( ) 2 So P(both bulbs are 75 watt│at least one is 75watt) 15 15 105 = = = 0.2174 69 69 105 QUESTION 28. When three experiments are performed, there are 3 different ways in which detection can occur on exactly 2 of the experiments: 1) #1 and #2 and not #3; 2) #1 and #2 and #3; 3)not#1, and #2 and #3. If the impurity is present, the probability of exactly 2 detections in three(independent) experiments is: (0.8)(0.8)(0.2) + (0.8)(0.2)(0.8) + (0.2)(0.8)(0.8) = 0.384 If the impurity is absent, the analogous probability is: 3(0.1)(0.1)(0.9) = 0.027 Thus, P(present│detected in exactly 2 out of 3): (0.384)(0.4) 𝑃(𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑 𝑖𝑛 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 2 ∩ 𝑝𝑟𝑒𝑠𝑒𝑛𝑡) = = 0.905 (0.384)(0.4) + (0.027)(0.6) 𝑃(𝑑𝑒𝑡𝑒𝑐𝑡𝑒𝑑 𝑖𝑛 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 2)

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QUESTION 29. Let A =1st functions, B = 2nd function, so, P(B) = 0.9, P(AUB) = 0.96, P(A∩B) = 0.75. Thus: 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) + 0.90 − 0.75 = 0.96 Therefore, P(A) = 0.81 This gives: 𝑃(𝐵|𝐴) =

𝑃(𝐵 ∩ 𝐴) 0.75 = = 0.926 𝑃(𝐴) 0.81

QUESTION 30. 4

𝐸(𝑥) = ∑ 𝑥. 𝑝(𝑥) 𝑥=0

= (0)(0.08) + (1)(0.15) + (2)(0.45) + (3)(0.27) + (4)(0.05) = 2.06 QUESTION 31. The variance is: 𝑉(𝑥) = ∑(𝑥 − 𝜇)2 . 𝑝(𝑥) 4

= ∑(0 − 2.06)2 (0.08) + … (4 − 2.06)2 (0.05) = 0.9364 𝑥=0

QUESTION 32. 4

𝐸(𝑥) = ∑ 𝑥. 𝑝(𝑥) 𝑥=0

(1)(0.20) + (2)(0.40) + (3)(0.30) + (4)(0.10) = 2.30 𝑉(𝑥) = ∑(𝑥 − 𝜇)2 . 𝑝(𝑥) = 0.81 Each lot weighs 5 lbs, so weight left = 100- 5X. Thus the expected weight left is 100-5E(x) = 88.5.

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QUESTION 33. X = Bin(6, 0.10) 𝑛 6 𝑃(𝑋 = 1) = ( ) (𝑝)𝑥 (1 − 𝑝)𝑛−𝑥 = ( ) (0.1)1 (0.9)5 = 0.3543 𝑥 1 QUESTION 34. Let S represent a telephone that is submitted for service while under warranty and must be replaced. Then p = P(S) = P(replaced│submitted). P(submitted) = (0.40)(0.20) = 0.08. Thus, X the number among the company’s 10 phones that must be replaced has a binomial distribution with n = 10, p =0.08, so p(2) = P(X=2) = = (

10 (0.08)2 (0.92)8 ) = 0.1478 2

QUESTION 35. X = number of flashlights that work. Let B – (battery has acceptable voltage). Then P(flashlight works) = P(both batteries work) = P(B)P(B) = (0.9)(0.9) = 0.81. X = Bin(10,0.81) 10 (0.81)9 (0.19) 10 ) + ( ) (0.81)10 = 0.407 9 10

𝑃(𝑋 ≥ 9) = 𝑃(𝑋 = 9) + 𝑃(𝑋 = 10) = ( QUESTION 36.

9

With X = temperature in C, temperature in F = 5 𝑋 + 32 9

9

So, 𝐸 (5 𝑋 + 32) = 5 (120) + 32 = 248 9 9 𝑉𝑎𝑟 [ 𝑋 + 32] = ( )2 (2)2 = 12.96; 𝜎 = 3.6 5 5 QUESTION 37. 𝑃(𝑋 ≤ 18) = 𝑃 ⌈𝑧 ≤

18.5 − 15 ⌉ = 𝑃(𝑍 ≤ 2.4) = Φ(2.4) = 0.9452 1.25

QUESTION 38. 𝜇 + 𝜎. (91𝑠𝑡 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 𝑓𝑟𝑜𝑚 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑛𝑜𝑟𝑚𝑎𝑙) = 30 + 5(1.34) = 36.7

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QUESTION 39. Since 1.28 is the 90th z percentile and -1.645 is the 5th z percentile (Z0.05 = 1.645), the given information implies that 𝜇 + 𝜎 (1.28) = 10.256 and 𝜇 + 𝜎(−1.645) = 9.671, from which σ(2.925) = -5.85, σ = 0.20.

QUESTION 40. 𝑃(2 ≤ 𝑋 ≤ 5) = 1 − 𝑒 (1)(5) − [1 − 𝑒 (1)(2) ] = 0.129 QUESTION 41. Given, μ = 12 cm, σ = 0.04cm, for n =16 11.99 − 12 12.01 − 12 ≤𝑍 ≤ ] 0.01 0.01

𝑃(11.99 ≤ 𝑋̅ ≤ 12.01) = 𝑃[

𝑃(−1 ≤ 𝑍 ≤ 1) = ∅(1) − ∅(−1) = 0.8413 − 0.1587 = 0.6826 QUESTION 42. Given , μ = 50, σ = 1.2, for n =9

𝑃(𝑋̅ ≥ 51) = 𝑃 (𝑍 ≥

51 − 50 ) = 𝑃(𝑍 ≥ 2.5) = 1 − 0.9983 = 0.0062 1.2 √9

QUESTION 43. The revenue from sales is: 𝑌 = 21.2𝑋1 + 21.35𝑋2 + 21.5𝑋3 𝐸(𝑌) = 21.2𝜇1 + 21.35𝜇2 + 21.5𝜇3 𝑉(𝑌) = (21.2)2 𝜎12 + (21.35)2 𝜎22 + (21.5)2 𝜎32 = 104,025 𝜎𝑌 = √104,025 = 322.53

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QUESTION 44. 𝐸(𝑋1 + 𝑋2 + 𝑋3 ) = 180 𝑉(𝑋1 + 𝑋2 + 𝑋3 ) = 45 𝜎𝑋1 +𝑋2 +𝑋3 = 6.708 𝑃(𝑋1 + 𝑋2 + 𝑋3 ≤ 200) = 𝑃 (𝑍 ≤

200 − 180 ) = 𝑃(𝑍 ≤ 2.98) = 0.9986 6.708

QUESTION 45. Zα/2 = 2.81 implies that α/2 = 1-ϕ(2.81) = 0.0025, so α = 0.005 and the confidence level is 100(1α) = 99.5% QUESTION 46. 4.85 ±

(1.96)(0.75) √20

= 4.85 ± 0.33 = (4.52, 5.18)

QUESTION 47. 𝑥̅ ± 𝑧0.025

𝑠 √𝑛

= 654.16 ± 1.96

164.43 √50

= (608.58; 699.74).

We are 95% confident that the true average CO2 levels in the population of homes with gas cooking appliances is between 608.58 ppm and 699.74 ppm. QUESTION 48. α=rejecting H0 whenH0 is true. 𝑃(𝑥̅ ≥ 10.1032 𝑜𝑟 ≤ 9.8968 when μ = 10. Since 𝑥̅ is normally distributed with standard deviation 𝜎/√𝑛 = 0.2/5 = 0.04, 𝛼 = 𝑃(𝑧 ≥ 2.58 𝑜𝑟 ≤ −2.58) = 0.005 + 0.005 = 0.01 QUESTION 49. P value = 0.001 < 0.05 so reject H0. The correct answer is C. Since all other values are lower than 0.05

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QUESTION 50. The test static value is: 𝑧=

(𝑥̅ − 𝑦̅) − 5000 2 2 √𝑠1 + 𝑠2 𝑚 𝑛

And H0 will be rejected at level 0.01 if z ≥ 2.33 𝑧=

(42,500 − 36,800) − 5000 2

2

√2200 + 1500 45 45

=

700 = 1.76 396.33

Which is less than 2.33, so we don’t reject H0 and conclude that the true average life for radials does not exceed that for economy brand.

QUESTION 51. ∑ 𝑥𝑖 = 25.7; ∑ 𝑦𝑖 = 14.40; ∑ 𝑥𝑖2 = 88.31, ∑ 𝑥𝑖 𝑦𝑖 = 46.856, ∑ 𝑦𝑖2 = 26.4324 𝑆𝑥𝑥 𝑆𝑦𝑦

(25.7)2 = 88.31 − = 5.75 8

(14.40)2 = 26.4324 − = 0.5124 8

𝑆𝑥𝑦 = 46.856 − 𝑟=

(25.7)(14.40) = 0.5960 8

0.5960 √5.75√0.5124

= 0.347

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Computational Tools Total Question 3-5 A. B.

Spreadsheets Flow charts

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Question 1. Which of the following statement is false? A. B. C. D.

Flowcharts use symbols to represent input/output, decision branches Pseudo code is an English-like description of a program Pseudo code uses symbols to represent steps in a program Structured programming breaks a program into logical steps or calls to subprograms

Question 2. In pseudo code, the following is a true statement for DOWHILE function: A. B. C. D.

DOWHILE is normally used for decision branching DOWHILE test condition must be false to continue the loop DOWHILE test condition tests at the beginning of the loop DOWHILE test condition tests at the end of the loop

Question 3. A spreadsheet contains the following formulas in the cells: ________________________________________________________________ A B C ________________________________________________________________ 1 A1+1 B1+1 2 A1^2 B1^2 C1^2 3 Sum (A1:A2) Sum (B1:B2) Sum (C1:C2) ________________________________________________________________ If 2 is placed in cell A1, the value of cell C3 is: A. B. C. D.

12 20 28 48

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Question 4. A spreadsheet contains the following:

______________________________________________________________________________ A B C D ______________________________________________________________________________ 1 3 4 5 2 2 A$2 3 4 4 6 If the formula from B2 is copied into D4, the equivalent formula in the cell D4 is: A. B. C. D.

A$2 C4 $C$4 C$2

Question 5. Transmission protocol: Serial, asynchronous, 8-bit ASCII, 1 Start, 1 Stop, 1 Parity bit, 9600 bps. The time (s) it takes to transfer a 1-Kbyte file is close to: A. B. C. D.

0.85 0.96 1.07 1.17

Question 6. Transmission protocol: Serial, synchronous, 8-bit ASCII, 1 parity bit, 9600 bps. The time (s) it takes to transfer a 1-Kbyte is close to: A. B. C. D.

0.85 0.96 1.07 1.17

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Question 7. The binary representation 10101 corresponds to the number A. B. C. D.

3 16 21 10,101

Question 8. For the base-10 number of 30, the equivalent binary number is: A. B. C. D.

1110 0111 1111 11110

Question 9. On a personal computer, the data storage device with the largest capacity is: A. B. C. D.

3 ½” diskette Hard disk Random access memory 1 GB USB flash drive

Question 10. All of the following statements are true of spreadsheets, EXCEPT: A. B. C. D.

A cell may contain label, value, formula or function A cell reference is made a numbered column and lettered row reference A cell content that is displayed is the result of a formula entered in the cell Line graphs, bar graphs and pie charts are typical graphs created from spreadsheets

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Question 11. In the structured programming shown below, line 2 is described as a(n) 1 2 3 4 A. B. C. D.

REAL X, Y X=3 Y = COS (X) PRINT Y

Assignment Command Declaration Function

Question 12. A program contains the following structured programming segment INPUT X Y=0 Z =X+1 FOR K = 1 TO Z Y = Y+Z NEXT K Y = Y – (2*Z-1) The final value of Y in terms of Z is: A. B. C. D.

X2 -1 X2 X2 +1 X2 +2X +1

Question 13. In a typical spreadsheet program, the cell directly below cell AB4 is: A. B. C. D.

AB5 AC4 AC5 BC4

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Question 14. A spreadsheet contains the following values as shown below. The formula B1+$A$1*A2 is entered into cell B2 and then coped into B3 and B4. The value displayed in cell B4 is: ______________________________ A B ______________________________ 1 3 111 2 4 3 5 4 6 ______________________________ A. B. C. D.

123 147 156 173

Question 15. Data are being collected automatically from an experiment at a rate of 14.4kbps. The time (min) it takes to completely fill a diskette with a capacity of 1.44 MB is close to: A. B. C. D.

1.67 10.0 13.3 14.0

Question 16. The acronym “ALU” is: A. B. C. D.

Arithmetic and Logic Unit Arithmetic Logic user Analog and Logic Unit Address Last Unit

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Question 17. A byte is defined as: A. B. C. D.

Eight bits typically used to encode a single character of text The smallest number of bits that can be used in arithmetic operation The basis unit of information accessed by a computer A binary digit that represents one of two equally accessible values or states

Question 18. A bit is defined as: A. A basic unit of a computer used to encode a single character of text B. The smallest portion of computer memory that can represent a distinct computer address C. A binary digit that represents one of two possible values or states D. Computer memory that can represent one of two possible values or states

Question 19. The time (min) it takes to transmit 400-Kbyte text file using a 28.8k modem in synchronous mode is close to: A. 1 B. 2 C. 11 D. 14

Question 20. The number of 8-bit bytes is in 2-MB of memory is close to: A. B. C. D.

16,000 16,256 2,000,000 2,097,152

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Question 21. A hard disk drive with three 5-cm diameter platters turns at 3000 rpm. The average latency for the drive (seconds) is nearly: A. B. C. D.

0.003 0.01 0.02 0.04

Question 22. A 12.7 diameter disk drive platter spins at 5600 rpm. The average access time is 11ms. The average (ms) rotational delay is close to: A. B. C. D.

5.17 5.36 5.50 5.64

Question 23. The number of bits in 5 GB hard drive is close to: A. B. C. D.

4.29 x 109 5.00 x 109 40.0 x 109 42.9 x 109

Question 24. The acronym RDRAM is defined as: A. B. C. D.

Rambus Dynamic Random Access Memory Rambus Direct Random Access Memory Redundant Dual Random Access Memory Random Duplex Random Access Memory

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Question 25. The number of cells in the range C5…Z30: A. B. C. D.

575 598 600 624

Question 26. A spreadsheet contains the following: ______________________________________________________________________________ A B C D ______________________________________________________________________________ 1 4 -1 3 0 2 3 A3+C1 -3 6 3 0.5 m 33 2 4 John 1 B2*B3 C4 5 The value of n in the following macro commands is close to: m=5 p = m*2 +6 n = D4 – 3*p^0.5 A. 4.1 B. 5.5 C. 10.5 D. 1041

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Question 27. Consider the following spreadsheet. The value of C1 is set to (A1+B1)/2. This formula is copied into the range of cells C1:C4. The value of cell C5 is set to Sum(C1:C4)*0.05. The number in cell C5 is: _______________________________________ A B C _______________________________________ 1 1 5 2 2 6 3 3 7 4 4 8 _______________________________________

A. B. C. D.

0.40 0.55 0.60 0.90

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Question 28. Consider the following algorithm: INPUT X N=0 T=0 DO WHILE N < X T = T +XN N=N+1 END DO OUTPUT T If the input value of X is 5, the output value of T is: A. 156 B. 629 C. 781 D. 3906

Question 29. Consider the following spreadsheet: ___________________________________________________________ A B C D ___________________________________________________________ 1 10 11 12 13 2 1 A2^2 B2*A$1 3 2 A3^2 B3*B$1 4 3 A4^2 B4*C$1 5 4 A5^2 B5*D$1 ___________________________________________________________

The values in column C are: A. B. C. D.

12, 1, 4, 9, 16 12, 10, 44, 108, 208 12, 19, 84, 207, 408 12, 19, 100, 250, 500

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Question 30. A spreadsheet evaluating the performance of a resistive element in a direct-current circuit is shown. The value displayed in cell C2 is: ____________________________________ A B C ____________________________________ 1 amps ohms watts 2 5.0 2.0 ____________________________________ A. B. C. D.

10.0 12.5 20.0 50.0

Question 31. Consider the following spreadsheet: the contents of column B are copied and pasted into columns C and D. The value in cell D4 is: ______________________________________________________ A B C D ______________________________________________________ 1 5 A$1 2 7 $B1 3 -3 B2+A2 4 0 A$3*B3 ______________________________________________________

A. 47 B. 60 C. 100 D. 150

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Question 32. In a spreadsheet, the formula $A$4 + B$2 + B2 is entered into cell C3. The contents of cell C3 are copied and pasted into cell D5. The formula in cell D5 is: A. $A$4 + C$2 + C4 B. $B$6 + C$4 + C4 C. $A$4 + C$4 + C4 D. $A$4 + B$2 + B2

Question 33. In a spreadsheet, the number in cell A4 is set to 6. Then, A5 is set to A4 +$A$4. This formula is copied into cells A6 and A7. The number shown in cell A7 is most nearly A. 12 B. 24 C. 36 D. 216

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Question 34. What is the value of X at the completion of the flow diagram shown?

A. 2 B. 4 C. 5 D. 12

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Question 35. A computer structured programming program contains the following program segment. What is the value of X after the segment is executed? X=2 T=3 IF X * 2 >T THEN T = X * 2 IF T > X THEN X = T * 2 IF T < X THEN X = T + 3 A. 2 B. 6 C. 7 D. 11

Question 36. A computer structured programming program contains the following program segment. What is the value of X after the segment is executed?

1

X=4 T=8 T=T–1 X=X+1 IF X < T THEN GOTO 1 A. 6 B. 7 C. 9 D. 11

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Question 37. A computer structured programming program contains the following program segment. What is the value of X after the segment is executed? X=4 T=8 DO WHILE T ≥ X T=T–2 X=X+2 ENDWHILE A. 4 B. 6 C. 8 D. 10

Question 38. A computer structured programming program contains the following program segment. What is the value of X after the segment is executed? X=0 FOR T = –1 TO 2 X=X+T NEXT T

A. B. C. D.

2 3 4 6

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Question 39. The numbers –3, 5, 2, –6, –1, 3 are in a file to be read and processed by the structured programming that follows. I=1 Y=0 WHILE I ≤ 3 Read a value from the file and set X equal to that value. If X < 0 GOTO 1 ELSE Y = Y + X*X 1I=I+1 ENDWHILE Z = Y/I

A. 7.3 B. 9.7 C. 19.5 D. 26.0

Question 40. Flash memory is classified as which type of memory? A. B. C. D.

ROM WORM RAM EPROM

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Ethics and Professional Practice Total Questions 3–5 A. B. C. D. E.

Codes of ethics Agreements and contracts Ethical and legal considerations Professional liability Public health, safety, and welfare

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QUESTION 1. Which of the following methods of advertising is most likely to violate an ethical standard for engineers? A. B. C. D.

Radio or television advertising Yellow pages Distribution of company calendars to clients Company brochures using self-laudatory language

QUESTION 2. Whistle blowing is best described as calling public attention to: A. B. C. D.

Your own previous unethical behavior Unethical behavior of employees under your control Secret illegal behavior by the employer Unethical or illegal behavior in a government agency

QUESTION 3. It is generally considered unethical to moonlight as a consulting engineer while you are working for your primary employer because A. You should not be competing with your primary employer B. You can’t do a good job for your primary employer if you come to work in the morning tired C. You might be tempted to use proprietary information from your current employer D. “Double dipping” (i.e. drawing two paychecks) is unfair to other engineers in the company

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QUESTION 4. Which of the following could you accept within most codes of ethical behavior? A. A trip to Bahamas from a vendor to learn about that vendor’s products B. A pen-and-pencil set sent by a blueprint reproduction company C. A smoked Thanksgiving turkey by a previous client in gratitude for a previous successful job D. A monetary incentive sent from a vendor in a country where such incentives are legal and common

QUESTION 5. A professional engineer who took the licensing examination in mechanical engineering: A. May not design in electrical engineering B. May design in electrical engineering if she feels competent C. May design in electrical engineering if she feels competent and the electrical portion of the design is insignificant and incidental to the overall job D. May design in electrical engineering if another engineer checks the electrical engineering work

QUESTION 6. An environmental engineer with five years of experience reads a story in the daily paper about a proposal being presented to the city council to construct a new sewage treatment plant near protected wetlands. Based on professional experience and the facts presented in the newspaper, the engineer suspects the plant would be extremely harmful to the local ecosystem. Which of the following would be an acceptable course of action? A. The engineer should contact appropriate agencies to get more data on the project before making a judgment B. The engineer should write an article for the paper’s editorial page urging the council not to pass the project C. The engineer should circulate a petition through the community condemning the project, and present the petition to the council D. The engineer should do nothing because he doesn’t have enough experience in the industry to express a public opinion on the matter

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QUESTION 7. An engineer is consulting for a construction company that has been receiving bad publicity in the local newspapers about its waste-handling practices. Knowing that this criticism is based on public misperception and the paper’s thirst for controversial stories, the engineer would like to write an article to be printed in the paper’s editorial page. What statement best describes the engineer’s ethical obligations? A. The engineer’s relationship with the company makes it unethical for him to take any public action on its behalf B. The engineer should request that a local representative of the engineering registration board review the data and write the article in order that an impartial point of view be presented C. As long as the article is objective and truthful, and presents all relevant information including the engineer’s professional credentials, ethical obligations have been satisfied D. The article must be objective and truthful, present all relevant information including the engineer’s professional credentials, and disclose all details of the engineer’s affiliation with the company

QUESTION 8. An engineering firm is hired by a developer to prepare plans for a shopping mall. Prior to the final bid date, several contractors who have received bid documents and plans contact the engineering firm with requests for information relating to the project. What can the engineering firm do? A. The firm can supply requested information to the contractors as long as it does so fairly and evenly. It can’t favor or discriminate against any contractor B. The firm should supply information to only those contractors that it feels could safely and economically perform the construction services C. The firm can’t reveal facts, data or information relating to the project that might prejudice a contractor against submitting a bid on the project D. The firm can’t reveal facts, data or information relating to the project without the consent of the client as authorized or required by law

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QUESTION 9. Without your knowledge, an old classmate applies to the company you work for. Knowing that you recently graduated from the same school, the director of engineering shows you the application and resume your friend submitted and asks your opinion. It turns out that your friend has exaggerated his participation in campus organizations, even claiming to have been an office in an engineering society that you are sure he was never in. On the other hand, you remember him as being highly intelligent student and believe that he could really help the company. How should you handle the situation? A. You should remove from the ethical dilemma by claiming that you don’t remember enough about the applicant to make an informal decision B. You should follow your instincts and recommend the applicant. Almost everyone stretches the truth a little in their resumes and the thing you are really being asked to evaluate his usefulness to the company. If you mention the resume padding, the company is liable to lose a good prospect C. You should recommend the applicant, but qualify your recommendation by pointing out that you think he may have exaggerated some details on his resume D. You should point out the inconsistencies in the applicant’s resume and recommend against hiring him

QUESTION 10. While working to revise the design of the suspension for a popular car, an engineer discovers a flaw in the design currently being produced. Based on the statistical analysis, the company determines that although this mistake is likely to cause a small increase in the number of fatalities seen each year, it would be prohibitively expensive to do a recall to replace the part. Accordingly, the company decides not to issue a recall notice. What should the engineer do? A. The engineer should go along with the company’s decision. The company has researched its options and chosen the most economic alternative B. The engineer should send an anonymous tip to the media, suggesting that they alert the public and begin an investigation of the company’s business practices C. The engineer should notify the National Transportation Safety Board, providing enough details for them to initiate a formal inquiry D. The engineer should resign from the company. Because of standard nondisclosure agreements, it would be unethical as well as illegal to disclose any information about the situation. In addition, the engineer should not associate with a company that is engaging in such behavior.

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QUESTION 11. If you check the calculations for a licensed (registered) friend who has gone into a consulting engineering business for himself/herself A. B. C. D.

You should be paid for your work Your friend’s client should be told of your involvement You do not need to be licensed or registered yourself Your friend assumes all the liability for your work

QUESTION 12. While supervising a construction project in a developing country, an engineer discovers that his client’s project manager is treating laborers in an unsafe and inhumane (but for that country, legal) manner. When he protests, the engineer is told by the company executives that the company has no choice in the matter if it wishes to remain competitive in the region he should just accept the way things are. What should ethics require the engineer to do? A. B. C. D.

Take no action-the company is acting in a perfectly legal manner Withdraw from the project, returning any fees he may already have received Report the company to the proper authorities for its human rights abuses Assist the laborers in organizing a strike to obtain better working conditions

QUESTION 13. An engineering professor with a professional engineering license and 20 years of experience in engineering education is asked to consult on a building design. Can the professor accept the request? A. Yes, but the professor should review and comment on only those portions of the project in which she is qualified by education and experience B. Yes, a professor is a subject matter expert and as such should be fully competent to review the design C. Yes, as a licensed professional engineer, the professor has demonstrated the competence in engineering and may review the design D. No, there is a tremendous difference between working in academia and having professional experience. The review should be conducted by a practicing engineer

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QUESTION 14. Two engineers submitted sealed bids to a prospective client for a design project. The client told engineer A how much engineer B had bid and invited engineer A to beat that amount. Engineer A really wants the projects and honestly believes that he can do a better job than engineer B. What should engineer A do? A. He should submit another quote, but only if he can perform the work adequately at the reduced price B. He should withdraw from consideration for the project C. He should remain in consideration for the project, but not change his bid D. He should bargain with the client for the cost of the work

QUESTION 15. A local engineering professor acts as technical advisor for the city council in a town. A few weeks before the council is scheduled to award a large construction contract, the professor is approached by one of the competing companies and offered a consulting position. Under what conditions would it be ethical to accept the job? A. Both the company and the council must know about and approve of the arrangement B. The professor should arrange not to begin work until after the council’s vote C. The professor may accept the job if the advisory position to the council is on a volunteer basis D. The professor must not participate in any discussion concerning the project for which the company is competing

QUESTION 16. Guidelines of ethical behavior among engineers are needed because: A. B. C. D.

Engineers are analytical and they don’t always think in terms of right or wrong All people, including engineers are inherently unethical Rules of ethics are easily forgotten It is easy for engineers to take advantage of clients

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QUESTION 17. Plan stamping is best defined as: A. The legal action of signing off on a project you didn’t design but are taking full responsibility for B. The legal action of signing off on a project you didn’t design or check but didn’t accept money for C. The illegal action of signing off on a project you didn’t design but did work D. The illegal action of signing off on a project you didn’t design or check

QUESTION 18. An engineer working for a design firm has decided to start a consulting business, but it will be a few months before she leaves. How should she handle the impending change? A. The engineer should discuss her plans with the current employer B. The engineer may approach the firm’s other employees while still working for the firm C. The engineer should immediately quit D. The engineer should return all of the pens, pencils and other equipment she has brought home over the years

QUESTION 19. During the day, an engineer works for a scientific research laboratory doing government research. During the night, the engineer uses some of the lab’s equipment to perform testing services for other consulting engineers. Why is this action probably unethical? A. The laboratory has not given its permission for the equipment use B. The government contract prohibits misuse and misappropriation of the equipment C. The equipment may wear out or be broken by the engineer and the replacement cost will be borne by the government contract D. The engineer’s fees to the consulting engineers can undercut local testing services’ fee because the engineer has a lower overhead

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QUESTION 20. During routine inspections, a field engineer discovers that one of the company’s pipeline is leaking hazardous chemicals into the environment. The engineer recommends that the line be shut down so that seals can be replaced and the pipe can inspected more closely. His supervisor commends him on his thoroughness and says the report will be passed on to the company’s maintenance division. The engineer moves on to his next job, assuming things will be taken care of in a timely manner. While working in the area again several months later, the engineer notices that the QUESTION hasn’t been corrected and in fact getting worse. What should the engineer do? A. Give the matter some more time. In a large corporate environment, it is understandable that some things take longer than people would like them to B. Ask the supervisor to investigate what action has been taken on the matter C. Personally speak to the director of maintenance and insist that this project be given high priority D. Report the company to EPA for allowing the situation to worsen without taking any preventive measures

QUESTION 21. A senior licensed professional engineer with 30 years of experience in geotechnical engineering is placed in charge of a multidisciplinary design team consisting of a structural group, a geotechnical group, and an environmental group. In this role, she is responsible for supervising and coordinating the efforts of the groups when working on a large interconnected project. In order to facilitate coordination, designs are prepared by the groups under the group leader, and they are submitted to her for review and approval. This arrangement is ethical as long as: A. She signs and seals each design segment only after being fully briefed by the appropriate group leader B. She signs and seals only those design segments pertaining to geotechnical engineering C. Each design segment is signed and sealed by the licensed group leader responsible for its preparation D. She signs and seals each design segment only after it has been reviewed by an independent consulting engineer who specializes in the field in which it pertains

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QUESTION 22. An engineer works at a large firm for several years during which he participates in the development of a new production technique. After leaving the company to start a consulting business, a competitor of his original employer asks for help with a similar question and the engineer is sure that the only solution is to use the process developed by his previous employer. Can the engineer ethically accept the job? A. No. This would constitute accepting payment from more than one party for the same project B. No. The engineer would have to use information obtained while working for his original employer C. Yes. Because he is no longer employed by the original company any nondisclosure agreements are invalid D. Yes. It is understood that consulting engineers often work for competing clients, and that some knowledge transfer is inevitable

QUESTION 23. A new junior engineer in a design company notices a detail in a design that she feels has the potential to be dangerous to end users. Her supervisors explain that this detail was incorporated in the design by the company to save manufacturing time. Furthermore, they assure her that although the analysis is technically correct, this shortcut has been used for several years and the company has never been accused of any wrongdoing. What should the engineer do? A. Go along with the advice of the more senior engineers. They have more experience in the field and are most likely right. Besides, if some harm does come from the design, they will take the blame B. Ask one of her college professors an expert in the field, to look at the plans and make a recommendation C. Bring the issue to the attention of the company’s upper nontechnical management D. Report the company’s violation to the state board and any other appropriate regulatory agencies

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QUESTION 24. Engineers are to uphold the health, safety and public………. A. B. C. D.

Trust Welfare Confidence Good

QUESTION 25. Information that is proprietary to a client: A. B. C. D.

May not be used in the work performed for the client May be divulged to third parties May be shared with other members of your design firm Should be kept absolutely confidential

QUESTION 26. After making a presentation for an international project, an engineer is told by a foreign official that his company will be awarded the contract, but only if it hires the official’s brother as an advisor to the project. The engineer sees this as form of bribery and informs his boss. His boss tells him that, while it might be illegal in the United States, it is a customary and legal business practice in the foreign country. The boss impresses upon the engineer the importance of getting the project, but leaves the details up to the engineer. What should the engineer do? A. He should hire the official’s brother, but insist that he perform some useful function for his salary B. He should check with other companies doing business in the country in QUESTION, and if they routinely hire relatives of government officials to secure work, then he should do the same C. He should withdraw his company from consideration for the project D. He should inform the government official that his company will not hire the official’s brother as a precondition for being awarded the contract, but invite the brother to submit an application for employment with the company FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION ANSWER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

D C B C C A D D C C B B A C D D D A D B C B C B D D

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Engineering Economics Total Questions 3-5 A. B. C. D.

Time value of money Cost, including incremental, average, sunk, and estimating Economic analyses Depreciation

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PART-I QUESTION 1 A small company borrowed $10,000 to expand its business. The entire principal of $10,000 will be repaid in two years, but quarterly interest of $330 must be paid every three months. The nominal interest (%) the company is paying is nearly: A. 3.3 B. 5.0 C. 6.6 D. 13.2

QUESTION 2 A deposit of $300 was made one year ago into an account paying monthly interest. If the account now has $320.52, the effective annual interest rate (%) is close to: A. 7 B. 10 C. 12 D. 15

QUESTION 3. A store charges 1.5% interest per month on credit purchases. This is equivalent to a nominal interest rate (%) of: A. 1.5 B. 15.0 C. 18.0 D. 21.0

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QUESTION 4. A store’s policy is to charge 3% interest every two months on the unpaid balance in charge accounts. The effective interest rate (%) is close to: A. 6 B. 12 C. 15 D. 19

QUESTION 5. The effective interest rate on a loan is 19.56%. If there are 12 compounding periods per year, the nominal interest rate (%) is close to: A. 1.5 B. 4.5 C. 9.0 D. 18.0

QUESTION 6. If 10% nominal annual interest is compounded daily, the effective annual interest rate (%) is nearest to: A. B. C. D.

10.05 10.38 10.50 10.75

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QUESTION 7. An individual wishes to deposit a certain quantity of money now so that he will have $500 at the end of five years. With interest at 4% per year, compounded semiannually, the amount ($) of the deposit is close to: A. B. C. D.

340 400 410 416

QUESTION 8. A retirement fund earns 8% interest compounded quarterly. If $400 is deposited every three months for 25 years, the amount ($) in the fund at the end of 25 years is nearest to: A. 50,000 B. 75,000 C. 100,000 D. 125,000

QUESTION 9. The repair costs for some handheld equipment are estimated to be $120 the first year, increasing by $30 per year in subsequent years. The amount a person needs to deposit ($) into a bank account paying 4% interest to provide for the repair costs for the next five years is close to: A. B. C. D.

500 600 700 800

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QUESTION 10. An engineer bought a machine with one year’s free maintenance. In the second year the maintenance cost is estimated to be $20. In subsequent years the maintenance costs will increase $20 per year. The amount ($) that must be set aside now at 6% interest to pay the maintenance costs on the machine for the first six years is close to: A. B. C. D.

101 164 229 284

QUESTION 11. One thousand dollars is borrowed for one year at an interest of 1% per month. If this same sum of money were borrowed for the same period at an interest rate of 12% per year, the saving ($) in interest charges would be close to: A. B. C. D.

0 3 5 7

QUESTION 12. An investor is considering buying a 20-year corporate bond with a face value of $1000 paying 6% interest per year in two semi-annual payments. The purchaser will receive $30 every six months and in addition he will receive $1000 at the end of 20 years along with the last $30 interest payment. If the investor thinks that he should receive 8% interest rate compounded semiannually, the amount ($) he is willing to pay is nearly: A. B. C. D.

500 600 700 800

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QUESTION 13. Annual maintenance costs for a particular section of highway pavement are $2000. The placement of a new surface would reduce the annual maintenance costs to$500 per year for the first five years and to $1000 per year for the next five years. The annual maintenance after 10 years would again be $2000. If the maintenance costs are the only saving, the maximum investment ($) that can be justified for the new surface, with an interest of 4% is close to: A. 5500 B. 7170 C. 10,000 D. 10,340

QUESTION 14. A project has an initial cost of $10,000, uniform annual benefits of $2400 and a salvage value of $3000 at the end of its 10-year useful life. At 12% interest rate, the net present worth ($) of the project is close to: A. B. C. D.

2500 3500 4500 5500 NPW = A (P/A, i, n) + S (P/F, i,n) - Cost

QUESTION 15. A person borrows $5000 at an interest rate of 18%, compounded monthly. The monthly payments of $167.10 are agreed. The length (months) of the loan is nearly: A. B. C. D.

12 20 24 40

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QUESTION 16. A machine costing $2000 to buy and $300 per year to operate will save labor expenses of $650 per year for eight years. The machine will be purchased if its salvage value at the end of eight years is sufficiently large to make the investment economically attractive. If an interest rate of 10% is used, the minimum salvage ($) value must be close to: A. B. C. D.

100 200 300 400

QUESTION 17. The amount of money deposited 50 years ago at 8% interest that would now provide a perpetual payment ($) of $10,000 per year is close to: A. 3,000 B. 8,000 C. 50,000 D. 125,000

QUESTION 18. With an interest rate of 2% per month, money will double (months) in value in: A. B. C. D.

20 22 24 35

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QUESTION 19. An industrial firm must pay a local jurisdiction the cost to expand its sewage treatment plant. In addition, the firm pays $12,000 annually toward the plant operating costs. The industrial firm will pay sufficient money into a fund that earns 5% per year to pay its share of the plant operating costs for ever. The ($) amount to be paid to the fund is close to: A. 15,000 B. 30,000 C. 60,000 D. 240,000

QUESTION 20. An engineer deposited $200 quarterly in her savings account for three years at 6% interest, compounded quarterly. Then for five years, she made no deposits or withdrawals. The amount ($) in the account after eight years is: A. B. C. D.

1200 1800 2400 3600

QUESTION 21. If $200 is deposited in a savings account at the beginning of each year for 15 years and the account earns interest at 6%, compounded annually the value ($) of the account at the end of 15 years will be most nearly: A. B. C. D.

4500 4700 4900 5100

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QUESTION 22. The maintenance expense on a piece of machinery is estimated as follows: _________________________________________________________ Year 1 2 3 4 Maintenance ($) 150 300 450 600 _________________________________________________________ If interest is 8%, the equivalent uniform annual maintenance cost is close to: A. B. C. D.

250 300 350 400

QUESTION 23. A manufacturer purchased $15,000 worth of equipment with a useful life of six years and a $2000 salvage at the end of the six years. Assuming a 12% interest rate, the equivalent uniform annual cost ($) is close to: A. B. C. D.

1500 2500 3500 4500

QUESTION 24. The equivalent uniform cost ($) for the machine based on 10% interest for the following data is nearly: Initial cost: $80,000 Salvage value: $20,000 Annual operating cost: $18,000 Useful life: 20 years A. B. C. D.

21,000 23,000 25,000 27,000

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QUESTION 25. Consider the following data: Initial cost: $80,000 Annual operating cost: $18,000 Useful life: 20 years Interest rate: 10% The salvage value ($) of the machine at the end of 20 years for the machine to have an equivalent uniform annual cost of $27,000 is close to: A. B. C. D.

10,000 23,000 30,000 40,000

QUESTION 26. The rate of return (%) for a $10,000 investment that will yield $1000 per year for 20 years is close to: A. 1 B. 4 C. 8 D. 12

QUESTION 27. An engineer invested $10,000 in a company. In return he received $600 per year for six years and his $10,000 investment back at the end of the six years. The rate of return (%) on the investment is close to: A. 6 B. 10 C. 12 D. 15

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QUESTION 28. An engineer made 10 annual end of year purchases of $1000 common stock. At the end of the tenth year, just after the last purchase, the engineer sold all the stock for $12,000. The rate of return (%) on the investment is nearly: A. 2 B. 4 C. 8 D. 10

QUESTION 29. Consider the following data: Initial cost: $80,000 Salvage value: $20,000 Annual operating costs: $18,000 Useful life: 20 years The machine will produce an annual savings in material of $25,700. The rate of return (%) is nearly: A. 6 B. 8 C. 10 D. 15

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QUESTION 30. Two mutually exclusive alternatives are being considered: ___________________________________________ Year A B ___________________________________________ 0 -$2500 -$6000 1 +$746 +$1664 2 +$746 +$1664 3 +$746 +$1664 4 +$746 +$1664 5 +$746 +$1664 ____________________________________________ The rate of return (%) on the difference between the alternatives is close to: A. 6 B. 8 C. 10 D. 12

QUESTION 31. A project will cost $50,000. The benefits at the end of the first year are estimated to be $10,000, increasing $1000 per year in subsequent years. Assuming a 12% interest rate, no salvage value, and an eight-year analysis period, the benefit-cost ratio is close to: A. B. C. D.

0.78 1.00 1.28 1.45

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QUESTION 32. Two alternatives are being presented below: _____________________________________________ A B _____________________________________________ Initial cost $500 $800 Uniform annual benefit $140 $200 Useful life years 8 8 _____________________________________________ The benefit-cost ratio of the difference between the alternatives, based on a 12% interest rate, is close to: A. B. C. D.

0.60 0.80 1.00 1.20

QUESTION 33. A piece of property is purchased for $10,000 and yields a $1000 yearly profit. If the property is sold after five years, the minimum price to break even, with interest at 6% is close to: A. B. C. D.

5000 6500 7700 8300

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QUESTION 34. The benefit-cost ratio for the following data: Project cost: $60,000,000 Gross income: $ 20,000,000 per year Operating costs: $5,500,000 per year Salvage value after 10 years: none The project life is ten years. Using 8% interest rate, the benefit-cost ratio is: A. B. C. D.

0.80 1.00 1.20 1.60

QUESTION 35. Given two machines: _________________________________________________ A B _________________________________________________ Initial costs $55,000 $75,000 Total annual costs $16,200 $12,450 _________________________________________________ With interest at 10% per year, at what service life (years) do these two machines have the same equivalent uniform annual cost? A. B. C. D.

4 5 6 8

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QUESTION 36. Two alternatives are being considered: _______________________________________________________ A B _______________________________________________________ Cost $1000 $2000 Useful life in years 10 10 Salvage value $100 $400 _______________________________________________________ The net annual benefit of alternative A is $150. If interest rate is 8%, what must be the net annual benefit ($) of alternative B for the two alternatives to be equally desirable? A. B. C. D.

150 200 225 275

QUESTION 37. A firm has determined that the two best paints for its machinery are Tuff coat at $45 per gallon and Quick at $22 per gallon. The Quick paint is expected to prevent rust for five years. Both paints take $40 of labor per gallon to apply, and both cover the same area. If a 12% interest rate is used, how long (years) must the Tuff coat paint prevent rust to justify its use? A. B. C. D.

5 6 7 8

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QUESTION 38. A municipal bond is being offered for sale for $10,000. It is a zero-coupon bond that pays no interest during its 15 year life. At the end, the owner of the bond will receive a single payment of $26,639. The bond yield (%) is close to: A. B. C. D.

4 5 6 7

QUESTION 39. Special tools for the manufacture of finished plastic products cost $15,000 and have an estimated $1000 salvage value at the end of an estimated three year useful life and recovery period. The third-year straight line depreciation ($) is close to: A. B. C. D.

3000 3500 4000 4500

QUESTION 40. A firm is considering purchasing $8000 of small hand tools for use on a production line. It is estimated that the tools will reduce the amount of required overtime work by $2000 the first year, with this amount increasing by $1000 per year thereafter. The payback period (years) for the hand tools is: A. B. C. D.

2.00 2.50 2.75 3.00

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QUESTION 41. An engineer is considering the purchase of an annuity that will pay $1000 per year for 10 years. The engineer feels he should obtain a 5% rate of return on the annuity after considering the effect of an estimated 6% inflation per year. The amount he would be willing to pay to purchase the annuity is close to: A. B. C. D.

1500 3000 4500 6000

QUESTION 42. An automobile costs $20,000 today. You can earn 12% tax-free on an auto purchase account. If you expect the cost of the auto to increase by 10% per year, the amount you would need to deposit ($) in the account to provide for the purchase of the auto five years from now is close to: A. B. C. D.

12,000 14,000 16,000 18,000

QUESTION 43. An engineer purchases a building lot for $40,000 cash and plans to sell it after five years. If he wants an 18% before tax rate of return, after taking the 6% annual inflation rate into account, the selling price ($) is nearly: A. 55,000 B. 65,000 C. 75,000 D. 125,000

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QUESTION 44. Consider the following cash flow: ___________________________ Year Cash flow ___________________________ 0 +P 1 0 2 0 3 -400 4 0 5 -600 ___________________________ Assuming a 12% interest rate, P is close to: A. B. C. D.

200 400 600 800

QUESTION 45. Jim would like to purchase a land that costs $1000. The amount he needs to deposit ($) in the savings account each month at 6% interest rate is: A. 41 B. 81 C. 120 D. 150

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QUESTION 46. Consider the following cash flows: ___________________________________ Year Cash flow ___________________________________ 1 +100 2 +100 3 +100 4 0 5 -F ___________________________________ With a 15% interest, the value of F is close to: A. B. C. D.

460 520 640 720

QUESTION 47. A county will build an aqueduct to bring water in two ways. It can build at a reduced size now for $300 million and be enlarged 25 years hence for an additional $350 million. An alternative is to construct the full sized aqueduct for $400 million. Both alternatives provide the needed capacity for the 50 years analysis period. Maintenance costs are negligible. At 6% interest rate which alternative should be selected? A. B. C. D.

Two-stage construction Single stage construction Either alternative is OK In sufficient information provided in the question

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QUESTION 48. A firm is trying to decide which of two weighing scales it should install to check a packagefilling unit. If both scales have same useful life period of 6 years, which one should be selected and the estimated savings will be with an interest rate of 8%: Alternatives

Cost

Scale A Scale B

$2000 $3000

A. B. C. D.

Benefits $450 $600

Salvage value 100 700

Scale A; $72 Scale B; $72 Scale A; $90 Scale B: $90

QUESTION 49. In the construction of an aqueduct to expand the water supply to a city, two alternatives are proposed. A tunnel through the mountain or a pipeline go around the mountain. Assuming an interest rate of 6%, which alternative must be selected? The following data apply: _______________________________________________ Tunnel Pipeline _______________________________________________ Initial cost ($) 5.5 million 5 million Maintenance 0 0 Useful life Permanent 50 years Salvage value 0 0 _______________________________________________ A. B. C. D.

Pipeline Tunnel Either option None of the above

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QUESTION 50. Consider the following mutually exclusive alternatives: _____________________________________________________ A B C _____________________________________________________ Initial cost ($) 2,000 4,000 6,000 Uniform annual benefit ($) 410 639 700 _____________________________________________________ Each alternative has a 20-year life and no salvage value. If the minimum attractive rate of return is 6%, which alternative should be selected? A. B. C. D.

A B C None of the above

QUESTION 51. A firm is trying to decide which of two weighing scales it should install to check a packagefilling operation in the plant. If both scales have a 6-year life, and with an interest rate of 8%, the payback period (years) for the alternative A is: ____________________________________________________________ Alternative Cost Annual benefit Salvage value ____________________________________________________________ A 2,000 450 100 B 3,000 600 700 ____________________________________________________________ A. B. C. D.

2.4 3.4 4.4 5.0

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QUESTION 52. Consider the following cash flows: ___________________________________________ Gross income from sales ($) 200 Purchase of tools ($) (Useful life 3 years) -60 All other expenditures ($) -140 ___________________________________________ The taxable income ($) is: A. B. C. D.

10 20 30 40

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PART-II QUESTION 1. A piece of machinery costs $900. After 5 years, the salvage value is $300. Annual maintenance costs are $50. If the interest rate is 8%, the equivalent uniform annual cost (EUAC) is most nearly: A. B. C. D.

$220 $300 $330 $350

QUESTION 2. A project is being considered wherein a small airfield will be built. Land for the project will cost $350,000, construction costs will be $800,000, and annual maintenance will be $25,000/year. Annual benefits expected from this project are $250,000 per year. The Benefit-Cost ratio assuming a rate of 6% and a life of 20 years is close to: A. B. C. D.

0.98 0.99 1.20 2.00

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QUESTION 3. The payback period for the following investment, assuming an interest rate of 6% is nearly: End of Year 0 1 2 3 4 5 6

A. B. C. D.

Investment A - $10,000 $0 $ 2,000 $ 4,000 $ 4,000 $ 2,270 $ 4,000

2 3 4 5

QUESTION 4. A company is trying to determine at what level they must operate their plant just to break even. Their fixed costs are $7500/month and their variable costs are $3.00 per unit. Revenue is expected to be $10/unit.

A. 500 B. 1000 C. 1500 D. 2000 QUESTION 5. The annual maintenance costs for a drilling machine are estimated at $750 the first year, increasing by $50 every year. Assuming a useful life of 8 years and a 6% inflation rate, the equivalent uniform annual cost (EUAC) for maintenance over the useful life is:

A. B. C. D.

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QUESTION 6. A company purchases a plastic injection system that will save the company $43,000 during the first year of operation, decreasing by $2,000 every year to $41,000 the second year, $39,000 the third year, and so forth. Given a MARR of 10% per year, the present worth of the savings over the 4-year life of the machine is closest to: A. B. C. D.

$125,500 $126,500 $127,500 $128,500

QUESTION 7. A company is considering two mutually exclusive alternative projects to enhance its production facility. The respective financial estimates for each project are as follows:

Project A

Project B

Initial Cost

75,000

105,000

Annual Savings

16,000

24,000

Salvage Value

9,000

0

If the useful life of Project A is 4 years, with a MARR of 15%, the useful life in years of Project B that makes both projects equally desirable is most nearly: A. B. C. D.

4 5 6 7

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QUESTION 8. An engineering department is considering purchase of an advanced computational fluid dynamics software system to enhance productivity. The initial cost of the software is $55,000 but is expected to result in efficiency savings of $25,000 the first year, with this amount decreasing by $5,000 per year thereafter. The payback period (years) for the software is closest to: A. B. C. D.

2.00 2.67 3.00 3.67

QUESTION 9. Which of the following alternatives is superior over a 30-year period if the interest rate is 7%? Type Life Initial cost Maintenance

A. B. C. D.

Alternate A Alternate B Brick Wood 30 years 10 years $1800 $450 $5/year $20/year

Alternative A Alternative B Either A or B None of the above

QUESTION 10. A corporation that pays 53% of its profit in income taxes invests $10,000in an asset that will produce $3000 annual revenue for eight years. If the annual expenses are $700, salvage after eight years is $500 and 9% interest is used. What is the after-tax present worth? Neglect depreciation. A. B. C. D.

-$1,000 -$2,500 -$3,700 -$5,000

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QUESTION 11. It costs $1000 for hand tools and $1.50 labor per unit to manufacture a product. Another alternative is to manufacture the product by an automated process that costs $15,000 with a $0.50 per unit cost. With annual production rate of 5,000 units, how long (years) it take to reach the breakeven point? A. B. C. D.

2.0 2.8 3.6 5.0

QUESTION 12. A loan of $10,000 is made today at an interest rate of 15% and the first payment of $3,000 is made 4 years later. The amount ($) that is still due on the loan after the first payment is most nearly: A. 7,000 B. 8,050 C. 8,500 D. 14,500

QUESTION 13. A subsurface remedial treatment technology costs $245,000 to construct initially with annual operation and maintenance costs of $9000 for a 5-yr operational life. Using an annual interest rate of 7%, and no equipment salvage value, the annualized cost for the remedial treatment technology is most nearly: A. B. C. D.

$58,000 $60,000 $65,000 $69,000

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QUESTION 14. A manufacturing process has present operating costs of $ 10.9 million annually and waste disposal costs of $ 3.3 million. The plant manager wishes to upgrade the process by installing new equipment at a capital cost of $ 23.3 million. This will reduce the annual operating and waste disposal costs to $ 7.1 and $ 1.4 million respectively. The saving (in million dollars) from the new process using present worth analysis is close to: Assume no salvage value. Use 12 percent interest for 10 years as a basis for calculations. A. B. C. D.

8.9 7.5 3.2 1.0

QUESTION 15. The total quantity of contaminated soils at the Pepper’s steel and alloy site was 120,000 tons. Estimate the savings (million dollars) from the land filling option versus stabilization for the management of the hazardous waste at this site. Each truck can carry 31,500 pounds to the nearest landfill site at a distance of 850 miles. The trucking cost per mile was $2.00 and the total stabilization cost was $67 per ton. A. B. C. D.

1.9 3.1 4.5 5.0

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QUESTION 16. A company must purchase a cyclone to control dust from a foundry operation. The lowest bid that would meet all requirements with installed cost of $40,000. The cyclone has a service life of 5 years. A bid was also received from another manufacturer that is guaranteed for 10 years and would lower maintenance costs by $1000 per year. The installed cost is $60,000. Both cyclones have zero salvage values. If the company currently receives a 12% return before taxes on all investments, the incremental ROI (%) is nearly:

A. B. C. D.

10 15 20 25

QUESTION 17. Company A and B brought identical venture scrubbers in 1992 that cost $75,000 each. In both applications, the service life was estimated to be 5 years with zero salvage value. The corporate income tax rate for both companies was 50%. Company A used straight-line depreciation and Company B used the MACRS method. The money Company B saved over the first 3 years of service based on its depreciation procedure is close to:

A. B. C. D.

$ 2,500 $ 4,000 $ 6,000 $ 9,000

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QUESTION 18. In the design of a cold storage warehouse, the specifications are for a maximum heat transfer through the warehouse of walls of 30,000 J/m2 of wall when there is a 30C temperature difference between the inside surface and outside surface of the insulation. The two insulation materials considered are:

Insulation material Cost per cubic meter Conductivity $ (J-m/m2-oC-hr) Rock wool 12.50 140 Foamed insulation 14.00 110

The savings ($) that can be achieved by selecting the low cost material per square meter is close to: A. B. C. D.

0.10 0.15 0.20 0.30

QUESTION 19. A firm is considering which of two mechanical devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of 5 years and no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year but will decline $50 annually, making the second year savings $350, the third year $300 and so forth. With interest at 7%, the device that should be purchased and the cost savings are close to: A. Device A; $30 B. Device B; $30 C. Device A; $10 D. Device B; $ 0

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QUESTION 20. Fairfax county will build an aqueduct to bring water in from the upper part of the state. It can be built with a reduced size now for $300 million and be enlarged 25 years hence for an additional $350 million (Alternate A). An alternative (B) is to construct the full-sized aqueduct now for $400 million. Both alternatives would provide the needed capacity for the 50-year analysis period. Maintenance costs are small and may be negligible. At 6% interest which alternative should be selected and the cost savings (million dollars) are close to:

A. B. C. D.

Alternate A; 20 Alternate A; 10 Alternate B; 20 Alternate B; 10

QUESTION 21. A purchasing officer received two quotes with the following information:

Manufacturer Cost Useful life Salvage value ($) (years) ($) Speedy 1500 5 200 Allied 1600 5 325

For a 5-year analysis period, which manufacturer’s equipment should be selected? Assume 7% interest and equal maintenance costs. A. B. C. D.

Speedy Allied Both offer same benefits Not enough information provided to evaluate

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QUESTION 22. A firm is trying to decide which of the two weighing scales it should install to check a package – filling operations in the plant. If both scales have lies equal to the 6-year analysis period, the cost savings ($) are close to: Assume an 8% interest rate. Alternatives

Cost ($)

Atlas scale Thumb scale

2000 3000

A. B. C. D.

Uniform annual Benefit ($) 450 600

Salvage Value ($) 100 700

20 40 70 90

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QUESTION 23. After purchasing a quarry and basic crushing equipment, the contractor is considering an alternative plan to improve the operation of the quarry. The alternative plan will produce an equal amount of crushed rock and Parameter Present Alternative equal revenue. plan plan First cost ($) 0 10,000 Salvage 0 1,000 Annual cost ($) 250,000 248,000 Life (years) 5

The benefit-cost ratio of the alternative plan (using a 10% rate of return on investment) when compared to the present plan is most nearly: A. B. C. D.

0.6 0.8 1.0 1.2

QUESTION 24. A contractor has been awarded the contract a 6-mile long tunnel in the mountains. During the 5year construction period, the contractor will need water from a nearby stream. He will construct a pipeline to convey the water to the main construction yard. An analysis of costs for various pipe sizes is as follows: Pipe size (in) Installed cost of pipeline and pump Cost per hour for pumping

2 $ 22,000

3 $ 23,000

4

6 $ 25,000

$30,000

$ 1.20

$ 0.65

$0.50

$0.40

The pipe and pump will have a salvage value at the end of 5 years, equal to the cost to remove them. The pump will operate 2000 hours per year. The lowest interest rate at which the contractor is willing to invest money is 7%. The pipe size that can be selected is: A. B. C. D.

2 3 4 6

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QUESTION 25. An investor paid $8000 to a consulting firm to analyze what he might do with a small parcel of land on the edge of town that can be bought for $30,000. In their report, the consultants suggested four alternatives: Alternatives

Total investment Uniform net annual benefit Salvage value Including land ($) ($) ($) Do nothing 0 0 0 Vegetable market 50,000 5,100 30,000 Gas station 95,000 10,500 30,000 Small motel 350,000 36,000 150,000

Assuming 10% is the minimum attractive rate of return, the best option is: A. B. C. D.

Do nothing Vegetable market Gas station Small motel

QUESTION 26. A piece of land may be purchased for $610,000 to be strip-mined for the underlying coal. Annual net income will be $200,000 per year for 10 years. At the end of the 10 years, the surface of the land will be restored to the original condition. The reclamation will cost $1.5 million more than the resale value of the land after it is restored. Using a 10% interest rate, the net present worth of the project is close: A. 40 B. 80 C. 120 D. 180

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QUESTION 27. A firm is considering which of the two devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful lives of 5 years with no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year, but will decline $50 annually, making the second year savings $350, the third year savings $300 and so forth. With interest rate at 7% which device should the firm purchase? A. B. C. D.

Device A Device B Either A or B Not enough information provided

QUESTION 28. Two pumps are being considered for purchase. If interest 7%, which pump should be bought? Pump A

Pump B

Initial cost

$7000

$5000

Salvage value

$1,500

$1,000

Useful life A. B. C. D.

12

6

Pump A Pump B Both are suitable Not enough information is provided

QUESTION 29. A firm is trying to decide which of two devices to install to reduce costs in a particular situation. Both devices cost $1000 and have useful life of 5 years and no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost savings of $400 the first year, bust savings will decline by $50 annually, making the second year savings $350, the third year savings $300 and so forth. With interest at 7%, which device should the firm purchase? A. Device A B. Device B C. Either Device A and B D. Neither Device A or B

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QUESTION 30. The cash flow for two alternatives is as follows:

Year A B 0 -1,000 -2,783 1 +200 +1,200 2 +200 +1,200 3 +1,200 +1,200 4 +1,200 +1,200 5 +1,200 +1,200 Based on payback period, which alternative should be selected? A. B. C. D.

Alternative A Alternative B Either A or B Neither A or B.

QUESTION 31. Consider the relocation and construction of a section of a rural highway. Data on costs are given in the table below. All values are in thousands of dollars. Use a 20-year analysis period with a 10% annual interest rate. Major maintenance will not be done in the 20th year. ________________________________________________________ First cost $ 6,000 Annual maintenance for first 10 years $50 Annual maintenance for second 10 years $75 Major maintenance every 10 years $300 Residual value $3,000 Annual road user costs $660 ________________________________________________________ The present value or present worth of the highway costs ($), ignoring user costs, is most nearly: A. B. C. D.

6,000,000 6,150,000 6,350,000 6,600,000

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QUESTION 32. A utility company can borrow at 6% interest. The company can buy a packaged wastewater plant for $10,000,000 payable up-front and estimates $200,000/year for maintenance. Another option is to lease the plant with a $1,000,000 down payment and $600,000 semi-annual payments (including maintenance) for the 20-year lease period. Assume the plant has a useful life for 20 years and a salvage value of zero. Which of the following statements is most accurate? A. B. C. D.

Buying is better by $2.6 million Buying is better by $4.9 million Leasing is better by $4.4 million Buying and leasing costs are about the same

QUESTION 33 A company purchases a plant for $14.5 million. The plant has a salvage value of $0.5 million at the end of 10 years. Based on the straight-line method for depreciation, the first year depreciation ($) is most nearly: A. B. C. D.

2.80 million 1.50 million 1.45 million 1.40 million

QUESTION 34. A subdivision developer asks your opinion on whether to construct roads all at once or stages. He finds he can put in the base course and pavement complete now for $210,000. As an alternative, the county engineer will permit him to install only base course now (cost estimated at $120,000), with the paving installed two years from now (cost estimated at $100,000). If the developer borrows at 10% interest, which do you recommend as the more economical alternative? A. B. C. D.

All at once Stage Construction Both options are OK None of the above

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QUESTION 35. A city finances its park acquisitions from a special tax of $0.02 a bottle, can or glass of all beverages sold in the city. It derives $100,000 a year from this tax. The city has an option to buy a 40-acre lakefront park site for $1,000,000 as soon as the money is available. How soon will this be if the tax receipts are invested annually at the end of each year at 6 percent interest? A. 4 B. 6 C. 8 D. 10

QUESTION 36. A young engineer decides to save for a down payment on a new car to be purchased 2 ½ years from now. She decides to deposit 450 per month into a savings account paying 6% nominal interest compounded monthly. She plans to make a series of 20 regular monthly deposits with the first deposit scheduled for one month from today. The account will then receive no more deposits but will receive monthly compound interest on the balance for another 10 months. The amount of money ($) in the account at the end of 30-month period is close to: A. B. C. D.

1000 1100 1200 1300

QUESTION 37. A new retirement community will need to add a second unit to the new water treatment plant when the population reaches 40,000. The community has just been built and is starting out now with zero population. An average of 6,000 new residents is expected to move into the community each year. The estimated birthrate for the community is 8 per 1,000 and the death rate is 14 per 1000 resulting in a negative increase of -0.6 percent. The time required before the second unit is needed for the new water treatment plant is nearly: A. B. C. D.

4 5 7 9

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QUESTION 38. A contractor buys a $10,000 truck for nothing down and $2,600 per year for five years. The interest rate she is paying is close to: A. 6 B. 8 C. 10 D. 12

QUESTION 39. A new civic auditorium is estimated to cost $5,000,000 and produce net annual revenue of $500,000. If funds to construct the auditorium are borrowed at 8% interest, how many annual payments of $500,000 are required to retire the debt? A. B. C. D.

10 20 30 40

QUESTION 40. A bridge authority collects $0.25 per vehicle and deposits the revenue once a month in an interest bearing account. Currently 600,000 vehicles a month cross the bridge and the number is expected to increase at a rate of 3,500 vehicles per month. The equivalent uniform monthly revenue ($) would be generated over the next five years if the interest in the account were 6% nominal compounded monthly is nearly: A. B. C. D.

100,000 125,000 150,000 175,000

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QUESTION 41. You are asked to compare current garbage pickup costs versus the cost of a new mechanical arm extending from the garbage truck which pick up garbage cans at curbside and empties them into the truck. The estimated costs are: New mechanical arm: $22,000 Salvage value in 10years: $2000 O&M first year: $500/year Increase in O&M per year: $100/year Interest rate: 6% The mechanical arm will replace one person on the crew who now draws $4,000/year with anticipated increase of $300/year. The cost savings ($) that can be realized by choosing an alternative is: A. 7,800 B. 9,800 C. 10,850 D. 12,440

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QUESTION 42. A county engineer has a choice of paving with either type A pavement or type B pavement. Type A pavement has a life expectancy of ten years, after which part of the material can be salvaged and reused. Type B pavement only lasts five years but is much less expensive. Which is the better alternative? Consider two sequential five-year installations of type B are compared to one ten year-life of type A.

Cost new Annual maintenance Estimated life i A. B. C. D.

Pavement cost ($) per mile Type A Type B 20,000 5,000 1,000 2,000 10 years 6%

5 years 6%

Type A Type B Both A and B provide same cost savings None of the above

QUESTION 43. A contractor has just bought a dragline for $50,000. He expects to keep the dragline for seven years and then sell it. His installment payment contract on the machine runs for five years. He paid $10,000 down and will pay the balance of $40,000 in 60 monthly installments with 1% per month interest on the unpaid balance. The amount of money ($) he needs to charge per cubic yard of earth moved in order to cover the following costs: P&I (Principal and interest) payments O&M costs: $2,000/month Resale value at the end of seven years: $15,000 i =1.5% month on invested capital Production is estimated to average 10,000 yd3 of material per month A. B. C. D.

0.150 0.210 0.268 0.286

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QUESTION 44. A bulldozer is scheduled for a major overhauling costing $10,000 at the end of the third year of its seven year economic life. The equivalent cost over the seven year period for an interest rate of 10% is nearly: A. B. C. D.

1,000 1,100 1,350 1,550

QUESTION 45. A marine dredge has an expected life of 12 years. The estimated maintenance costs for years 3 through 6 years are estimated at $15,000 per year payable in this case at the beginning of the year. The equivalent annual cost ($) over the 12-year life of the dredge with an interest rate of 7% is nearly: A. B. C. D.

3,000 6,000 7,500 9,000

QUESTION 46. An airport runway is expected to incur no maintenance costs for the first five years of its life. In year 6 maintenance should cost $1,000, in year 7 maintenance cost is $2,000 and each year thereafter until resurfacing the runway is expected to increase in maintenance by $1,000 per year. If the resurfacing is expected after 15 years of services, the equivalent uniform annual maintenance cost is incurred if i=7% is close to: A. B. C. D.

2,700 2,800 3,100 3,300

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QUESTION 47. Two roofs are under consideration for a building needed for 20 years. Their anticipated costs and lives are: Cost new Replacement cost Life of roof Salvage value@ 20 years Interest rate (%)

Roof C 50,000 20 years 0

Roof D 25,000 Rise 10%/year 10 years 0

12

12

Based on the annual worth, the roof that provides the lowest cost option is: A. B. C. D.

Roof C Roof D Both provide similar costs None of the above

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QUESTION 48 A consulting firm has a choice of two word processing machines. Each machine will permit a reduction in clerical staff. Estimated costs, lives and salary savings are tabulated below. Assume i=15% Processor A Processor B Cost new -8,000 -17,000 Salary savings +3,500 +4,800 Life (yr) 5 6 Salvage value 0 +4,000 The processor that must be selected is: A. B. C. D.

Processor A Processor B Both A and B None of the above

QUESTION 49. An income of $1,000 per year is expected from a certain enterprise at the end of years 1 through 6. These funds will be invested at 10 percent as soon as received and remain invested at that interest until the end of year 10. The amount ($) of money the account worth at the end of 10 th year is:

A. B. C. D.

10,300 11,300 12,600 14,900

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QUESTION 50. A contractor is arraigning financing for a project he is about to construct. He needs to borrow $1,000 at the beginning of the first month, $2,000 at the beginning of the second month, and increase the amount borrowed by $1,000 through the beginning of the tenth month. He expects payment for the entire job from the owner at the end of the twelfth month. The bank agrees to loan the money at 1% interest compounded monthly on the amount borrowed. The amount ($) the contractor have to repay at the end of the twelfth month is nearly:

A. B. C. D.

34,800 44,800 54,400 58,400

QUESTION 51. A $1000 cost resulted in a lump sum payment of $2,500 at the end of ten years. The rate of return on this investment is:

A. B. C. D.

6.6 7.6 8.6 9.6

QUESTION 52. A bus company is for sale for $150,000. The net income this year is $50,000 but is expected to drop $6,000 per year next year and each year thereafter. At the end of ten years the franchise will be terminated and the company assets sold for $50,000. The rate of return (%) on this investment is:

A. 8 B. 10 C. 15 D. 18

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QUESTION 53. The purchase of an existing ready mixed concrete plant in good operating condition requires a capital investment of $3,700,000 (plant, trucks, property etc). Calculation of O&M costs of producing concrete are $20.52 per cubic yard of concrete (this includes material costs for cement and aggregate as well as O&M costs for the equipment). If the concrete is sold for $45.00 per cubic yard, what annual volume (yd3) of concrete must the company sell to breakeven if their before tax MARR is 20 percent? Assume a 20 year plant life with no salvage value.

A. B. C. D.

27,000 31,000 38,000 44,000

QUESTION 54 A small, but profitable consulting engineering firm is considering hiring another engineer for $30,000 salary +$30,000 benefits per year. Currently their fixed costs total $740,000 per year and their variable costs average $11,200 per engineering contract. Their income per engineering contract averages $19,300. If the new engineer is hired and his costs are all fixed costs, how many more contracts must be acquired for the company to breakeven?

A. 4 B. 6 C. 8 D. 10

QUESTION 55. A water and sewer utility department is considering the acquisition of a bell holder which will dig small bell holes for pipe joints. The equipment costs $5,000 with no predictable salvage after ten years. The machine can dig five holes an hour and requires one operator at $8 an hour. Fuel, maintenance and overheads add another $2 per hour. At present the bell holes are hand dug by laborers at $5 per hour averaging two holes per person per hour. How many holes per year is the breakeven point if i=10 percent?

A. B. C. D.

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QUESTION 56. A dozer is used to push load a fleet of five scrapers. If the dozer is out of service, production stops. Production is usually at 1,000 yd3/hr with costs at $900 per hour and income at $1.20 per cubic yard. With the dozer out of service, the costs continue at $900 per hour but production stops. A standby unit may be rented at $35 per hour. Should the unit be rented? Assume 95.5% availability of the dozer.

A. B. C. D.

Yes rent the dozer No, the unit is expensive to rent Buy a new dozer Not enough information provided.

QUESTION 57. Find the after-tax present worth of taxable income of $410,000 per year for ten years with i-10%. The tax payer is in the 30% tax bracket.

A. B. C. D.

23,000 33,000 43,000 53,000

QUESTION 58. A $1,000 bond paying 6% semiannually ($30) have five years to go until maturity. The current market for these bonds is 8%. That is investors are willing to buy the bond if they can receive a nominal interest rate of 8 percent on their investment. How much will the bond be worth?

A. B. C. D.

919 939 949 999

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QUESTION 59. An investor group holds 3,000acres of undeveloped land which they purchased at $1,000 per acre five years ago. The land is assessed at $500 per acre and the tax rate is $10 per acre per year. At the end of the fourth year they put in some access trails at a cost of $50,000. What must they sell (Million $) it for at the end of the fifth year to realize a before-tax rate of return of 15 % on their investment?

A. B. C. D.

4.3 6.3 8.3 9.9

QUESTION 60. A ten-unit apartment house can be constructed for $200,000. The mortgage company will furnish an 80 percent mortgage at 8 percent for 25 years. Operating expenses are estimated at $3,000 per year. If 90% occupancy is expected, what annual rent ($) is required to return 15% on the investment the first year before any income tax considerations and with zero depreciation?

A. B. C. D.

1,400 2,400 3,200 3,600

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QUESTION 61. An automobile is purchased for $8000. Anticipated annual maintenance costs and salvage values are shown below:

Year

1 2 3 4 5

Annual maintenance cost ($) 0 1000 2000 3000 4000

End of year salvage value ($) 6000 5000 4000 3000 2000

For a i=15%, the economic life of the automobile is:

A. B. C. D.

2 4 6 8

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QUESTION 62. A recreation facility is being designed for use by the public. At this point two maintenance programs are being considered with cots and income as shown: Maintenance Level A

Maintenance Level B

First cost ($)

100,000

100,000

O&M costs

50,000/year

13,000/year

Expected life

20 years

5 years

Income

60,000/year

60,000/yr Decreasing 10,000/year

Which level of maintenance should be selected if i=8%? A. B. C. D.

Maintenance A Maintenance B Either A or B Neither A or B

QUESTION 63. A water service line is required for a new subdivision. Estimate on the costs of the alternative installations are shown. Funding may be through utility revenue bonds currently selling at 6 percent interest. Alternate A

Life 10 years

B

40 years

Capital cost 100,000 replacement increases $50,000 each 10 years $300,000

Annual maintenance $10,000 increasing $5,000/year for 10 years until replacement $10,000 increasing $1,000 per year

No salvage value is anticipated in either case. Which of the two proposals is more economical if owned by a tax exempt municipality? A. B. C. D.

Proposal A Proposal B Either A or B None of the above

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SOLUTIONS

PART-I QUESTION 1 The interest paid per year is:

$330 𝑞𝑢𝑎𝑟𝑡𝑒𝑟

𝑥 4 𝑞𝑢𝑎𝑟𝑡𝑒𝑟𝑠 = $1320

$1320

The nominal interest rate is: $10,000 𝑥 100 = 13.2%

QUESTION 2 The amount of interest earned = $320.52 − $300.00 = $20.52 The effective annual interest rate =

$20.52 $300

𝑥 100 = 6.84%

QUESTION 3 The nominal interest rate is the annual interest rate without compounding. The nominal interest rate = 1.5% 𝑥 12 = 18%

QUESTION 4 𝑟

The effective interest rate can be calculated as: 𝑖𝑒 = (1 + 𝑚)𝑚 − 1 r = 3% for every two months = 18% for one year m = 6 time periods 𝑖𝑒 = (1 +

18% 6 ) − 1 = 0.194 𝑜𝑟 19.4% 6

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QUESTION 5 𝑟

The effective interest rate can be calculated as: 𝑖𝑒 = (1 + 𝑚)𝑚 − 1 (1 + (1 +

𝑟 𝑚 ) = 1 + 𝑖𝑒 𝑚

𝑟 ) = (1 + 𝑖𝑒 )1/𝑚 𝑚

Or 𝑟 = (1 + 𝑖𝑒 )1/𝑚 − 1 𝑚 𝑟 = (1 + 0.1956)1/12 − 1 = 0.015 𝑚 𝑟 = 0.015 𝑥 12 = 18% QUESTION 6 𝑟

The effective interest rate can be calculated as: 𝑖𝑒 = (1 + 𝑚)𝑚 − 1 = (1 +

10% 365 ) − 1 = 0.1052 𝑜𝑟 10.52% 365

QUESTION 7 Given F = $500 i = 4% compounded semi-annual i = 5 years Therefore, i = 4%/2 = 2% semi-annually n = 5 x 2 = 10 time periods 𝑃 𝑃 𝑃 = 𝐹 [ , 𝑖, 𝑛] = $500[ , 2%, 10] 𝐹 𝐹 500[0.8203] = $410

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QUESTION 8 Given A = $400 i = 8% compounded quarterly or 2% per quarter n = 25 years or 100 time periods 𝐹 𝐹 𝐹 = 𝐴 [ , 𝑖, 𝑛] = $400 [ , 2%, 100] 𝐴 𝐴 $400[312.23] = $125,000 QUESTION 9 𝑃 𝑃 𝑃 = 𝐴 [ , 𝑖, 𝑛] + 𝐺 [ , 𝑖, 𝑛] 𝐴 𝐺 𝑃 𝑃 = $120 [ , 4%, 5] + $30 [ , 4%, 5] 𝐴 𝐺 = $120[4.452] + $30[8.555] = $791 QUESTION 10 Given G = $20 i = 6% n= 6 𝑃 𝑃 𝑃 = 𝐺 [ , 𝑖, 𝑛] = $20 [ , 6%, 6] 𝐺 𝐺 = $20[11.459] = $229 QUESTION 11 At i = 1% per month, 𝐹 = $1000 (1 + 0.01)12 = $1126.83 At i = 12% per year, 𝐹 = $1000 (1 + 0.12)1 = $1120.00 Savings in interest charges: $1126.83 − $1120 =$6.83 QUESTION 12 𝑃 𝑃 𝑃𝑊 = $30 [ , 4%, 40] + $1000[ , 4%, 40] 𝐴 𝐹 = $30[19.793] + $1000[0.2083] = $802 QUESTION 13 Benefits are $1500 per year for first five years and $1000 per year for the next five years. The benefits may be considered as $1000 per year for 10 years plus an additional benefit of $500 for first five years. Maximum investment: 𝑃 𝑃 $1000 [ , 4%, 10] + $500 [ , 4%, 5] 𝐴 𝐴 $1000[8.111] + $500[4.452] = $10,337 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 14 𝑁𝑃𝑊 = 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 − 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 𝑁𝑃𝑊 = 𝑃𝑊 𝑜𝑓 𝐴𝑛𝑛𝑢𝑎𝑙 𝐵𝑒𝑛𝑒𝑓𝑖𝑡 + 𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 𝑃 𝑃 $2400 [ , 12%, 10] + $3000 [ , 12%, 10] − $10,000 = $4526 𝐴 𝐹 QUESTION 15 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 = 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 𝑃 $5000 = $167.10[ , 1.5%, 𝑛] 𝐴 𝑃 $5000 [ , 1.5%, 𝑛] = = 29.92 𝐴 $167.10 From 1.5% interest tables, and in P/A column, n = 40 months QUESTION 16 𝑁𝑃𝑊 = 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 − 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 To get minimum salvage value, we equate the equation to zero 𝑃 𝑃 [$650 − $300] [ , 10%, 8] + 𝑆 [ , 10%, 8] − $2000 = 0 𝐴 𝐹 $350[5.335] + 𝑆[0.4665] = $2000 𝑆 = $285 QUESTION 17 The amount of money needed now to begin perpetual payments is: 𝐴 $10,000 𝑃= = = $125,000 𝑖 0.80 Compute P, that would need to have been deposited 50 years ago as: 𝑃 𝑃 = $125,000 [ , 8%, 50] 𝐹 = $125,000[0.0213] = $2663 QUESTION 18 𝐹 2 = 1 [ , 𝑖, 𝑛] 𝑃 𝐹 2 = 1 [ , 2%, 𝑛] 𝑃 From 2% interest tables, n = 35 months QUESTION 19 𝑃=

𝐴 $12,000 = = $240,000 𝑖 5%

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QUESTION 20 𝐹 𝐹 𝐹𝑊 = $200 [ , 1.5%, 12] [ , 1.5%, 20] 𝐴 𝑃 𝐹𝑊 = $200[13.041][1.347] = $3513 QUESTION 21 𝐹 𝐹 𝐹′ = 𝐴 [ , 𝑖, 𝑛] = $200 [ , 6%, 15] 𝐴 𝐴 = $200[23.276] = $4655.20 𝐹 𝐹 𝐹 = 𝐹 ′ [ , 𝑖, 𝑛] = $4655.20 [ , 6%, 1] 𝑃 𝑃 $4655.20[1.06] = $4935 QUESTION 22 𝐴 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝐴𝑛𝑛𝑢𝑎𝑙 𝐶𝑜𝑠𝑡𝑖𝑛𝑔 (𝐸𝑈𝐴𝐶) = $150 + $100 [ , 8%, 4] 𝐺 = $150 + $150[1.404] = $361 QUESTION 23 𝐴 𝐴 𝐸𝑈𝐴𝐶 = $15,000 [ , 12%, 6] − $2000 [ , 12%, 6] 𝑃 𝐹 = $15,000[0.2432] − 2000[0.1232] = $3402 QUESTION 24 𝐴 𝐴 𝐸𝑈𝐴𝐶 = $80,000 [ , 10%, 20] + 18,000 − $20,000[[ , 10%, 20] 𝑃 𝐹 = $80,000[0.1175] + $18,000 − $20,000[0.0175] = $27,050 QUESTION 25 𝐸𝑈𝐴𝐶 = 𝐸𝑈𝐴𝐵 𝐴 𝐴 $27,000 = $80,000 [ , 10%, 20] + $18,000 − 𝑆 [ , 10%, 20] 𝑃 𝐹 $27,000 = $80,000[0.1175] + $18,000 − 𝑆[0.0175] 𝑆 = $22,857 QUESTION 26 𝑃 𝑁𝑃𝑊 = $1000 [ , 𝑖, 20] − $10,000 = 0 𝐴 𝑃 $10,000 [ , 𝑖, 20] = = 10 𝐴 $1,000 From the choices provided, chose choice B or C and look up in the interest rate tables to check for n = 20, P/A column, , the value is 10. If not correct, select the next choice. The correct interest rate is 8%.

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QUESTION 27 $600

The rate of return = $10,00 𝑥 100 = 6% QUESTION 28 𝐹 𝐹 = 𝐴 [ , 𝑖, 𝑛] 𝐴 𝐹 $12,000 = $1000 [ , 𝑖, 10] 𝐴 𝐹 $12,000 [ , 𝑖, 10] = = 12 𝐴 $1000 𝐹 In 4% interest tables, [𝐴 , 𝑖, 10] = 12.006 Therefore, i = 4% QUESTION 29 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = 𝑃𝑊 𝑜𝑓 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑃 𝑃 $80,000 = [$25,700 − $18,000] [ , 𝑖, 20] + $20,000[ , 𝑖, 20] 𝐴 𝐹 Try 8% interest rate $80,000 = $7,700[9.9818] + $20,000[0.2145] = $79,889 Therefore i =8% QUESTION 30 The key word is “difference between the alternatives” Incremental costs:$6000 − $2500 = $3500 Incremental benefits:$1664 − $746 = $918 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡𝑠 = 𝑃𝑊 𝑜𝑓 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 𝑃 $3500 = $918[ , 𝑖, 5] 𝐴 𝑃 $3500 [ , 𝑖, 5] = = 3.81 𝐴 $918 Therefore, i = 10% QUESTION 31 𝐵𝑒𝑛𝑒𝑓𝑖𝑡 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 = 𝐶𝑜𝑠𝑡 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 𝑃 𝑃 $10,000 [𝐴 , 12%, 8] + $1000[𝐺 , 12%, 8] = $50,000 $10,000[4.968] + $1000[14.471] = = 1.28 $50,000

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QUESTION 32 𝐵𝑒𝑛𝑒𝑓𝑖𝑡 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 = 𝐶𝑜𝑠𝑡 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 𝑃 $60[𝐴 , 12%, 8] $60[4.968] = = = 1.00 $300 $300 QUESTION 33 𝐹 𝐹 𝐹 = $10,000 [ , 6%, 5] − $1000 [ , 6%, 5] 𝑃 𝐴 = $10,000[1.338] − $1000[5.637] = $7743 QUESTION 34 𝐵 𝐸𝑈𝐴𝐵 $20,000,000 − $5,000,000 = = = 1.62 𝐴 𝐶 𝐸𝑈𝐴𝐶 $60,000,000[[𝑃 , 8%, 10] QUESTION 35 [𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡]𝐴 = [𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡]𝐵 𝑃 𝑃 $55,000 + $16,200 [ , 10%, 𝑛] = $75,000 + $12,450 [ , 10%, 𝑛] 𝐴 𝐴 𝑃 $75,000 − $55,000 [ , 10%, 𝑛] = = 5.33 𝐴 $16,200 − $12,450 From 10% interest rate tables, n = 8 years QUESTION 36 At breakeven, [𝑁𝑒𝑡 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑊𝑜𝑟𝑡ℎ]𝐴 = [𝑁𝑒𝑡 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑊𝑜𝑟𝑡ℎ]𝐵 𝑃 𝑃 $150 [ , 8%, 10] + $100 [ , 8%, 10] − $1000 𝐴 𝐹 𝑃 𝑃 = [𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐵𝑒𝑛𝑒𝑓𝑖𝑡] [ , 8%, 10] + $400 [ , 8%, 10] − $2000 𝐴 𝐹 𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝐵𝑒𝑛𝑒𝑓𝑖𝑡 = $278 QUESTION 37 [𝐸𝑈𝐴𝐶] 𝑇𝐶 = [𝐸𝑈𝐴𝐶]𝑄𝑢𝑖𝑐𝑘 𝐴 𝐴 [$45 + $40] [ , 12%, 𝑛] = [$22 + $40][ , 12%, 5] 𝑃 𝑃 𝐴 $17.20 [ , 12%, 𝑛] = = 0.202 𝑃 $85 From 12% interest rate tables, n =8 years

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QUESTION 38 Given P =$10,000 F =$26,639 n = 15 years i=? 𝐹 = 𝑃[1 + 𝑖]𝑛 $26,639 = $10,000[1 + 𝑖]15 i = 6.75% QUESTION 39 Straight-line depreciation:𝐷 =

𝐶−𝑆 𝑛

$15,000 − $1000 = $4666 3

QUESTION 40 The annual benefits are $2000, $3000 and $4000. The payback period is the time when $8000 of benefits is received. This will occur in 2.75 years. QUESTION 41 𝑑 = 𝑖 + 𝑓 + [𝑖 𝑥 𝑓] = 0.05 + 0.06 + [0.05 + 0.06] = 0.113 𝑜𝑟 11.3% 𝑃 𝑃 = 𝐴 [ , 11.3%, 10] 𝐴 Since the interest rate is not given, we need to use the equation in page 127, given A finding P (1 + 0.113)10 − 1 = $1000 [ ] = $5816 0.113(1 + 0.113)10 QUESTION 42 Cost of automobile after five years: 𝐹 = 𝑃[1 + 𝑖𝑛𝑓𝑙𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒]𝑛 = $20,000[1 + 0.10]5 = $32,210 Amount of deposit now to have $32,210 available in five years is: 𝑃 𝑃 = 𝐹[ , 𝑖, 𝑛] 𝐹 𝑃 = $32,210 [ , 12%, 5] = $32,210[0.5674] = $18,276 𝐹 QUESTION 43 𝐹 𝐹 Selling price: $40,000 [𝑃 , 18%, 5] [𝑃 , 6%, 5] = $40,000[2.288][1.338] = $122,500

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QUESTION 44 𝑃 𝑃 𝑃 = $400 [ , 12%, 3] + $600 [ , 12%, 5] 𝐹 𝐹 = $400[0.7118] + $600[0.5674] = $625.16 QUESTION 45 Given F = $1000 n = 12 years i = ½% A=? 𝐴 1 𝐴 = $1000 [ , %, 12] 𝐹 2 = $1000[0.0811] = 81.10 QUESTION 46 𝐹 = 𝐹1 + 𝐹2 + 𝐹3 𝐹 𝐹 𝐹 = $100 [ , 15%, 4] + $100 [ , 15%, 3] + $100[ , 15%, 2] 𝑃 𝑃 𝑃 $100[1.749] + $100[1.521] + $100[1.322] = $459.20 QUESTION 47 For the two-stage construction: 𝑃 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = $300 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 + +$350 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 [ , 6%, 25] 𝐹 = $300 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 + $81.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 = $381.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 For the single-stage construction: 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = $400 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 The two-stage construction has a smaller present worth of cost and is the preferred construction plan QUESTION 48 Scale A: 𝑃 𝑃 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 − 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 = $450[ , 8%, 6 + $100 [ , 8%, 6] − $2000 𝐴 𝐹 = $450[4.623] + $100[0.6302] − $2000 = $143 Scale B: 𝑃 𝑃 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 − 𝑃𝑊 𝑜𝑓 𝐶𝑜𝑠𝑡 = $600 [ , 8%, 6] + $700 [ , 8%, 6] − $3000 𝐴 𝐹 = $600[4.623] + $700[0.6302] − $3000 = $215 The criteria is to maximize the PW of benefits minus the PW of cost, the preferred option is Scale B.

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QUESTION 49 Tunnel: 𝐸𝑈𝐴𝐶 = 𝑃𝑖 $5.5 𝑚𝑖𝑙𝑙𝑖𝑜𝑛[0.06] = $330,000 Pipeline: 𝐴 𝐸𝑈𝐴𝐶 = $5 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 [ , 6%, 50] 𝑃 = $5𝑚𝑖𝑙𝑙𝑖𝑜𝑛[0.0634] = $317,000 For fixed output, minimize EUAC, select pipeline QUESTION 50 𝑃 𝑃𝑊 𝑜𝑓 𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 = (𝐴𝑛𝑛𝑢𝑎𝑙 𝐵𝑒𝑛𝑒𝑓𝑖𝑡) [ , 6%, 20] 𝐴 PW of benefits: Alternate A:$410[11.470] = $4703 Alternate B:$639[11.470] = $7329 Alternate C:$700[11.470] = $8209 QUESTION 51 Scale A: 𝑃𝑎𝑦 𝑏𝑎𝑐𝑘 𝑝𝑒𝑟𝑖𝑜𝑑 =

𝐶𝑜𝑠𝑡 $2000 = = 4.4 𝑦𝑒𝑎𝑟𝑠 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝐴𝑛𝑛𝑢𝑎𝑙 𝐵𝑒𝑛𝑒𝑓𝑖𝑡 $450

Scale B: 𝑎𝑦 𝑏𝑎𝑐𝑘 𝑝𝑒𝑟𝑖𝑜𝑑 =

𝐶𝑜𝑠𝑡 $3000 = = 5 𝑦𝑒𝑎𝑟𝑠 𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝐴𝑛𝑛𝑢𝑎𝑙 𝐵𝑒𝑛𝑒𝑓𝑖𝑡 $600

Chose Scale A based on payback period QUESTION 52 Annual depreciation charge: 𝐶 − 𝑆 [$60 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 − 0] = = $20 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑛 3 Taxable income: $200 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 − $140 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 − $20 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 = $40 𝑚𝑖𝑙𝑙𝑖𝑜𝑛

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PART-II QUESTION 1. EUAC = −$900( A/P,8%,5) + $300( A/F,8%,5) − $50 (These factors are in the NCEES FE Reference Handbook.) EUAC = -( $900)(0.2505) + ($300)(0.1705) - $50 = -$225.45 + $51.15 - $50 = $224.30 Answer is (A). QUESTION 2. Present Worth of Benefits = A(P/A,i.n) = $250,000(P/A,6%,20) = $250,000(11.4699) = $2,867,475 Present Worth of Costs = $350,000 + $800,000 + A(P/A,i,n) = $350,000 + $800,000 + $25,000(P/A,6%,20) = $1,436,475 Then B/C = $2,867,475/$1,436,475 > 1.0 and project should be considered. QUESTION 3. Thus at the end of the 0th year, we are $ 10,000 in the hole. For year 1, we add an additional present value of $0 * (P/F, 6%,1) = $0 For the end of year 2, we would add $2000 * (P/F, 6%,2) = $1780 At the end of year 3, we would add $4000 * (P/F, 6%,3) = $3358

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At the end of year 4, we add $4000 * (P/F, 6%,4) = $3168 At the end of year 5, add $2270 * (P/F, 6%,5) = $1696 At the end of year 6, add $4000 * (P/F, 6%,6) = $2820 Thus, adding the additional present values year by year, we see that after the 5th year, we would have our money back, including the effect of interest and the fact that someone else was enjoying our money instead of us. QUESTION 4. The Fixed costs are given as $7500/month, and a variable cost of $ 3.00/unit is given Solving for X, the number of units required to break even: Costs = Benefits $7500 + $3.00X = $ 10.00X X = 1071 units Thus we must make at least 1071 units, just to break even. QUESTION 5. EUAC = $750 + $50 (A/G, 6%, 8) = $750 + $50 (3.1952) = $910 The correct answer is (b) $900. QUESTION 6. P = $43,000 (P/A, 10%, 4) – $2,000 (P/G, 10%, 4) = $43,000 (3.1699) – $2,000 (4.3781) = $127,550 The correct answer is (c) $127,500.

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QUESTION 7. PA = –$75,000 + $16,000(P/A, 15%, 4) + $9,000(P/F, 15%, 4) = –$75,000 + $16,000(2.85498) + $9,000(0.57175) = –$24,175 PB = –$24,175 = –$105,000 + $24,000 (P/A, 15%, n) (–$24,175 + $105,000) / $24,000 = (P/A, 15%, n) 3.36771 = (P/A, 15%, n) The correct answer is (b) 5. QUESTION 8. The payback period is the time when total income to date is equal to the total costs. Costs = $55,000 Since the income is stated as $25,000 per year, one can assume that savings occur uniformly throughout the year.

Costs = Income = $25,000 + $20,000 + $15,000 + … Therefore, the payback period is 2.67 years. The correct answer is (b) 2.67. QUESTION 9. 𝐴 𝐸𝑈𝐴𝐶 (𝐴) = 1800 ( , 7%, 30) + 5 𝑃 = 1800(0.080) + 5 = $150 𝐴 𝐸𝑈𝐴𝐶 (𝐵) = 450 ( , 7%, 10) + 20 𝑝 = 450(0.1424) + 20 = $84 Based on the above Alternative B is attractive, therefore select Alternative B.

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QUESTION 10. P P p P = −$10,000 + $3000 ( , 9%, 8) (1 − 0.53) − $700 ( , 9%, 8) (1 − 00.53) + $500 ( , 9%, 8) A A F = −10,000 + 3000(5.5348)(0.47) − 700(5.5348)(0.47) + 500(0.5019) = −$3766 QUESTION 11. Cumulative cost (hand tools) = $1000 +$1.50 x, where x is the number of units Cumulative cost (automated) = $15,000 +$0.50 x Set both costs equal to get the breakeven: $1000 +$1.50 x = $15,000 +$0.50 x X = 14,000 units. t (breakeven) = 14,000/5000 = 2.8 years

QUESTION 12. Loan due: 𝐹 $10,000 ( , 15%, 4) − $3,000 𝑃 ($10,000)(1 + 0.15)4 − $3,000 $10,000(1.7490) − $3,000 = $14,500 QUESTION 13.

Given: i = 7% n = 5 years P = $ 245,000 From the interest tables, A = 0.2439 Annualized cost

=

0.2439 (245,000) + 9000

=

$68,756

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The present worth factor is: ((1 + 𝑖)𝑛 − 1) (1 + 0.12)10 − 1 𝑃𝑊𝐹 = = = 5.650 𝑖 𝑥 (1 + 𝑖)𝑛 0.12 𝑥 1.1210 For the present process: PW = (10.9 +3.3)(5.65) = $80.233 million For the new process: PW = 23.3 + (7.1+1.4)(5.650) = $ 71.327 million On a present worth basis, the new process saves $ 8.9 million.

QUESTION 15.

Stabilization cost estimate: 120,000 tons x $67/ton = $ 8,040,000 Trucking cost estimate: 120,000 x 2000 pounds/ton/31,500 pounds/truck = 7619 trucks 7619 trucks x 850 miles/truck x $ 2.00/mile =

$ 12,952,381

The total savings are close to $ 5 million QUESTION 16.

Depreciation on carbon-steel cyclone = $40,000/5 = $8000/year Depreciation on stainless steel cyclone = $60,000/10 = $6000/year Total yearly savings with stainless cyclone = ($8000 - $6000) + $1100 = $3100/year Incremental investment = $60,000 - $40,000 = $20,000 Incremental ROI = $3100/$20,000 = 15.5%

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Depreciation claimed by Company A: 𝑑𝐴 =

$75,000 𝑥 3 = $45,000 5

Depreciation claimed by Company B: The MACRS depreciation in the first three years totals 76% of the initial cost. The depreciation charged during those years is 𝑑𝐵 = $75,000 𝑥 0.76 = $57,000 Since corporate tax rate for each company is 50%, Company B saved: 0.50 ($57,000 - $45,000) = $6000 QUESTION 18.

The basic equation for heat conduction through the wall is: 𝑄=

𝐾∆𝑇 𝐿

Where, Q = Heat transfer (J/hr/m2 of wall) K = conductivity (J.m/m2-oC.hr) ∆T = difference in temperature between two surfaces, oC L = thickness of insulating material, m Required insulation thickness: 𝑅𝑜𝑐𝑘𝑤𝑜𝑜𝑙: 30,000 =

140(30) ; 𝐿 = 0.14𝑚 𝐿

Foamed insulation: 30,000 = 110

30 ; 𝐿 = 0.11𝑚 𝐿

Cost of insulation per square meter of wall: Rockwool: $12.50 𝑥 0.14 𝑚 = $1.75 Foamed insulation: $14.00 𝑥 0.11 𝑚 = $1.54 Cost savings: $1.75 − 1.54 = $0.20

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QUESTION 19. The analysis period can conveniently be selected as the useful of the devices or 5 years. Since both devices cost $1000, there is a fixed input (cost) of $1000 regardless of whether A or B is chosen. The appropriate decision criterion is to choose the alternative that maximized the present worth of benefits. PW of benefits A 𝑃 = 300 ( , 7%, 5) = 300(4.100) = $1230 𝐴 PW of benefits B 𝑃 𝑃 = 400 ( , 7%, 5) − 50 ( , 7%, 5) 𝐴 𝐺 = 400(4.100) − 50(7.647) = $1257.65 Device B has the present worth of benefits and is therefore, the preferred alternative. The cost savings are approximately $30. QUESTION 20. This problem illustrates staged construction. The aqueduct may be built in a single stage or in a smaller first stage followed many years later by a second stage to provide the additional capacity when needed. For the Two-stage construction: 𝑃 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = $300 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 + 350𝑚𝑖𝑙𝑙𝑖𝑜𝑛 ( , 6%, 25) 𝐹 = $300 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 + 81.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 = $381.6 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 For the single stage construction: 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = $400 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 The two stage construction has a smaller present worth of cost and is preferred construction plan. The cost savings are $20 million.

QUESTION 21. For the fixed output, the criterion is to minimize the present worth of cost. Speedy: 𝑃 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = 1500 − 200 ( , 7%, 5) 𝐹 = 1500 − 200(0.7130) = $1357

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Allied: 𝑃 1600 − 325 ( , 7%, 5) 𝐹 = 1600 − 325(0.7130) = $1368 Since it is only the difference between alternatives that are relevant, maintenance costs may be left out of the economic analysis. Although the PW of cost for all alternatives are nearly identical, we would nevertheless, choose the one with minimum present worth of cost unless there were other tangible or intangible differences that would change the decision. Buy the speedy equipment.

QUESTION 22. Atlas scale 𝑃 𝑃 𝑃𝑊 𝑜𝑓 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 − 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = 450 ( , 8%, 6) + 100 ( , 8%, 6) − 2000 𝐴 𝐹 = 450(4.623) + 100(0.6302) − 2000 = $143 Thumb Scale 𝑃 𝑃 𝑃𝑊 𝑜𝑓 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 − 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = 600 ( , 8%, 6) + 700 ( , 8%, 6) − 3000 𝐴 𝐹 = 600(4.623) + 700(0.6302) − 3000 = $215 The salvage value of the scale, it should be noted is simply treated as another benefit of the alternative. Since the criterion is to maximize the present worth of benefits minus the present worth of cost, the preferred alternative is the Thumb scale. The cost savings are $70.

QUESTION 23. Benefit = reduction in annual costs = 250,000 − 248,000 = $2,000 4 10% 𝐶𝑜𝑠𝑡𝑠 = 9000 ( ) + 1,000 (0.10) 𝑝 5 𝑦𝑟 = 2,374 + 100 = 2,474 Alternatively, 4 10% 4 10% 𝐶𝑜𝑠𝑡𝑠 = 10,000 ( ) − 1,000 ( ) 𝑝 5 𝑦𝑟 𝑝 5 𝑦𝑟

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2,638 − 164 = 2,474 𝐵 2,000 = = 0.81 𝐶 2,474 QUESTION 24. We can compute the present worth of cost for each alternative. For each pipe size, the present worth of cost is equal to the installed cost of the pipeline and pump plus the present worth of 5 years of pumping costs.

Installed cost of pipeline and pump 1.20x2000 hr (P/A, 7%,5) 0.65 x 2000hrx(P/A, 7%,5) 0.50 x 2000 hr x 4.100 0.40 x 2000hr x 4.1000 Present worth of cost

2” 22,000 9,840

3” 23,000

4” 25,000

6” 30,000

5,330 4,100 3280 31840

28330

29100

33280

Based on the above, select 3” pipe size QUESTION 25. Alternative A represents the “do-nothing” alternative. Generally, one feasible alternative in any situation is to remain in the present status and do nothing. In this problem, the investor could decide that the most attractive alternative is not to purchase the property and develop it. This is clearly a do-nothing decision. We note, however that if he does nothing, the total venture would not be a very satisfactory one. This is because the investor spent $8000 for professional advice on the possible uses of the property. But because the $8000 is a past cost, it is a sunk cost. The only relevant costs in an economic analysis are present and future costs; past events and past, or sunk costs are gone and cannot be allowed to affect future planning. The past should not deter the investor from making the best decision now, regardless of the costs that brought him to this situation and point of time. This problem is one of neither fixed input nor fixed output, so our criterion will be to maximize the present worth of benefits minus the present worth of cost or simply, stated maximize net present worth.

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𝑁𝑃𝑊 = 0 Alternative B, Vegetable market 𝑃 𝑃 𝑁𝑃𝑊 = −50,000 + 5100 ( , 10%, 20) + 30,000 ( , 10%, 20) 𝐴 𝐹 = 50,000 + 5100(8.514) + 30,000(0.1486) = −$2120 Alternative C, gas station 𝑃 𝑃 𝑁𝑃𝑊 = −95,000 + 10,500 ( , 10%, 20) + 30,000 ( , 10%, 20) 𝐴 𝐹 −95,000 + 89,400 + 4460 = −$1140 Alternative D, small motel 𝑃 𝑃 −350,000 + 36,000 ( , 10%, 20) + 150,000 ( , 10%, 20) 𝐴 𝐹 −350,000 + 306,500 + 22,290 = −$21.210 The criterion is to maximize net present worth. In this situation, one alternative has NPW equal to zero and three alternatives have negative values for NPW. We will select the best of the four alternatives, namely the do-nothing Alternative A, with NPW equal to zero.

QUESTION 26. The investment opportunity may be described by the following cash flow: Year 0 1-10 10

Cash flow (thousands) -$610 +200 (per year) -1500 𝑃 𝑃 𝑁𝑃𝑊 = −610 + 200 ( , 10%, 10) − 1500 ( , 10, % 10) 𝐴 𝐹 = −610 + 200(6.145) − 1500(0.3855) = $41

Since NPW is positive, the project is desirable

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QUESTION 27. Device A 𝐸𝑈𝐴𝐵 = $300 Device B 𝐴 𝐸𝑈𝐴𝐵 = 400 − 50 ( , 7%, 5) 𝐺 = 400 − 50(1.865) = $306.75 To maximize EUAB select Device B. QUESTION 28. The annual cost for 12 years for pump A 𝐴 𝐸𝑈𝐴𝐶 = (𝑃 − 𝑆) ( , 𝑖, 𝑛) + 𝑆𝑖 𝑃 𝐴 = (7000 − 1500) ( , 7%, 12) + 1500(0.07) 𝑃 = 5500(0.1259) + 105 = $797 The annual cost for 6 years of Pump B 𝐴 𝐸𝑈𝐴𝐶 = (5000 − 1000) ( , 7%, 6) + 1000(0.07) 𝑃 = 4000(0.2098) + 70 = $909 Based on the annual cost select Pump A.

QUESTION 29. Device A 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = $1000 𝑃 𝑃𝑊 𝑜𝑓 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 = 300 ( , 7%, 5) = 1230 𝐴 𝐵 𝑃𝑊 𝑜𝑓𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 1230 = = = 1.23 𝐶 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 1000

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Device B 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 = $1000 𝑃 𝑃 𝑃𝑊 𝑜𝑓 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 = 400 ( , 7%, 5) − 50( , 7%, 5) = 1258 𝐴 𝐺 𝐵 𝑃𝑊 𝑜𝑓𝐵𝑒𝑛𝑒𝑓𝑖𝑡𝑠 1258 = = = 1.26 𝐶 𝑃𝑊 𝑜𝑓 𝑐𝑜𝑠𝑡 1000 To maximize the benefit-cost ratio, select Device B.

QUESTION 30. Payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment. In the first 2 years, only $400 of the $1000 cost is recovered. The remaining $600 cost recovered in the first half of Year 3. Thus the payback period of Alternative A is 2.5 years. Alternative B Since the annual benefits are uniform, the payback period is simply: =

$2783 = 2.3 𝑦𝑒𝑎𝑟𝑠 $1200

To minimize the payback period, choose alternative B. QUESTION 31 Calculate the present worth for the annual maintenance 𝑃𝐴𝑀 𝑃 = (𝑎𝑛𝑛𝑢𝑎𝑙 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑐𝑒 𝑓𝑜𝑟 𝑓𝑖𝑟𝑠𝑡 10 𝑦𝑒𝑎𝑟𝑠) ( , 10%, 10𝑦𝑟) 𝐴 𝑃 𝑃 + (𝑎𝑛𝑛𝑢𝑎𝑙 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑓𝑜𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 10 𝑦𝑒𝑎𝑟𝑠) ( , 10%, 10𝑦𝑟) ( , 10%, 10𝑦𝑟) 𝐴 𝐹 = $50,000(6.1446) + $75,000(6.1446)(0.3855) = $485,000 Calculate the present worth for the major maintenance

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𝑃 𝑃𝑀𝑀 = (𝑚𝑎𝑗𝑜𝑟 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒) ( , 10%, 10𝑦𝑟) 𝐹 = $300,000(0.3855) = $116,000 Calculate the present worth of residual value 𝑃 𝑃𝑅𝑉 = (𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑣𝑎𝑙𝑢𝑒) ( , 10%, 20𝑦𝑟) 𝐹 = $3,000,000(0.1486) = $446,000 Calculate the sum, noting that the present worth of the first cost P FC = $6,000,000 and the residual value RRV is a deduction. 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ = 𝑃𝐹𝐶 + 𝑃𝐴𝑀 + 𝑃𝑀𝑀 − 𝑃𝑅𝑉 = $6,000,000 + 485,000 + 116,000 − 446,000 = $6,155,000 QUESTION 32 𝑃 𝑃𝐵 = $10,000,000 + $200,000 ( , 20,6%) 𝐴 = $12.29 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑃 𝑃𝐿 = $1,000,000 + $600,000 ( , 40, 3%) 𝐴 = $14.87 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑃𝐵 − 𝑃𝐿 = 12.29 − 14.87 = $2.58 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 PB is better by almost $2.6 million QUESTION 33. Straight-line depreciation: 𝑑=

𝑑=

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑠𝑡 − 𝑠𝑎𝑙𝑣𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 𝑡𝑖𝑚𝑒 (𝑦𝑒𝑎𝑟𝑠)

14,500,000 − 500,000 = $1,400,000 10

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QUESTION 34. The present worth (cost in this case) of each alternative may be determined, and the alternative with the lowest cost selected. Present cost of future pavement cost: 𝑃 𝑃 = 𝐹 ( , 𝑖, 𝑛) 𝐹 𝑃 = 100,000 ( , 10%, 2) = 82,640 𝐹 Present cost of base course: $120,000 Total present cost of alternative: = $82,640 + $120,000 = $202,640 In other words, the developer can either allocate $210,000 (alternative A) for the base and pavement installation all at once now, or allocate $ $202,640 (alternative B) for stage construction. The developer saves (210,000 – 202,640) = $7,360 by utilizing stage construction. QUESTION 35. 𝐴 𝐴 = 𝐹 ( , 𝑖, 𝑛) 𝐹 𝐴 100,000 = 1,000,000 ( , 6%, 𝑛) 𝐹 Then, 𝐴 100,000 ( , 6%, 𝑛) = = 0.1000 𝐹 1,000,000 Find in the 6% table, 𝐴 ( , 6%, 8) = 0.1010 𝐹 This is close to 0.1000. Therefore, the time required is 8 years.

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QUESTION 36. With 6% interest, = 0.5% per month First find the lump sum future amount in the account at the end of the first 20-month period: (1 + 𝑖)𝑛 − 1 (1.005)20 − 1 𝐹1 = 𝐴 [ ] = 50 [ ] = $1,049 𝑖 0.005 Second, find the future amount that will accumulate as a result of the lump sum balance + monthly payments without any additional deposits for the final ten-month period. 𝐹2 = 𝑃(1 + 𝑖)𝑛 = 1,049(1 + 0.005)10 = $1,103 QUESTION 37. The population must reach 40,000 before the second unit is needed, so that F =40,000. The number of new residents expected each year is 6,000, so A =6,000. The residents that are there should have a natural rate of increase of i = -0.006 To solve for the number of years: 𝐹 [(1 + 𝑖)𝑛 − 1] = 𝐴 𝑖 40,000 (0.994)𝑛 − 1 = 6,000 −0.006 Solving for n yields 6.8 years QUESTION 38. 𝐴 𝐴 = ( , 𝑖, 5) 𝑃 𝐴 2600 = 10,000 ( , 𝑖, 5) 𝑃 𝐴 0.26 = ( , 𝑖, 5) 𝑃 From the answer choices, look up the interest rate tables, for n =5, the interest rate is close to 10%.

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QUESTION 39. 𝑃 𝑃 = 𝐴 ( , 𝑖, 𝑛) 𝐴 𝑃 5,000,000 = 500,000 ( , 8%, 𝑛) 𝐴 𝑃 10 = ( , 8%, 𝑛) 𝐴 From the 8% interest rate tables, for 20 years it is close to 9.8181 and for 21 years it is 10.0168. The time required therefore is close to 20 years. QUESTION 40. The monthly costs for the problem in terms of A and G are determined as follows: 𝐴 = 600,000

𝐺 = 3,500

𝑣𝑒ℎ𝑖𝑐𝑙𝑒𝑠 $0.25 $150,000 𝑥 = 𝑚𝑜𝑛𝑡ℎ 𝑣𝑒ℎ𝑖𝑐𝑙𝑒 𝑚𝑜𝑛𝑡ℎ 𝑣𝑒ℎ𝑖𝑐𝑙𝑒𝑠 $0.25 $875 𝑥 = 𝑚𝑜𝑛𝑡ℎ 𝑣𝑒ℎ𝑖𝑐𝑙𝑒 𝑚𝑜𝑛𝑡ℎ

𝑛 = 5𝑦𝑒𝑎𝑟𝑠 𝑥

12𝑚𝑜𝑛𝑡ℎ𝑠 = 60 𝑚𝑜𝑛𝑡ℎ𝑠 𝑦𝑒𝑎𝑟

The cash flow: 𝐴1 = $120,000 𝐴 𝐴2 = 875 ( , 0.5%, 60) = $24,500 𝐺 120,000 + 24,500 = $174,500 QUESTION 41. Mechanical Arm: Salvage:

-$22,000 𝑃 +2,000 ( , 6%, 10) = +1,120 𝐹

O&M: 𝑃 −500 ( , 6%, 10) = −3,680 𝐴 Increase in O&M:

NPW of mechanical arm:

𝑃 −100 ( , 6%, 10) = −2,960 𝐺 -$27,520

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Manual loading: 𝑃 −4,000 ( , 6%, 10) = −29,440 𝐴 Increase in wages: 𝑃 −300 ( , 6%, 10) = −8,800 𝐺 -$38,320

NPW of manual loading:

The cost savings from mechanical arm is close to $10,800 QUESTION 42. Type A Cost new:

$-20,000 𝑃

Annual maintenance:𝑃 = −1,000 (𝐴 , 6%, 10) = −7,360 𝑃

Less salvage: 𝑃 = 2,500 (𝐹 , 6%, 10) = +1,396 NPW:

$-25,964

Type B Cost new first application:

-$5,000 𝑃

Second application:𝑃 = −5,000 (𝐹 , 6%, 10) = −3,736 𝑃

Annual maintenance:𝑃 = −2,000 (𝐴 , 6%, 10) = −14,720 NPW: $-23,456 Based on the NPW, type B is less expensive and the preferred option. QUESTION 43. Find the present worth of all costs, then find equivalent monthly cost and divide by monthly production to obtain dollar per cubic yard charges. The monthly payments for the dragline in 60 months are: 𝐴 𝐴1 = 40,000 ( , 1%, 60) = $890/𝑚𝑜𝑛𝑡ℎ 𝑃 The present worth of all costs and resale value with i=1.5% per month is: Down payment:

$-10,000 𝑃

Monthly payments:−890 (𝐴 , 1.5%, 60) = 𝑃

O&M:−2,000 (𝐴 , 1.5%, 84) =

−$35,050 −$95,140

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

238 𝑃

Resale:15,000 (𝐹 , 1.5%, 84) =

+4,290

Total:

$135,900

Monthly equivalent: 𝐴 𝐴2 = $135,900 ( , 1.5%, 84) = $2,857 𝑃 The charge per cubic yard: $2,857 = $0.286 𝑦𝑑3 10,000

QUESTION 44. The problem can be solved by converting the lump sum to an equivalent lump sum at the beginning of the periodic series, and then converting again to a periodic series as follows: 𝑃 𝐴 𝐴 = 𝐹 ( , 𝑖, 𝑛) ( , 𝑖, 𝑛) 𝐹 𝑃 𝑃 𝐴 $ 1,543 𝐴 = 10,000 ( , 10%, 3) ( , 10%, 7) = 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑎𝑛𝑛𝑢𝑎𝑙 𝑐𝑜𝑠𝑡 𝐹 𝑃 𝑦𝑒𝑎𝑟 QUESTION 45. The problem may be solved by converting the series to an equivalent lump sum at the beginning of the series, then converting the lump sum to another equivalent lump sum at the of the desired equivalent periodic series, and then converting that last lump sum into the desired period series with n equal to proper n for the whole problem. 𝑃 𝑃 𝐴 𝐴2 = 𝐴1 ( , 𝑖, 𝑛) ( , 𝑖, 𝑛) ( , 𝑖. 𝑛) 𝑎 𝐹 𝑃 𝑃 𝑃 𝐴 𝐴1 = 15,000 ( , 7%, 4) ( , 7%, 1) ( , 7%, 12) 𝐴 𝐹 𝑃 𝑃 𝑃 𝐴 ( , 7%, 4) = 3.3872; ( , 7%, 1) = 0.9346; ( , 7%, 12) = 0.1259 𝐴 𝐹 𝑃 𝐴1 = $5,979 𝑎𝑛𝑛𝑢𝑎𝑙 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑣𝑒𝑟 𝑎 12 𝑦𝑒𝑎𝑟 𝑝𝑒𝑟𝑖𝑜𝑑.

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QUESTION 46. 𝐹 𝐴 $2,719 𝐴 = 1,000 ( , 7%, 11) ( , 7%, 15) = 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑎𝑛𝑛𝑢𝑎𝑙 𝑐𝑜𝑠𝑡 𝐺 𝐹 𝑦𝑒𝑎𝑟 𝐹 𝐴 ( , 7%, 11) = 68.3371; ( , 7%, 15) = 0.03979 𝐺 𝐹 QUESTION 47. Roof C 𝐴 𝐴1 = −50,000 ( , 12%, 20) = −6,695 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑃 Roof D 𝐴 𝐴2 = −25,000 ( , 12%, 20) = −3,348 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑃 𝐹 𝑃 𝐴 𝐴3 = −25,000 ( , 10%, 10) ( , 12%, 10) ( , 12%, 10) = −2,796 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟 𝑃 𝐹 𝑃 Net annual worth: -$6,144 per year Based on the cost analysis, Roof D is the best option. QUESTION 48. The procedure is to calculate the net annual worth for i=15 percent and select the alternative with the greatest net annual worth. 𝐴

Processor A:(𝑃 , 15%, 5) = 0.2983 𝐴 𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑊𝑜𝑟𝑡ℎ = −8,000 ( , 15%, 5) + 3,500 = +$1,114 𝑃 𝐴 ( , 15%, 5) = 0.2983 𝑃 𝐴

𝐴

Processor B: (𝑃 , 15%, 6) = 0.2642; (𝐹 , 15%, 6) = 0.1142 𝐴 𝐴 𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑊𝑜𝑟𝑡ℎ = −17,000 ( , 15%, 6) + 4,800 + 4,000 ( , 15%, 6) = +$765 𝑃 𝐹 The results indicate that processor A has the greatest positive net annual worth.

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QUESTION. 49. 𝑃 𝐹 𝐹 = 𝐴 ( , 𝑖, 𝑛1 ) ( , 𝑖, 𝑛2 ) 𝐴 𝑃 𝑃 𝐹 = $1,000 ( , 10%, 6) ( , 10%, 10) = $11,300 𝐴 𝑃 *QUESTION 50. P F F = G ( , i, n1 ) ( , i, n2 ) G P 𝑃 𝐹 𝐹 = 1,000 ( , 1%, 11) ( , 1%, 14) = $58,400 𝑓𝑢𝑡𝑢𝑟𝑒 𝑣𝑎𝑙𝑢𝑒 𝐺 𝑃

QUESTION 51. PW of benefits: 𝑃 1 = 2,500 ( , 𝑖, 𝑛) = 2,500 ( ) (1 + 𝑖)10 𝐹 PW of costs: 1,000 NPW: 2,500 (

1 ) − 1,000 = 0 (1 + 𝑖)10

Solving for i gives a value of 9.6%.

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QUESTION 52. From the choices, try i=15% 𝑃1 = −$150,000 𝑃 𝑃2 = −𝐺 ( , 15%, 10) = 6000𝑥16.979 = −$101,890 𝐺 𝑃 𝑃3 = 𝐴 ( , 15%, 10) = 50,000𝑥5.0187 = +$250,950 𝐴 𝑃 𝑃4 = 𝐹 ( , 15%, 10) = 50,000𝑥0.2472 = +$12,360 𝐹 𝑃𝑇𝑜𝑡𝑎𝑙 = 𝑃1 + 𝑃2 + 𝑃3 + 𝑃4= + $11,420 The rate of return when PW of income equals the PW of cost. Therefore, the interest rate is 18%.

QUESTION 53. The annual fixed cost of the investment is: 𝐴 𝐴 = 3,700,000 ( , 20%, 20) = $759,830 𝑃 To solve the breakeven: 759,830 + 20.52𝐵 = 45𝐵 𝐵 = $31,040𝑦𝑑3 QUESTION 54. The existing breakeven is: 𝐵(𝑒𝑥𝑖𝑠𝑡𝑖𝑛𝑔) =

740,000 = 91.4 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑠 (19,300 − 11,200)

The new breakeven is: 𝐵(𝑛𝑒𝑤) =

740,000 + 60,000 98.9 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑠 19,300 − 11,200

Additional contracts for breakeven = 98.8 – 91.4 = 7.4 or 8 contracts FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 55. Annual capitalization cost of bell holder: 𝐴 𝐴 = 5,000 ( , 10%, 10) = $813.50 𝑃 There is no salvage value in ten years. Per hole operating costs for bell holer are: Operator: $8/hr Fuel etc: $2/hr Total: 10/hr/5holes/hr = $2/hole

Total cost for bell holer/hole: =

$2 813.50 +( )/(𝑌ℎ𝑜𝑙𝑒𝑠. 𝑦𝑟) ℎ𝑜𝑙𝑒 𝑦𝑟

Total cost for hand labor/hole: $5 ℎ𝑟 = $2.50/ℎ𝑜𝑙𝑒 $2ℎ𝑜𝑙𝑒𝑠/ℎ𝑟 Breakeven point: $2.50 = $2.00 + $813.50/𝑌 Y = 1627 holes/yr QUESTION 56. Without standby, $1.20 1000𝑦𝑑3 $1,146 𝐼𝑛𝑐𝑜𝑚𝑒 = 0.955 𝑥 𝑥 = 𝑦𝑑3 ℎ𝑟 ℎ𝑟 𝐶𝑜𝑠𝑡 = −900/ℎ𝑟 𝑁𝑒𝑡 = 246/ℎ𝑟 With standby

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$1.20 1000𝑦𝑑3 $1,200 𝐼𝑛𝑐𝑜𝑚𝑒 = 𝑥 = 𝑦𝑑3 ℎ𝑟 ℎ𝑟 𝐶𝑜𝑠𝑡 =

$900 $35 + = −$935/ℎ𝑟 ℎ𝑟 ℎ𝑟 𝑁𝑒𝑡 = $265/ℎ𝑟

The results indicate that a standby unit would be profitable any cost less than $1200-$1146 = $54 per hour. Since this one is available at $35 per hour it is profitable to have on hand.

QUESTION 57. Find the after-tax cash flow as: = 𝑡𝑎𝑥𝑎𝑏𝑙𝑒 𝑖𝑛𝑐𝑜𝑚𝑒 𝑥 (1 − 𝑡𝑎𝑥 𝑟𝑎𝑡𝑒) = 10,000(1 − 0.3) 𝑃 𝐴𝑓𝑡𝑒𝑟 𝑡𝑎𝑥 𝑐𝑎𝑠ℎ 𝑓𝑙𝑜𝑤 = 10,000 𝑥 0.7 ( , 10%, 10) = $43,010 𝐴 QUESTION 58. Find the present worth of the remaining ten coupons representing $30 income every six months plus redemption value of $1,000 after five years at 8 percent. Since the payments are semin annual, n = 5x2 = 10 and i =8/2 = 4% PW of 10 semi annual payments: 𝐴 𝑃 𝑃1 = 𝐴 ( , 𝑖, 𝑛) = $30 ( , 4%, 10) = $243 𝑃 𝐴 PW of the $1,000 redemption value in 5 years: 𝑃 𝑃 𝑃2 = 𝐹 ( , 𝑖, 𝑛) = $1,000 ( , 4%, 10) = $676 𝐹 𝐹 Total = $919. The bond should sell for $919 for an 8% nominal interest rate.

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QUESTION 59. Future value of cost 𝑝𝑢𝑟𝑐ℎ𝑎𝑠𝑒 𝑝𝑟𝑖𝑐𝑒 𝐹1 = 3000 𝑎𝑐𝑟𝑒𝑠 𝑥 𝑇𝑎𝑥𝑒𝑠 𝐹2 = 3,000 𝑎𝑐𝑟𝑒𝑠 𝑥

1,000 𝐹 𝑥 ( , 15%, 5) = −$6,034,100 𝑎𝑐𝑟𝑒 𝑃

$10 𝐹 𝑥 ( , 15%, 5) = −$202,300 𝑎𝑐𝑟𝑒 𝑃

𝐹 𝐴𝑐𝑐𝑒𝑠𝑠 𝑟𝑜𝑎𝑑 𝐹3 = $50,000 𝑥 ( , 15%, 1) = −$57,500 𝐴 Selling price required after 5 year for 15% return = $6,293,900

QUESTION 60. Amount invested: 0.20 𝑥 $200,000 = $40,000 15% 𝑟𝑒𝑡𝑢𝑟𝑛 = 0.15 𝑥$40,000 = $6,000 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑎𝑦𝑚𝑒𝑛𝑡𝑠 𝑓𝑖𝑟𝑠𝑡 𝑦𝑒𝑎𝑟 = 0.08𝑥$160,000 = 12,800 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑠𝑡𝑠 = $3,000 𝐼𝑛𝑐𝑜𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = $21,800 𝐴𝑛𝑛𝑢𝑎𝑙 𝑟𝑒𝑛𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =

$21,800 $2,422 = 10𝑥0.9 𝑦𝑒𝑎𝑟

QUESTION 61.

Year Annual worth of purchase Annual maintenance Annual salvage Net Annual $8,000(A/P,15%,n) $1000(A/G,15%,n) F(A/F,15%,n) worth 1 -9200 0 +6000 -3200 2 -4920 -464 +2324 -3060 3 -3504 -908 +1152 -3260 4 -2800 -1326 +600 -3526 5 -2384 -1728 +296 -3816

The economic life is two years based on the information in the last column. FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 62. Plan A 𝐴 𝐴𝑛𝑛𝑢𝑎𝑙 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑠𝑡 = −100,000 ( , 8%, 20) = −10,190 𝑃 𝐴𝑛𝑛𝑢𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒, 𝑂&𝑀 = 60,000 − 50,000 = +10,000 𝑁𝑒𝑡 = −190 Plan B 𝐴 𝐴𝑛𝑛𝑢𝑎𝑙 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑠𝑡 = −100,000 ( , 8%, 5) = −25,050 𝑃 𝐴 𝐴𝑛𝑛𝑢𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 = 60,000 − 10,000 ( , 8%, 5) = +41,535 𝐺 𝐴𝑛𝑛𝑢𝑎𝑙 𝑂&𝑀 𝑐𝑜𝑠𝑡𝑠 = −13,000 𝑁𝑒𝑡 = 3,485 Chose Plan B. QUESTION 63. Find the present worth of each proposal and compare. Proposal A 𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑠𝑡 = 100,000 𝑃 𝑅𝑒𝑝𝑎𝑙𝑐𝑒 𝑖𝑛 10 𝑦𝑒𝑎𝑟 = 𝐹 ( , 6%, 10) = 150,000𝑥0.5584 = 83,760 𝐹 𝑃 𝑅𝑒𝑝𝑎𝑙𝑐𝑒 𝑖𝑛 20 𝑦𝑒𝑎𝑟𝑠 = 200,000 𝑥 ( , 6%, 20) = 62,360 𝐹 𝑃 𝑅𝑒𝑝𝑎𝑙𝑐𝑒 𝑖𝑛 30 𝑦𝑒𝑎𝑟𝑠 = 250,000 ( , 6%, 30) = 43,525 𝐹 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑃 = 𝐴 ( , 6%, 40) = 10,000 𝑥 15.046 = 150,460 𝑦𝑒𝑎𝑟 𝐴 𝐴 𝑃 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝐺 ( , 6%, 10) ( , 6%, 40) = 302,583 𝐺 𝐴 𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 40 𝑦𝑒𝑎𝑟𝑠 𝑜𝑓 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 = $742,688 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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Proposal B 𝐹𝑖𝑟𝑠𝑡 𝑐𝑜𝑠𝑡 = $300,000 𝑃 𝑀𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 = 𝐴 ( , 6%, 40) = 10,000 𝑥 15.046 = 150,460 𝐴 𝑃 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝐺 ( , 6%, 40) = 1,000 𝑥 185.957 = 185,957 𝐺 𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 40 𝑦𝑒𝑎𝑟𝑠 𝑜𝑓 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 = $636,417 The totals indicate that Proposal B is the preferred alternative under these circumstances.

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Electricity, Power, and Magnetism Total Questions 3-5

A. B. C. D. E.

Charge, current, voltage, power, and energy Current and voltage laws (Kirchhoff, Ohm) Equivalent circuits (series, parallel) AC circuits Motors and generators

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Question 1. A point charge of 0.3 x 10-3 C is at origin. The magnitude of the electrical field intensity (kV/m) at a point located 2-m in x-direction, 3-m in y-direction and 4-m in z-direction from the origin is: A. 5 B. 50 C. 93 D. 930

Question 2. A 15µC point charge is located on the y-axis at (0, 0.25). A second charge of 10µC is located on the x-axis at (0.25, 0). If the two charges are separated by air, the force (N) between the charged particle is: A. B. C. D.

0.09 0.34 3.49 10.8

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Question 3. A thin metal plate, 20cm x 20cm carries a total charge of 24µC. The magnitude of the electrical field (N/C) 2.5 cm away from the center of the plate is close to: A. B. C. D.

1.27 x 103 3.70 x 106 3.39 x 107 4.34 x 108

Question 4. A parallel plate capacitor separated by an air gap of 1cm and with an applied voltage across the plates of 500 V. The force (N) on an electron mass of 18.2 x 10-31 kg inserted in the space is nearly: The mass of an electron: 9.1 x 10-31 kg A. B. C. D.

1.6 x 10-14 3.2 x 10-14 9.1 x 10-31 1.6 x 10-19

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Question 5. Two long wires, bundled together have a magnetic flux density around them One wire carries a current of 5 A and the other a current of 1A in the opposite direction. The magnitude of the flux density (T) at a point 0.2m away is close to: A. B. C. D.

2π x 10-6 4 x 10-6 4π x 10-6 16π x 10-6

Question 6. For a coil of 100 turns wound around a toroidal core of iron with a relative permeability of 1000. The current needed to produce a magnetic flux density of 0.5T in the core is close to: R = 5 cm r = 1 cm

A. 1.2 B. 12 C. 39 D. 390

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Question 7. A current passes through a 0.2 H inductor. The current increases in linear fashion from a zero at t=0 to 20A at t=20s. The amount of energy stored in the inductor at 10s is nearly: A. 0 B. 1 C. 10 D. 40

Question 8. The expression for the complex number 3 + j4 in polar form is: A. B. C. D.

3∟36.87o 5 ∟36.87o 3 ∟53.13o 5 ∟53.13o

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Question 9. If z1 = 5∠25𝑜 and z2 = 3∠40𝑜 then z1 +z2 is: A. B. C. D.

4.07 ∠81.72𝑜 5.81 ∠47.21𝑜 7.64 ∠32.57𝑜 7.94 ∠30.60𝑜

Question 10. If z1 = 24.2 ∠32.3𝑜 and z2 = 16.2 ∠45.8𝑜 , z1 x z2 and z1/z2 is: A. B. C. D.

40.4∠78.1𝑜 ; 8.01 ∠13.5𝑜 392∠78.1𝑜 ; 1.49 ∠−13.5𝑜 398∠39.1𝑜 ; 8.03 ∠1.4𝑜 241∠268.5𝑜 ; 135 ∠38.5𝑜

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Question 11. A heating element consists of two wires of different materials connected in series. At 20C, they have resistances 600Ω and 300Ω and average temperature coefficients of 0.001 /C and 0.004/C, respectively. The total resistance (Ω) of the heating element at 50C is close to: A. B. C. D.

900 950 980 990

Question 12. A solid copper conductor at 20C has the following parameters: Resistivity: 1.77 x 10-8Ω.m Diameter: 5mm Length: 5000m The resistance (Ω) of the conductor is: A. B. C. D.

0.02 4.55 12.1 18.9

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Question 13. A power line is made of copper having a resistivity of 1.83 x 10-6Ω.m. The wire diameter is 2cm. The resistance (Ω) of 5km of power line is: A. 0.03 B. 30 C. 67 D. 303

Question 14. A 10kV power line 5km long has a total resistance of 0.7Ω. The current flowing is 10A. The power (W) dissipated in the line resistance is: A. 7 B. 14 C. 70 D. 700

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Question 15. Two terminals are connected to 20V source and a current of 10A is measured through a 10Ω resistance. The current (A) that would flow through the resistance if a 30V source is connected across the terminals is: A. B. C. D.

12 15 17 20

Question 16. A current of 10A flows through a 1mm diameter wire. The average number of electrons per second that pass through the cross-section of the wire is: A. B. C. D.

1.6 x 1018 6.2 x 1018 1.6 x 1019 6.3 x1019

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Question 17. The internal resistance of 9V battery that delivers 100 A when the terminals are shorted is close to: A. B. C. D.

0.09 1.0 11 90

Question 18. The reactances of a 10mH inductor and a 0.2µF capacitor are equal if the frequency (kHz) is: A. 3.56 B. 7.12 C. 14.76 D. 21.55

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Question 19. The resonance frequency (rad/s) of a circuit if 3H, 5Ω and 8µF are connected in series is close to: A. 0 B. 20 C. 100 D. 200

Question 20. A two-pole AC motor operates on a three-phase, 60Hz 200 V line-to-line supply. synchronous speed (rpm) is: A. B. C. D.

The

1000 1800 2400 3600

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Question 21. A three-phase synchronous generator is operating at 1200 rpm with an output frequency of 60 Hz. The number of poles in the generator is: A. B. C. D.

2 4 6 8

Question 22. A two-pole induction motor operates on a three phase, 60Hz, 240 V line-to-line supply. The motor speed is 3420 rpm. The slip (%) is: A. 5 B. 7 C. 11 D. 15

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Question 23. A four-pole induction motor operates on a three-phase 60Hz, 240Vrms line-to-line supply. The slip is 2%. The operating speed (rpm) is: A. B. C. D.

1240 1660 1760 1800

Question 24. A charge is placed at the center of the line joining two equal charges Q. The system of 3 charges will be in equilibrium, if Q is equal to: Q

A. – 2

Q

B. – 4 C. D.

𝑄 2 𝑄 4

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Question 25. If A and B are mutually perpendicular vectors give by A = 5i + 7j – 3kk and B = 2i + 2j – ck, the value of C is: A. B. C. D.

-8 -6 -2 0

Question 26. Four two-ohms resistors are connected together along the edges of a square. A 10V battery of negligible resistance is connected across a pair of the diagonally opposite corners of the square. The power (J) dissipated in the circuit is: A. B. C. D.

20 30 40 50

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Question 27. A 2 µF conductance is charged to 500 V and then the plates are joined through a resistance. The heat (J) produced in the resistance is: A. B. C. D.

0.25 0.50 0.75 1.0

Question 28. The ratio of L/R, where L is inductance an R is resistance have the same dimensions as: A. B. C. D.

Velocity Acceleration Time Force

Question 29. A varying current in a circuit changes from 10A to zero in 0.5s. If the average emf induced in the circuit is 220V, the inductance (H) is close to: A. 5 B. 8 C. 9 D. 11

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Question 30. A current is passed through two resistances (R1 and R2) connected in series. The voltage difference across the first resistance is 3 V and in the second resistance is 4.5. If the R 1 is 2 ohms, the R2 is: A. B. C. D.

3 5 7 9

Question 31. The dimensions of resistance x capacitance are same as: A. B. C. D.

Frequency Energy Time period Current

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Question 32. The electrical potential at the center of a charged conductor is: A. B. C. D.

Zero Twice that on the surface Half that on the surface Same as on the surface

Question 33. Four bulbs each marked 40W, 250V are connected in series with a 250V source. The total power (W) output is: A. 10 B. 40 C. 80 D. 160

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Question 34. Two capacitors 2µF and 4 µF are connected in series to a 100V battery. The energy (J) stored in the capacitors is: A. B. C. D.

0.0033 0.0067 0.057 0.067

Question 35. A current passes through a long straight wire. At 5cm distance, the magnetic field is B, the magnetic field at 20cm will be: A. B B B. 2 C. 2B B D. 4

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Question 36. If M, L, T and I are mass, length, time and electrical current, respectively, the dimensional formula for capacitance is: A. B. C. D.

M-1L2T-4I2 M-1L-2T4I2 ML2T4I2 ML2T-4I-2

Question 37. Three capacitors 3µF, 10µF and 15µF are connected in series to a 100V source. The charge (µC) on 15µF capacitance is: A. 22 B. 100 C. 200 D. 2,800

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Question 38. Two charges 5µC and 10µC are placed 20cm apart. The electric field at midpoint between the two charges is: A. B. C. D.

4.5 x 106 N/C directed towards 5µC 4.5 x 106 N/C directed towards 10µC 13.5 x 106 N/C directed towards 5µC 13.5 x 106 N/C directed towards 10µC

Question 39. Two equal charges placed 3cm apart attract each other with a force of 40N. The magnitude of each charge is: 1

Assume 4πε =9 x 109 in SI Units 0

A. 0.2 B. 2 C. 20 D. 200

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Question 40. The electric intensity in air at 20 cm from a point charge is 4.5 x 105 N/C, the magnitude of the charge is: A. B. C. D.

0.1 0.2 1 2

Question 41. Two long parallel wires separated by 1m carry a current of 3A. The force (N/m) between the wires is: µ0 = 4π x 10-7 in SI units A. 2 x 10-7 B. 5 x 10-7 C. 12 x 10-7 D. 18 x 10-7

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Question 42. Four resistances 10, 5, 7, and 3 ohms connected to form sides of a rectangle AB, BC, CD and DA, respectively. Another resistance 10 ohms is connected across the diagonal AC. The equivalent resistance between A and B is: A. 2 B. 5 C. 7 D. 10

Question 43. A wire carries a current of 2A. The magnetic induction (T) at distance 5m from the wire is: A. 4 x 10-8 B. 8 x 10-8 C. 12 x 10-8 D. 16 x 10-8

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Question 44. Electrical charges of 1µC, -1µC and 2µC are placed in the corners for an equilateral triangle ABC having a length of 10cm on each side. The resultant force (N) on the charge C is: A. B. C. D.

0.9 1.8 2.7 3.6

Question 45. Two long parallel wires carry a current of 5A each in opposite direction. If the wires are separated by 0.5m, the force (N) between them is: A. 10-5 attractive force B. 10-5 repulsive force C. 2 x 10-5 attractive force D. 2 x 10-5 repulsive force

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Question 46. A capacitor 4µF is charged to 80V, while another capacitor of 6µF is charged to 30V, when they are connected together, the energy (mJ) lost by 4µF capacitor is: A. B. C. D.

2.5 3.2 4.6 7.8

Question 47. A 4A current passes through a 50 cm long wire with 1mm2 cross sectional area when connected to a 2V battery. The resistivity (ohm.m) is: A. B. C. D.

1 x 10-6 2 x 10-7 3 x 10-7 4 x 10-7

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Question 48. An electron moves with a velocity of 2 x 105 m/s along a positive x-direction in the presence of a magnetic induction B (T) = i + 4j – 3k. The magnitude of the force experienced by the electron is: A. B. C. D.

1.18 x 10-13 1.28 x 10-13 1.61 x 10-13 1.72 x 10-13

Question 49. A capacitance of 10µF is charged to 40V and a second capacitance of 15µF is charged to 30V. If they are connected in parallel, the amount of charge (µC) that flows from 10µF to 15µF is: A. 30 B. 60 C. 90 D. 120

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Question 50. The time (s) required to produce a potential difference of 20V across a capacitor of 1000µF when it is charged at a steady rate of 200µC/s is: A. 50 B. 100 C. 150 D. 200

Question 51. A resistance R is cut into 20 pieces. Half of them joined in series and the rest in parallel. If these two combinations joined in series, the equivalent resistance will be: A. R R B. 2 C. D.

101R 200 201R 200

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Question 52. The dimensional equation for magnetic flux is: A. B. C. D.

ML2T-2I-1 ML2T-2I-2 ML-2T-2I-1 ML-2T-2I-2

Question 53. A current of 30A flows in a straight wire placed in a magnetic field of 4 x 10-4 T. The magnetic induction (T) 2 cm away from the wire is: A. B. C. D.

1 x 10-4 3 x 10-4 5 x 10-4 6 x 10-4

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Question 54. The temperature coefficient of resistivity of a material is 0.0004/K. When the temperature increased by 50C, the resistivity increases by 2 x 10-8 Ω.m. The initial resistivity (Ω.m) of the material is: A. 50 x 10-8 B. 100 x 10-8 C. 150 x 10-8 D. 200 x 10-8

Question 55. A magnetic flux of 500µWb passing through a 200 turn coil reversed in 20ms. The average emf (V) induced in the coil is: A. 2.5 B. 5.0 C. 7.5 D. 10.0

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Question 56. A capacitor of 3pF is connected in series with an inductance L to produce a frequency of 1MHz, the inductance is close to: A. 8.44 B. 84.4 C. 844 D. 8440

Question 57. A 4µF capacitor is connected to a 200V battery. It is disconnected from the battery and now connected with a 2µF capacitor. The loss of energy (J) during the process is: A. B. C. D.

0 2.6 x 10-2 4.0 x 10-2 5.3 x 10-2

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Question 58. Two charges 2C and 6C are separated by a finite distance. If a charge of -4C is added to each of them, the initial force of 12 x 103N changes to: A. B. C. D.

1 x 103 2 x 103 3 x 103 4 x 103

Question 59. A 1H inductance in series with a 220V and at a frequency 50Hz. The inductive reactance (Ω) is: A. 2π B. 50π C. 100π D. 1000π

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Question 60. Two unit negative charges are placed on a straight line. A positive charge “q” is placed at the mid-point between the charges. If the three charges are in equilibrium, the value of “q” (C) is: A. B. C. D.

0.25 0.50 0.75 1.0

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Statics Total Questions 8–12 A. B. C. D. E. F. G.

Resultants of force systems Concurrent force systems Equilibrium of rigid bodies Frames and trusses Centroids Moments of inertia Static friction

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Question 1. Find the maximum force (N) acting on the 100 kg block as shown in the figure is close to: Assume the static friction is 0.30. A. B. C. D.

100 180 280 380

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Question 2. The maximum static friction force (N) acting on a 100-kg block as shown, if P = 500N is close to: Assume the forces are applied with the block initially at rest. Static friction: 0.20 Kinetic friction: 0.17 A. 65 B. 135 C. 185 D. 215

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Question 3. The single threaded screw of the vise has a mean diameter of 1inch and has 5 square threads per inch. The coefficient of static friction in the threads is 0.20. A 60-lb pull applied normal to the handle at A produces a clamping force of 1000 lbs between the jaws of the vice. The frictional moment (lb-in) developed at B due to the thrust of the screw against the body of the jaw is: Assume helix angle α = 3.64o and friction angle = 11.31o

A. B. C. D.

106 166 266 466

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Question 4. The single threaded screw of the vise has a mean diameter of 1inch and has 5 square threads per inch. The coefficient of static friction in the threads is 0.20. A 60-lb pull applied normal to the handle at A produces a clamping force of 1,000 lbs between the jaws of the vice. The moment acting on the screw is 266 lb-in. The force applied normal to the handle at A required to loosen the vise is:

A. B. C. D.

10 30 50 60

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Question 5. A flexible cable supports a 100-kg load is passed over a fixed circular drum and subjected to a force P to maintain equilibrium. The coefficient of static friction μ between the cable and the fixed drum is 0.30. For α = 0, the maximum and minimum values P may have in order not to raise or lower the load is: A. B. C. D.

1600; 600 1200; 400 1000; 800 600; 1500

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Question 6. A flexible cable supports a 100-kg load is passed over a fixed circular drum and subjected to a force P to maintain equilibrium. The coefficient of static friction μ between the cable and the fixed drum is 0.30. For P = 500 N, the minimum value which the angle α may have before the load begins to slip is: A. B. C. D.

20 40 60 90

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Question 7. The area moment of inertia (mm)4 of the area under the parabola about the x-axis is nearly: A. B. C. D.

5 10 15 20

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Question 8. The area moment of inertia (104)(mm4) about the x-axis of the semi-circular area is: A. B. C. D.

10 20 30 40

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Question 9. The resultant of the two forces 800 lb and 600 lb in the figure shown is: A. B. C. D.

125 525 725 925

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Question 10. The 500-N force F is applied to the vertical pole as shown. The force in terms of unit vectors i and j is: A. B. C. D.

(220i – 430j)N (120i – 540j)N (250i – 433j)N (433i – 250j)N

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Question 11. The magnitude of moment about the base point O of the 600-N force is close to: A. B. C. D.

2,600 3,250 4,570 8,460

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Question 12. The rigid structural member is subjected to a couple consisting of 100-N forces. If this couple is replaced by an equivalent couple consisting of the two forces P and –P, each of which has a magnitude of 400N. The proper angle θ is close to: A. B. C. D.

51.3 75.5 12.4 17.6

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Question 13. The resultant force acting on the plate and the angle θ is: A. 50; 15 B. 100; 63 C. 150; 63 D. 150; 46

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Question 14. The turnbuckle is tightened until the tension in cable AB is 2.4 kN The magnitude of the moment about point O is: A. B. C. D.

2.52 5.62 8.34 9.89

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Question 15. A force of 40 lb is applied at A to the handle of the control lever which is attached to the fixed shaft OB. The magnitude of the moment about O is: A. B. C. D.

100 200 300 400

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Question 16. Determine the tension T (lb) in the cable which supports the 1,000 lb load with the pulley arrangement as shown in the figure. Each pulley is free to rotate about its bearing and the weights of all parts are small compared with the load. A. 50 B. 150 C. 250 D. 450

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Question17. A simple pendulum weighs 4 lb and has a torsional spring with a spring constant of 100lb-in/rad (see figure). When θ=0, the torsional spring is undeformed. The potential energy (in-lb) of the system when θ = 45o is close to:

A. B. C. D.

1.0 1.5 2.0 2.5

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Question 18. A rope is used to lift a 100-lb weight using the locked pulley as shown in figure. If μ = 0.2, the P necessary to begin lifting the weight is nearly: A. B. C. D.

230 300 370 460

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Question 19. The jackscrew having 5 threads/in is to be used to lift a weight. The maximum magnitude of force P which produces the uplifting moment is 500 lb. If μo = 0.3, the weight (lb) that can be lifted is: A. B. C. D.

23,000 32,500 47,300 65,200

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Question 20. A block weighing 100 lb is pulled by a rope. The coefficient of friction is 0.6. The tension (lb) needed in the rope to start the block moving is: A. B. C. D.

20 40 60 80

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Question 21. A circular cylinder has a right circular cone attached to it as shown in the figure. If the body weighs 490 lb/ft3, the moment of inertia (lb-ft.s2) for the cylinder in AA’ axis is close to: A. B. C. D.

0.50 1.15 2.50 4.45

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Question 22. For the triangle shown, the centroidal moment of inertia (in4) about x-axis is: A. B. C. D.

12 24 36 48

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Question 23. The moment of area (in4) of the rectangle about the y-axis is: A. B. C. D.

10 25 35 45

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Question 24. The moment of inertia (in4) about the y-axis for the shaded area as shown in figure for a = 2in and b = 10in is close to: A. B. C. D.

12 24 36 48

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Question 25. For the beam shown in figure, the reaction at support B (lb) is: A. 50 B. 100 C. 150 D. 350

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Question 26. The beam shown in figure has an applied load of 200 lb at its center and a triangular distributed load whose ordinate value at point A is 100 lb/ft. The reaction at support B (lb) is: A. B. C. D.

300 600 750 975

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Question 27. The area is measured to be 6.25 in2. The area revolved about x-axis, generates body which is later submerged in a circular cylinder tank of water. The tank’s radius measures 5in and the water rises 1in when the body is submerged. The 𝑦̅ (in) is: A. B. C. D.

1 2 3 4

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Question 28. The volume (in3) generated by revolving the area shown in figure about the x-axis is: A. B. C. D.

135 225 270 400

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Question 29. The centroid 𝑥̅ in the figure is: A. B. C. D.

1 2 3 4

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Question 30. The centroid 𝑥̅ 𝑎𝑛𝑑 𝑦̅ for the area for the figure is: A. B. C. D.

3.9 and 3.7 4.1 and 1.3 3.6 and 1.8 4.0 and 2.0

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Question 31. The centroid of an area bounded by the curve y = -x2 + 4, the y-axis as shown in the figure: A. B. C. D.

0.25, 1.00 0.50, 1.28 0.75, 2.34 1.25, 3.21

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Question 32. Find the centroid of the line from point A to point B is: A. B. C. D.

1 2 3 4

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Question 33. The area under the curve y = x2 between the interval x =0 to x = 6 is: A. B. C. D.

12 24 48 72

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Question 34. Two forces are applied to a cantilever beam as shown in the figure. The moment (lb-ft) is: A. B. C. D.

250 500 750 999

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Question 35. A force of 100 lb is applied to a beam as sown in figure. The beam weighs 50lb. The reaction at B (lb) is: A. 35 B. 105 C. 145 D. 203

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SOLUTIONS

QUESTION 1 The friction force on the block acts down the plane. The weight mg = 100 (9.81) = 981 N. The equations of equilibrium give: 𝑁 − 981 𝐶𝑜𝑠20 = 0 𝑁 = 922 𝑁 𝐹𝑚𝑎𝑥 = 𝜇𝑁 = 0.30 𝑥 921𝑁 = 277𝑁 QUESTION 2 A balance of forces in both x and y directions give: ∑ 𝐹𝑦 = 0; 𝑃𝑐𝑜𝑠20 + 𝐹 − 981 𝑠𝑖𝑛20 = 0 ∑ 𝐹𝑥 = 0; 𝑁 − 𝑃𝑠𝑖𝑛20 − 981 𝑐𝑜𝑠 20 = 0 Given P =500N 𝑁 = 1093 𝑁 The maximum static friction force is: 𝐹𝑚𝑎𝑥 = 𝜇𝑠 𝑁 = 0.20 𝑥 1093𝑁 = 219𝑁 QUESTION 3 The moment applied about the screw-axis is: 60 𝑥 8 = 480 𝑙𝑏 − 𝑖𝑛 in the clock wise direction. The frictional moment MB due to the frictional forces acting on the collar at B is in the counterclock wise direction to oppose the impending motion. 1000(16) − 10𝑇 = 0 𝑜𝑟 𝑇 = 1600 𝑀 = 𝑇𝑟 𝑟𝑎𝑛(𝛼 + 𝜙) 480 − 𝑀𝐵 = 1600(0.5) tan(3.64 + 11.31) 𝑀𝐵 = 266 𝑙𝑏 − 𝑖𝑛 QUESTION 4 Using the equation: 𝑀 = 𝑇𝑟 𝑟𝑎𝑛(𝛼 − 𝜙) 𝑀′ − 266 = 1600 𝑥 0.5 tan(11.31 − 3.64) 𝑀′ = 374 𝑙𝑏 − 𝑖𝑛 The force on the handle required to loosen the vise is: 𝑀′ 374 𝐹= = = 46.8 𝑙𝑏 𝑑 8 QUESTION 5 With α = 0, the angle of contact is β = π/2 rad.

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For impending upward motion of the load, 𝑇2 = 𝑃𝑚𝑎𝑥 ; 𝑇1 = 981𝑁 𝑃𝑚𝑎𝑥 = 𝑒 (0.30)𝜋/2 981 𝑃𝑚𝑎𝑥 = 1572 𝑁 For impending downward motion of the load: 𝑇2 = 981𝑁; 𝑇1 = 𝑃𝑚𝑖𝑛 981 = 𝑒 (0.30)𝜋/2 𝑃𝑚𝑖𝑛 𝑃𝑚𝑖𝑛 = 612 𝑁 QUESTION 6 With T2 = 981N, T1 = P = 500N 981 = 𝑒 0.30 𝛽 500 𝛽 = 2.25 𝑟𝑎𝑑 𝛽 = 2.25 𝑥

360 = 128.7 𝑜𝑟 2𝜋

128.7 − 90 = 38.70 QUESTION 7 The constant K =4/9 is obtained first by substituting x=4 and y=3 into the equation for the parabola. The moment of inertia of the strip about the x-axis is: y2dA 𝑦2 𝑑𝐴 = (4 − 𝑥)𝑑𝑦 = 4(1 − )𝑑𝑦 9 𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴 3

𝑦2 72 ∫ 4𝑦 (1 − ) 𝑑𝑦 = 𝑜𝑟 14.40 9 5 0 2

QUESTION 8 The moment of inertia of the semi-circular area is one-half of the complete circle. 1 𝜋𝑟 4 𝜋 204 𝐼𝑥 = = = 2𝜋 (10)4 𝑚𝑚4 2 4 8 4𝑟 4(20) 80 𝑟= = = 𝑚𝑚 3𝜋 3𝜋 3𝜋 𝐼 ̅ = 𝐼 − 𝐴𝑑 2 202 𝜋 80 2 = 2𝜋 (10)4 − ( ) = 1.755(10)4 𝑚𝑚4 2 3𝜋 Transfer from the centroid x0-axis to the x-axis 𝐼 = 𝐼 + 𝐴𝑑2 202 𝜋 80 4 4 𝐼𝑥 = 1.755(10) 𝑚𝑚 + (15 + )2 2 3𝜋 = 36.4 (10)4 𝑚𝑚4 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 9 tan 𝛼 =

𝐵𝐷 = (6 sin 60) /(3 + 6 cos 60) = 0.866 𝐴𝐷 𝛼 = 40.9

The law of cosines gives: 𝑅 2 = (600)2 + (800)2 − 2(600)(800) cos 40.9 𝑅 = 525 𝑙𝑏 QUESTION 10 𝐹 = (𝐹𝑐𝑜𝑠 𝜃)𝑖 − 9𝐹 sin 𝜃) 𝑗 (500 cos 60)𝑖 − (500 sin 60)𝑗 𝐹 = (250 𝑖 − 433𝑗)𝑁 QUESTION 11 The moment arm to the 600N force is: 𝑑 = 4 cos 40 + 2 sin 40 = 4.35𝑚 𝑀 = 𝐹𝑑 = 600 𝑥 4.35 = 2610 𝑁. 𝑚 QUESTION 12 The original couple is counter-clockwise when the plane of the forces is viewed from above. 𝑀 = 100 𝑥 0.1 = 10 𝑁. 𝑚 The forces P and –P produce a counter-clockwise couple. 10 = (0.04)(400 𝐶𝑜𝑠 𝜃) 10 𝜃 = cos −1 = 51.3 16 QUESTION 13 Point O is selected as a convenient reference point for the force-couple system. 𝑅𝑥 = 40 + 80 cos 30 − 60 cos 45 𝑅𝑥 = 66.9𝑁 𝑅𝑦 = 50 + 80 sin 30 + 60 cos 45 𝑅𝑦 = 132.4𝑁 𝑅 = √(66.9)2 + (132.4)2 = 148.3 𝑁 132.4 𝜃 = tan−1 = 63.2 66.9 QUESTION 14 Writing the force as a vector: 0.8 𝑖 + 1.5𝑗 − 2𝑘 𝑇 = 2.4 [ ] √(0.8)2 + (1.5)2 + (2)2 = 0.731 𝑖 + 1.371 𝑗 − 1.829 𝑘 The moment of the force at the point O is: 𝑀0 = 𝑟 𝑥 𝑇 = (1.6𝑖 + 2𝑘)(0.731 𝑖 + 1.371 𝑗 − 1.829 𝑘) = −2.74𝑖 + 4.39𝑗 + 2.19 𝑘 𝐾𝑁. 𝑚 The magnitude of the moment is: √(2.74)2 + (4.39)2 + (2.19)2 = 5.62 𝐾𝑁. 𝑚

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QUESTION 15 The couple may be expressed in vector notation 𝑀=𝑟𝑥𝐹 𝑟 = 8𝑗 = 5𝑘 𝑎𝑛𝑑 𝐹 = −40𝑖 𝑙𝑏 𝑀 = (8𝑖 + 5𝑘)(−40𝑖) = −200𝑗 + 320 𝑘 𝑙𝑏 − 𝑖𝑛 The magnitude of the moment is: = √(−200)2 + (320)2 = 377 𝑙𝑏 − 𝑖𝑛 QUESTION 16 ∑ 𝑀0 = 0; 𝑇1 𝑟 − 𝑇2 𝑟 = 0 𝑜𝑟 𝑇1 = 𝑇2 ∑ 𝐹𝑦 = 0; 𝑇1 + 𝑇2 − 1000 = 0 2𝑇1 = 1000 𝑜𝑟 𝑇1 = 500𝑙𝑏 𝑇2 500 𝑇3 = 𝑇4 = = = 250𝑙𝑏 2 2 For pulley C, the angle θ=30, in no way affects the moment of T about center of the pulley, so the moment equilibrium requires: 𝑇 = 𝑇3 𝑜𝑟 𝑇 = 250𝑙𝑏 QUESTION 17 The potential energy of the spring is: 𝑃𝐸 = For θ = π/4,

1 𝐾𝜃 2 2

1 𝑃𝐸 = 4(−10 cos 45) + (100)(𝜋/4)2 2 = −28.3 + 30.8 = 2.5 𝑙𝑏 − 𝑖𝑛 QUESTION 18 The tension force P must be greater than the weight since frictional forces have to overcome. The angle of contact is: 𝜋 𝜃 = 240 𝑥 = 4.19 𝑟𝑎𝑑 180 𝑃 = 𝑒 4.19 𝑥 0.2 9𝑏𝑒𝑙𝑡 − 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 100 𝑃 = 231 𝑙𝑏 QUESTION 19 Since the screw moves through 5 turns and travels 1 inch, it moves 0.2 inches per turn. Therefore 0.2 is the pitch of the thread. The pitch (l) is related to the angle by: 𝑙 0.2 tan 𝛼 = = = 0.0637 2𝜋𝑟 0.52𝜋 𝛼 = 3.6 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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𝜙 = tan−1 0.3 = 16.7 (𝑠𝑐𝑟𝑒𝑤 𝑡ℎ𝑟𝑒𝑎𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛) 0.5 (500)(1) = 𝑤 ( ) tan(16.7 + 3.6) 12 𝑤 = 32,400 𝑙𝑏𝑠 QUESTION 20 The force F must overcome the frictional force: 𝐹 = 𝜇𝑁 𝐹 = 0.6 𝑥 100 = 60𝑙𝑛 QUESTION 21 For the cylinder: 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝜋(4)2 (10) = 502 𝑖𝑛3 𝑜𝑟 0.291 𝑓𝑡 3 𝑤𝑒𝑖𝑔ℎ𝑡 = 490 𝑥 0.291 = 142.5 𝑙𝑏 𝑊 142.5𝑙𝑏 𝑀𝑎𝑠𝑠 = = = 4.43 𝑙𝑏. 𝑠 2 /𝑓𝑡 𝑔 32.2 𝑀 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 = [3𝑅 2 + 4𝐿2 ] 12 4.43 4 2 10 2 [3( ) + 4 ( ) = 1.15 𝑙𝑏. 𝑓𝑡. 𝑠 2 12 12 12 QUESTION 22 The moment of inertia for the triangle is: 𝐿𝐻 3 𝐼𝑥 = 12 H = L = 6in 𝐼𝑥 =

6 𝑥 63 = 108 𝑖𝑛4 12

QUESTION 23 The moment of inertia for the rectangle is: 𝐼𝑦 =

𝐴𝐿2 3

(5)(3)(3)2 = 45𝑖𝑛4 3 QUESTION 24 From the figure, the differential area parallel to the Y-axis is: 𝑑𝐴 = 𝑦𝑑𝑥 3 Since x = ky 𝑑𝑥 = 3𝑘𝑦 2 𝑑𝑦, 𝑑𝐴 = 3𝑘𝑦 3 𝑑𝑦 The moment of area becomes: 𝑏

𝐼𝑦 = ∫ 𝑥 2 𝑑𝑥 = ∫ (𝑘 2 𝑦 6 )( 3𝑘𝑦 3 )𝑑𝑦 𝑎

= 3𝑘 3 ∫ 𝑦 9 𝑑𝑦 = x =a and y= b; k = a/b3

3𝑘 3 𝑏10 10

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3𝑏10 𝑎3 3𝑏𝑎3 𝐼𝑦 = = 10𝑏 9 10

Given a= 2” and b = 10”

𝐼𝑦 = 24 𝑖𝑛4 QUESTION 25 The figure shows two resultant forces, F1 and F2, where F1 is due to the triangular load on the left of x = 3ft and F2 is due to the triangular load to the right of x=3ft. Their magnitudes are: 1 1 𝐹1 = 𝑥 50 𝑥 3 = 75 𝑙𝑏𝑠 𝑎𝑛𝑑 𝐹2 = 𝑥 50 𝑥 12 = 300 𝑙𝑏𝑠 2 2 The beam reaction at B is: ∑ 𝑀𝐴 = 0 15𝐵 − 2(75) − 7(300) = 0 𝑜𝑟 𝐵 = 150 𝑙𝑏𝑠 QUESTION 26 1 𝐹 = (100)(12) = 600 𝑙𝑏 2 F equals the area under the triangle whose base equals the length of the loading and the height equals the maximum magnitude of the load. ∑ 𝑀𝐴 = 0 12𝐵 − 49600) − 6(200) = 0 𝑜𝑟 𝐵 = 300𝑙𝑏𝑠 QUESTION 27 The volume of the body equals the volume of water displaced. Therefore, 𝑉 = 𝜋𝑟 2 ℎ = 𝜋(25)(1) = 25𝜋 𝑖𝑛3 Since, 𝑉 = 2𝜋𝑦̅𝐴 25𝜋 = = 2" 2𝜋(6.25) QUESTION 28 The area of the semi-circle is: 1 𝜋(4)2 𝐴 = 𝜋𝑟 2 = = 25.1 𝑖𝑛2 2 2 The generated volume about the x-axis is: 𝑉 = 2𝜋𝑦̅𝐴 4𝑟 𝑦̅ = = 1.70𝑖𝑛 3𝜋 Therefore, 𝑉 = 2𝜋(25.1)(1.70) = 268 𝑖𝑛3 QUESTION 29 Let the body be represented by 7in x 2in x 4in block and three negative volumes, each being a circular cylinder of radius 1/2in and height 2”. 𝑉 = 𝑉1 − 𝑉2 − 𝑉3 − 𝑉4 1 2 1 1 = 7(4)(2) − 𝜋( ) (2) − 𝜋( )2 (2) − 𝜋( )2 (2) = 51.3𝑖𝑛3 2 2 2 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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𝑉𝑥̅ = 𝑉1 ̅̅̅ 𝑥1 − 𝑉2 ̅̅̅ 𝑥2 − 𝑉3 ̅̅̅ 𝑥3 − 𝑉4 ̅̅̅ 𝑥4 = 56(2) − 1.57(1) − 1.57(1) − 1.57(1) = 104.1 𝑥̅ = 2.03 𝑖𝑛 QUESTION 30 It is convenient to divide the area into three parts. A semi-circular area A1, a 4in x 8in rectangle A2, and a negative area A3 representing the hole of radius of 1in. We include the hole in A2 and subtract the contribution of A3. 𝐴 = 𝐴1 + 𝐴2 − 𝐴3 𝜋 2 = (4) + 4 𝑥 8 − 𝜋(1)(1) = 54 𝑖𝑛2 2 𝐴𝑥̅ = 𝐴1 ̅̅̅ 𝑥1 + 𝐴2 ̅̅̅ 𝑥2 − 𝐴3 ̅̅̅ 𝑥3 = 25.12(4) + 32(4) − 3.14(6) = 209 𝑥̅ = 3.87 𝑖𝑛𝑐ℎ𝑒𝑠 𝐴𝑦̅ = 𝐴1 ̅̅̅ 𝑦1 + 𝐴2 ̅̅̅ 𝑦2 − 𝐴3 ̅̅̅ 𝑦3 4 4 = 25.12(4 + ( ) + 32(2) − 3.14(2) = 201 3 𝜋 𝑦̅ = 3.72 𝑖𝑛𝑐ℎ𝑒𝑠 QUESTION 31 Let the element of area be: 𝑑𝐴 = 𝑦𝑑𝑥 = (4 − 𝑥 2 )𝑑𝑥 2 𝑥3 𝐴 = ∫ (4 − 𝑥 2 )𝑑𝑥 = [4𝑥 − ] 𝑎𝑡 (0,2) = 5.33 3 0 2 1 1 𝑥̅ = ∫ 𝑥𝑑𝐴 = ∫ (4𝑥 2 − 𝑥 3 )𝑑𝑥 = 1.25 𝐴 5.33 0 2 1 1 𝑦̅ = ∫ 𝑦𝑑𝐴 = ∫ (4 − 𝑥 2 )2 𝑑𝑥 = 3.21 𝐴 5.33 0 QUESTION 32 The differential element dL is: 𝑑𝐿 = √𝑑𝑥 2 + 𝑑𝑦 2 = 𝑑𝑥√(1 +

𝑑𝑦 2 ) 𝑑𝑥

Since 𝑑𝑦 = 1 𝑑𝑥 𝑑𝐿 = √2𝑑𝑥

𝑦 = 𝑥 − 2; 6

𝐿 = √2 ∫ 𝑑𝑥 = 6√2 0

1 𝑥̅ = ∫ 𝑥 𝑑𝐿 𝐿

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1 2 (𝑥 ) 1 6 = ∫ 𝑥√2 𝑑𝑥 = 6 𝑎𝑡 (0,6) = 3 𝐿 0 2 QUESTION 33 6

𝐴 = ∫ 𝑥 2 𝑑𝑥 0

𝑥3 (0,6)𝑜𝑟 3 216 = 72 3 QUESTION 34 ∑ 𝑀𝐴 = 0 0 = 𝑀 + 0.707(100)(5) − 12(50) M = 246 lb-ft QUESTION 35 ∑ 𝑀𝐴 = 0 −5(50) − 8(100) + 10𝐵 = 0 B = 105 lb

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Dynamics, Kinematics and Vibrations Total Questions 9-14 A. B. C. D. E. F. G. H. I. J. K.

Kinematics of particles Kinetic friction Newton’s second law for particles Work-energy of particles Impulse-momentum of particles Kinematics of rigid bodies Kinematics of mechanisms Newton’s second law for rigid bodies Work-energy of rigid bodies Impulse-momentum of rigid bodies Free and forced vibrations

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Question 1. The velocity (m/s) of a falling particle is described by the equation v = 32 + t + 6t2 The acceleration (m/s2) at t =2 is: A. B. C. D.

9.8 25 32 58

Question 2. The location of a particle moving in x-y plane is given by the parametric equations x = t2 +4t y = ¼ t4 -60t where, x and y are in meters and t in seconds. The particle velocity (m/s) at t = 4 s is: A. B. C. D.

8.95 11.3 12.6 16.0

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Question 3. The position of a particle is defined by the equation: s = 2 sin ti + 4 cos tj (t in radians) The magnitude of the particle velocity at t = 4 rads is close to: A. B. C. D.

2.61 2.75 3.30 4.12

Question 4. The position of a particle is defined by the equation: s = 2 sin ti + 4 cos tj (t in radians) The magnitude of the particle’s acceleration at t = π is close to: A. B. C. D.

2.00 2.56 3.14 4.00

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Question 5. A particle motion is represented by the following equation: S = 20t + 4t2 - 3t3 The maximum speed (m/s) is reached by the particle is close to: A. B. C. D.

21.8 27.9 34.6 48.0

Question 6. A particle has a tangential acceleration at, when it moves around a point in a curve with instantaneous radius of 1m. at = 2t – sin t +3 cot t (in m/s2) The instantaneous angular velocity (rad/s) of the particle is: A. B. C. D.

t2 + cos t + 3 ln │csc t│ t2 - cos t + 3 ln │csc t│ t2 - cos t +3 ln │sin t│ t2 +cos t +3 ln │sin t│

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Question 7. For a reciprocating pump, the radius of curvature r =0.3m and the rotational speed is 350 rpm. The tangential velocity (m/s) on the crank corresponding to an angle of 35o from the horizontal is: A. B. C. D.

0 1.15 11.0 15.0

Question 8. A motorist travelling at 70 km/h sees a traffic light in an intersection 250m ahead turn red. The light’s red cycle is 15s. The motorist wants to enter the intersection without stopping his vehicle when the light is green. The uniform deceleration (m/s2) of the vehicle will just put the motorist in the intersection when the light turns green is: A. B. C. D.

0.18 0.25 0.37 0.95

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Question 9. The position (radians) of a car travelling around a curve is described by the following equation: Θ = t3 – 2t2 – 4t + 10 The angular acceleration (rad/s2) at t = 5s is: A. 4 B. 6 C. 16 D. 26

Question 10. A 5 kg block begins from rest and slides down an inclined surface. After 4s, the block has a velocity of 6 m/s. If the angle of inclination is 45o, the distance travelled by the block in 4s is: A. 3 B. 6 C. 9 D. 12

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Question 11. A 100 kg block is pulled along a smooth, flat surface by an external 500 N force at an angle of 30o. If the coefficient of friction between the block and the surface is 0.15, the acceleration (m/s2) experienced by the block due to external force is: A. B. C. D.

3.23 3.80 4.33 5.00

Question 12. A 5 kg block begins from rest and slides down an inclined surface. After 4s, the block has a velocity of 6 m/s. If the angle of inclination is 45o, the coefficient of friction between the plane and the block is close to: A. B. C. D.

0.15 0.22 0.78 0.89

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Question 13. A block-spring system oscillates once every 3s with negligible friction. The spring constant is 6 N/m, the approximate mass (kg) of the block is: A. B. C. D.

1.37 5.47 13.7 23.7

Question 14. A mass of 10kg is suspended from a vertical spring with a spring constant of 10N/m. The period (s) of vibration is: A. B. C. D.

0.30 0.60 0.90 6.35

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Question 15. A variable force (40N) cos θ is attached to the end of a spring whose spring constant is 50 N/m. At what angle, will the spring deflect 20 cm from the equilibrium position? A. B. C. D.

-14 25 64 76

Question 16. A spring has a constant of 50N/m. The spring is hung vertically, and a mass is attached at the end. The spring end displaces 30 cm from its equilibrium position. The same mass is removed from the first spring and attached to a second spring, and the displacement was 25 cm. The spring constant (N/m) for the second spring is: A. B. C. D.

46 56 60 90

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Question 17. A 6kg sphere moving at 3m/s collides with a 10kg sphere travelling 2.5m/s in the same direction. The 6kg ball comes to a complete stop after the collision. The new velocity (m/s) of the 10kg ball after the collision is: A. B. C. D.

0.5 2.8 4.3 5.8

Question 18. A 2kg ball of clay moving at 40m/s collides with a 5kg ball of clay moving at 10m/s directly towards the first ball. The final velocity (m/s) if both balls stick together after collision is: A. B. C. D.

4.29 42.9 53.7 65.8

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Question 19. A 60,000kg railcar moving at 1km/h is instantly coupled to a stationary 40,000kg railcar. The speed (km/h) of the coupled cars is: A. B. C. D.

0.40 0.60 0.88 0.99

Question 20. A 50kg cylinder has a height of 3m and a radius of 50cm. The cylinder sits on the x-axis and is oriented with its major axis parallel to the y-axis. The mass moment of inertia (kg.m2) about the x-axis is: A. 4.1 B. 16 C. 41 D. 150

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Question 21. Traffic travels at 100 km/h around a banked highway curve with a radius of 1000m. The banking angle necessary such that friction will not be required to resist the centrifugal force is: A. B. C. D.

1.4 2.8 4.5 46

Question 22. A 1530kg car is towing 300 kg trailer. The coefficient of friction between all tires and the road is 0.80. The car and the trailer are travelling at 100 km/h around a banked curve of radius 200m. The necessary banking angle such that the friction will not be necessary to prevent skidding is close to: A. 8 B. 21 C. 36 D. 78

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Question 23. A 1530kg car is towing 300 kg trailer. The coefficient of friction between all tires and the road is 0.80. How fast (km/h) can the car and travel around an unbanked curve of radius of 200m without either the car or trailer skidding? A. 40 B. 70 C. 108 D. 145

Question 24. An automobile travels on a perfectly horizontal unbanked circular track of radius r. The coefficient of friction between the tires and the track is 0.3. If the car’s velocity is 10m/s, the smallest radius (m) possible without skidding is: A. B. C. D.

10 34 50 68

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Question 25. A car with a mass of 1,530 kg tows a trailer of 200 kg at 100 km/h. The total momentum (N.s) of the car-trailer combination is: A. B. C. D.

16,000 22,000 37,000 48,000

Question 26. A 60,000 kg railcar moving at 1km/h is coupled to a secondary stationary railcar. If the velocity of the two cars after coupling is 0.2 m/s and the coupling is completed in 0.5s, the average impulsive force (N) on the 60,000 kg railcar is: A. 520 B. 990 C. 3,100 D. 9,300

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Question 27. A 1,500 kg car travelling at 100 km/h is towing a 250kg trailer. The coefficient of friction between the tires and the road is 0.8 for both the car and the trailer. The energy (kJ) dissipated by the brakes if the car and trailer are braked to a complete stop is close to: A. B. C. D.

125 375 579 675

Question 28. A 3,500kg car travelling at 65km/h skids and hits a wall 3s later. The coefficient of friction between the tires and the road is 0.60. The speed (m/s) of the car when it hits the wall is: A. B. C. D.

0.14 0.40 5.1 7.5

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Question 29. Keeping the bank angle same to increase the maximum speed with which a vehicle can travel on a curved road by 10 percent the radius of curvature of road has to be changed from 20m to: A. B. C. D.

16 18 24 31

Question 30. The physical quantities having the same dimensional formula is: A. B. C. D.

Angular momentum and torque Torque and entropy Entropy and power Power and angular momentum

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Question 31. A ball of mass M moving with a velocity V collides elastically with another ball of the same mass M, but moving in the opposite direction with the same velocity V. After the collision: A. B. C. D.

The velocities are exchanged between the balls Both balls come to rest Both the balls move at right angles to the original line of motion One ball comes to rest and the other ball travels back with a velocity 2V.

Question 32. A vehicle is moving with a speed of 35 m/s on a curvy road having a radius of 490 m. the banking angle is: A. ¼ B. ½ 4 C. 5 D.

7 8

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Question 33. A 2 kg body and a 3 kg body have equal momentum. If the kinetic energy of a 3kg body is 10 J, the kinetic energy in 2 kg body will be: A. B. C. D.

7 15 23 45

Question 34. All of the following have the same dimensional formulas, EXCEPT: A. B. C. D.

Acceleration, gravitational field strength Torque, angular momentum Pressure, modulus of elasticity Velocity, speed

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Question 35. A spring with a force constant K is cut into two equal parts. The force constant in each part is: A. K B. 2K C. 3K D. 4K

Question 36. A 2 kg mass moving on a smooth frictionless surface with a velocity of 10m/s collides with another 2 kg mass at rest in an elastic collision. After collision, if they move together: A. B. C. D.

Both travel with a velocity of 5m/s in the same direction Both travel with a velocity of 10m/s in the same direction Both travel with 5m/s in the opposite direction Both travel with 10m/s in the opposite direction

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Question 37. A car with a mass of 300 kg is moving with a velocity of 25m/s was brought to rest in 15 m distance. The force (N) applied on the brakes is: A. B. C. D.

2,500 4,500 6,250 7,500

Question 38. The coefficient of friction between a body and a surface of an inclined plane at 45o is 0.50. The acceleration of the body (m/s2) is: 4.9

A. √2 B. 4.9 C. 49√2 19.6 D. √2

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Question 39. Which of the following parameters has no dimensions? A. B. C. D.

Angular velocity Momentum Angular momentum Strain

Question 40. The displacement of a particle moving in a straight line is given by x = 2t2 + t + 5, where x in meters and t in s. The acceleration (m/s2) at t =2 s is: A. 4 B. 8 C. 10 D. 15

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Question 41. The moment of inertia of a solid sphere of mass M and radius R about a tangent is: A. B. C. D.

2 5 7 5 2 3 5 3

𝑀𝑅 2 𝑀𝑅 2 𝑀𝑅 2 𝑀𝑅 2

Question 42. The modulus of elasticity is dimensionally equivalent to: A. B. C. D.

Stress Surface tension Strain Coefficient of viscosity

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Question 43. A solid wooden block resting on a frictionless surface is hit by a bullet. The bullet gets embedded. During the process: A. B. C. D.

Only kinetic energy is conserved Only momentum is conserved Both momentum and kinetic energy is conserved Neither momentum nor kinetic energy are conserved

Question 44. The angular velocity of a radial tire is 70 rad/s. If the radius of the tire is 0.5m, the linear velocity (m/s) is: A. 1 B. 10 C. 30 D. 70

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Question 45. A weight of 50 N is placed on a smooth surface. If the force required to move the body on the smooth surface is 30N, the coefficient of friction is: A. B. C. D.

0.35 0.61 1.20 1.67

Question 46. A motor cycle is travelling on a curved track of radius 500 m. If the coefficient of friction between the tires and the road is 0.5.The maximum speed (m/s) to avoid skidding is: A. 10 B. 50 C. 250 D. 500

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Question 47. A wheel of mass 10 kg has a moment of inertia of 160kg.m2 on its own axis. The radius of gyration (m) is close to: A. 4 B. 5 C. 6 D. 10

Question 48. A force F equal to 3i + cj + 2k acting on a particle causes a displacement of -4i + 2j + 3k in it’s own direction. If the work done is 6 J, the value of c is: A. B. C. D.

0 1 6 12

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Question 49. A constant torque of 1,000 N-m turns a wheel with a moment of inertia 200 kg-m2 about an axis through its center. The angular velocity (rad/s) after 3 s is: A. 1 B. 5 C. 10 D. 15

Question 50. A thin metal disc with a mass of 2 kg and a radius 0.25m rolls down an inclined surface. If the kinetic energy is 4J at the foot of the inclined plane, the linear velocity (m/s) is: A. 1 B. 3 C. 7 D. 20

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Question 51. The dimensions of impulse are: A. B. C. D.

MLT-2 M2LT-1 MLT-1 ML2T-1

Question 52. A body is projected along a rough horizontal surface with a velocity of 6 m/s. if the body comes to rest after 9m, the coefficient of friction is: A. B. C. D.

0.2 0.4 0.5 0.6

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Question 53. An elevator is moving with a retardation of 5 m/s2. The percentage change in the weight of a person in the elevator is: A. 25 B. 50 C. 75 D. 100

Question 54. Which of the following are conserved in an elastic collision are: A. B. C. D.

Momentum, kinetic energy and temperature Momentum and kinetic energy Momentum Kinetic energy

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Question 55. The relation F = ma cannot be derived from Newton’s second law, if: A. B. C. D.

Force depends on time Momentum depends on time Acceleration depends on time Mass depends on time

Question 56. A force 100N acts on a 2 kg body for 10s. The change in momentum (N.s) is: A. 100 B. 250 C. 500 D. 1000

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Question 57. The kinetic energy of a body is four times its momentum. The velocity (m/s) is: A. 2 B. 4 C. 8 D. 16

Question 58. A van is moving with a speed of 72 km/h on a straight road having a friction of 0.5. The minimum radius of curvature (m) for safe driving of the van is: A. 4 B. 20 C. 40 D. 80

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Question 59. The unit of moment of inertia is: A. B. C. D.

Kg/m2 Kg/m2 N/m2 N.m2

Question 60. A ball moving with a speed of 90m/s collides directly with another identical ball moving with a speed of V m/s. The second ball comes to rest after collision. If the coefficient of restitution is 0.8, the speed (m/s) of second ball is: A. B. C. D.

10 23 80 90

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Question 61. The physical quantity which has the same dimensional formula of [Energy/(mass x length)] is: A. B. C. D.

Force Power Pressure Acceleration

Question 62. A body is moving with a uniform acceleration is 0.34 m/s covers a distance of 3.06m. If the change in the velocity of the body is 0.18 m/s during this distance, the uniform acceleration (m/s2) is: A. B. C. D.

0.01 0.02 0.03 0.04

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Question 63. A constant force acts on 0.9 kg mass at rest for 10 s. If the body moves a distance of 250m, the magnitude of the force (N) is: A. 1.0 B. 4.5 C. 8.5 D. 36.0

Question 64. A mass of 5 kg rests of a rough horizontal surface (µ = 0.2). The mass is pulled through a distance of 10m by a horizontal force of 25N. The kinetic energy (J) required is: A. 50 B. 100 C. 150 D. 200

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Question 65. The average force that must act on a 5 kg mass to reduce the speed from 65 cm/s to 15 cm/s is: A. 12 B. 25 C. 50 D. 100

Question 66. A force of 5N making an angle θ with the horizontal displaces a body by 0.4m along the horizontal direction. If the body gains 1 J of kinetic energy, the horizontal component of the force (N) is: A. B. C. D.

1.5 2.5 3.5 4.5

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Question 67. A body of mass m1 moving with a velocity of 10 m/s collides with another body at rest with a mass m2. After collision, the velocities of the two bodies are 2 m/s and 5 m/s, respectively, along the direction of motion. The ratio of m1/m2 is: A. B. C. D.

5 12 5 8 8 5 12 5

Question 68. The moment of inertia of a sphere of mass M and radius R about an axis passing through its center is 2/5 MR2. The radius of gyration of the sphere about a parallel axis to the above and tangent to the sphere is: A. B.

7 5𝑅 3 5𝑅 7

C. √5 𝑅 D.

8 5

𝑅

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Question 69. A wheel spins at 1,200 rpm and made to slow down to 4 rad/s. The number of revolutions the wheel makes before coming to rest is: A. B. C. D.

143 272 314 722

Question 70. The dimensional formula of a product of two physical quantities P and Q is ML2T-2. The dimensional formula of P/Q is MT-2. P and Q are: A. B. C. D.

Force, velocity Momentum displacement Force, displacement Work, velocity

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Question 71. The initial velocity of a particle is 4i + 3j, its uniform acceleration is 0.4i + 0.3j. The velocity after 10s is: A. 3 B. 4 C. 5 D. 10

Question 72. A body is thrown vertically upwards with a velocity of 50m/s. The percentage of the kinetic energy converted to potential energy after 4s is: A. 4 B. 24 C. 50 D. 96

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Question 73. A block of weight 200N is pulled along a rough horizontal surface at constant speed by a force 100N acting at 30o above the horizontal. The coefficient of kinetic friction between the block and the surface is: A. B. C. D.

0.43 0.58 0.75 0.83

Question 74. A mass of 6kg under a force causes a displacement by S = t2/4. The work (J) done by the force in 2s is: A. 3 B. 6 C. 9 D. 12

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Question 75. The velocities of three particles of masses 20, 30 and 50 kg are 10i, 10j and 10k, respectively. The velocity of the center of the mass of the three particles is: A. B. C. D.

2i + 3j + 5k 10(i + j + k) 20i + 30j + 5k 2i + 30j + 50k

Question 76. A weight of 64 N is pushed with a minimum force to start to move on a horizontal force. If the coefficients of static and dynamic frictions are 0.6 and 0.4, respectively, the acceleration (m/s 2) of the body is: (g = acceleration due to gravity) A. 0.2g 𝑔 B. 64 C. 0.64g 𝑔 D. 3 2

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Question 77. A mass of 50 kg is pulled on an inclined surface 12m long and 2m high by a constant force of 100N at 2m/s. The work (J) done by friction when it reaches to the top is: A. 50 B. 100 C. 150 D. 200

Question 78. A body moves along a 5m circular path. The coefficient of friction is 0.5. The angular velocity (rad/s) required so that the body does not skid is: A. B. C. D.

1 2 3 4

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Question 79. Two springs with spring constants of 1,000 N/m and 2,000 N/m are stretched by the same force. The ratio of their potential energies is: A. B. C. D.

2:1 1:2 4:1 1:4

Question 80. A mass of 4kg is moving with a momentum of 8 kg/m.s. A force of 0.2N acts in the direction of motion for 10 s. The increase in the kinetic energy (J) is: A. B. C. D.

2.5 4.5 7.5 9.5

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Question 81. A mass of 2kg starts moves in uniform acceleration. The velocity of the mass is 20 m/s in 4s. The power (W) exerted on the body in 2s is: A. 50 B. 100 C. 150 D. 200

Question 82. The diameter of a flywheel is 1m with a mass of 20kg. If the angular speed is 120 rpm, the angular momentum (kg.m2/s) is: A. B. C. D.

13.4 31.4 41.4 43.4

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Question 83. The equation of motion of a projectile is given by x = 36t and 2y = 96t – 9.8t2. The angle of projection is: 4

A. 𝑠𝑖𝑛−1 (5) 3

B. 𝑠𝑖𝑛−1 (5) 4

C. 𝑠𝑖𝑛−1 (3) 3

D. 𝑠𝑖𝑛−1 (4)

Question 84. Two bodies with masses of 200g and 500g have velocities of 10i m/s and 3i + 5j m/s. The velocity (m/s) of the center of the mass is: A. 5i – 25j 5 B. 7 𝑖 – 25𝑗 C. 5i +

25

7 5

j

D. 25i – 7 𝑗

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Question 85. A 2kg ball moving at 24 m/s collides inelastically head on with 4kg ball moving at 48 m/s. If the coefficient of restitution is 2/3, their velocities after impact are: A. B. C. D.

-56, -8 -28, -4 -14, -2 -7, -1

Question 86. A mass of 2 kg is initially at rest on a horizontal frictionless surface. A force F = (9 – x2)i N is applied, when the block is at rest. The kinetic energy of the block when x = 3 is: A. B. C. D.

15 18 20 24

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Question 87. A block of wood resting on an inclined plant at an angle of 30o starts moving down. If the coefficient of friction is 0.2, the velocity (m/s) after 5 s is: A. B. C. D.

13 16 18 20

Question 88. A 2kg block slides on a horizontal force with a speed of 4m/s and compresses a spring till the block is motionless. The kinetic friction force is 15N and the spring constant is 10,000 N/m. The spring compresses to: A. 2.5 B. 5.5 C. 8.5 D. 10.5

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Question 89. Angular momentum of the particle rotating with a central force is constant due to: A. B. C. D.

Constant torque Constant force Constant linear momentum Zero torque

Question 90. A particle is projected at 60o to the horizontal with a kinetic energy of K. The kinetic energy at the highest point is: A. 0 𝐾 B. 2 𝐾

C. 4 D. K

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Question 91. A sphere of mass m moving with a constant velocity “u” collides with another stationary sphere of same mass. If e is the coefficient of restitution, the ratio of final velocities of the spheres is: A. B.

1−𝑒 1+𝑒 1−𝑒 𝑒 𝑒

C. 1+𝑒 D. None of the above

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SOLUTIONS QUESTION 1 𝑎=

𝑑𝑣 = 1 = 12𝑡 𝑑𝑡

At t =2s 𝑎 = 1 + 12𝑥2 = 25

𝑚 𝑠2

QUESTION 2 𝑑𝑥 𝑑 = (𝑡 2 + 4𝑡) = 2𝑡 + 4 𝑑𝑡 𝑑𝑡 𝑑 𝑡4 𝑣𝑦 = ( − 60𝑡) = 𝑡 3 − 60 𝑑𝑡 4

𝑣𝑥 =

At t =4s, 𝑣𝑥 = 12

𝑚 𝑚 ; 𝑣𝑦 = 4 𝑠 𝑠

𝑉 = √𝑣𝑥 2 + 𝑣𝑦 2 = √122 + 42 = 12.6

𝑚 𝑠

QUESTION 3 𝑆(𝑡) = 2 sin 𝑡 𝑖 + 4 cos 𝑡 𝑗 𝑑𝑠 𝑣(𝑡) = = 2 𝑐𝑜𝑠𝑡 𝑖 = 4 sin 𝑡 𝑗 𝑑𝑡 At t =4 rad 𝑣(4𝑟𝑎𝑑) = 2 cos(4𝑟𝑎𝑑)𝑖 − 4 sin (4𝑟𝑎𝑑)𝑗 = −1.31 𝑖 + 3.03 𝑗 𝑚 |𝑣| = √(−1.31)2 + (3.03)2 = 3.30 𝑠 QUESTION 4 𝑣9𝑡) = 2 𝑐𝑜𝑠𝑡 𝑖 − 4 sin 𝑡𝑗 𝑑𝑣 𝑎= = −2 sin 𝑡 𝑖 − 4 𝑐𝑜𝑠𝑡 𝑗 𝑑𝑡

t=π

𝑎(𝜋) = −2 sin 𝜋 𝑖 − 4 cos 𝜋 𝑗 0 + 4𝑗 𝑚 |𝑎| = √02 + 42 = 4 2 𝑠 QUESTION 5

𝑣𝑚𝑎𝑥

𝑆 = 20𝑡 + 4𝑡 2 − 3𝑡 3 𝑑𝑠 𝑣= = 20 + 8𝑡 − 9𝑡 2 𝑑𝑡 𝑑𝑣 = 8 − 18𝑡 = 0 𝑜𝑟 𝑑𝑡 8 𝑡= = 0.444𝑠 18 𝑚 = 20 + 8(0.444) − 9(0.444)2 = 21.8 𝑠

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QUESTION 6 𝑎=

𝑑𝑣 𝑜𝑟 𝑣 = ∫ 𝑎 𝑑𝑡 𝑑𝑡

= ∫(2𝑡 − sin 𝑡 + 3 cos 𝑡) 𝑑𝑡 = 𝑡 2 + cos 𝑡 + 3 ln|sin 𝑡| 𝑣𝑡 𝑡 + cos 𝑡 + 3 ln|sin 𝑡| 𝜔= = = 𝑡 2 + cos 𝑡 + 3 ln|sin 𝑡| 𝑟 1 2

QUESTION 7 𝑣𝑡 = 𝑟𝜔 𝑟𝑒𝑣 𝑟𝑎𝑑 1𝑚𝑖𝑛 𝑟𝑎𝑑 𝜔 = 350 𝑥 2𝜋 𝑥 = 36.65 1𝑚𝑖𝑛 𝑟𝑒𝑣 60𝑠 𝑠 𝑣𝑡 = 0.3 𝑥 36.65

𝑟𝑎𝑑 𝑚 = 11 𝑠 𝑠

QUESTION 8 The initial speed is: 𝑘𝑚 𝑚 1ℎ 𝑥 1000 𝑥 = 19.44 𝑚/𝑠 ℎ 𝑘𝑚 3600𝑠 The distance travelled under a constant deceleration is: 1 𝑥 = 𝑥0 + 𝑣0 𝑡 − 𝑎𝑡 2 2 Initial distance x0 = 0 𝑣0 𝑡 − 𝑥 19.44 𝑥 15 − 250 𝑚 𝑎=2 = 2[ ] = 0.37 2 2 𝑡 15 𝑥 15 𝑠 QUESTION 9 Angular acceleration: 𝑑2𝜃 𝜔 = 2 = 6𝑡 − 4 𝑑𝑡 At t = 5s 𝑟𝑎𝑑 𝜔 = 6𝑥5 − 4 = 26 2 𝑠 QUESTION 10 𝑣(𝑡) = 𝑣0 + 𝑎0 𝑡 6𝑚 0𝑚 − 𝑠 𝑣(𝑡) − 𝑣0 𝑚 𝑎0 = = 𝑠 = 1.5 2 𝑡 4𝑠 𝑠 1 2 𝑆(𝑡) = 𝑆0 + 𝑣0 (𝑡) + 𝑎0 𝑡 2 1 = 0 + 0 + 𝑥 1.5 𝑥 42 = 12𝑚 2 𝑣0 = 70

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QUESTION 11 The frictional force 𝐹 = 𝜇𝑁 = 𝜇[𝑚𝑔 − 𝐹𝑦 ] = 0.15[100𝑥9.81 − 500 𝑥 sin 30] = 110𝑁 𝑚𝑎𝑥 = 𝐹𝑥 − 𝐹𝑦 [500 sin 30 − 110] 𝑚 𝑎= = 3.23 2 100 𝑠 QUESTION 12 ∑ 𝐹𝑥 = 𝑚𝑎𝑥 = 𝑚𝑔 − 𝐹𝑓 𝑚𝑎𝑥 = 𝑚𝑔 sin 45 − 𝜇𝑚𝑔 cos 45 𝜇=

9.81 sin 45 − 1.5 = 0.78 9.81 cos 45

QUESTION 13 The angular frequency is: 𝑘 2𝜋 𝜔=√ = 𝑚 𝑇 𝑘 4𝜋 2 = 2 𝑚 𝑇 𝑁 2 6 (3) 𝑚 = 𝑚 2 = 1.37 𝑘𝑔 4𝜋 QUESTION 14 𝑚 10 = 2𝜋√ = 6.3 𝑠 𝑘 10

𝑇 = 2𝜋√ QUESTION 15 From Hook’s law:

𝐹 − 𝑘𝑥 𝑁 20𝑐𝑚 40𝑁 cos 𝜃 = 50 𝑥 = 0.25 𝑚 100𝑐𝑚 𝜃 = 76 QUESTION 16 From Hook’s law: 𝐹 = 𝑘1 𝑥1 = 𝑘2 𝑥2 𝑘1 𝑥1 30 𝑁 𝑘2 = = 50 𝑥 = 60 𝑥2 25 𝑚 QUESTION 17 Using conservation of momentum equation: 𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣′1 + 𝑚2 𝑣′2 𝑚 𝑚 6𝑘𝑔 𝑥 3 + 10 𝑘𝑔 𝑥 2.5 = 6𝑘𝑔 𝑥 0 + 10 𝑘𝑔 𝑥 𝑣′2 𝑠 𝑠 𝑚 𝑣′2 = 4.3 𝑠 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 18 𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣′ 𝑚 𝑚 2 𝑥 40 + 5 𝑥 (−10 ) = (2 + 5)𝑣′ 𝑠 𝑠 𝑚 ′ 𝑣 = 4.29 𝑠 QUESTION 19 Using the conservation of momentum equation: 𝑚1 𝑣1 + 𝑚2 𝑣2 = (𝑚1 + 𝑚2 )𝑣′ 𝑘𝑚 𝑘𝑚 60000 𝑘𝑔 𝑥 1 + 40,000 𝑘𝑔 𝑥 0 = (60000 + 40000)𝑣′ ℎ ℎ 𝑘𝑚 𝑣 ′ = 0.6 ℎ QUESTION 20 The mass moment of inertia in x-direction is: 𝑚 𝐼𝑥 = [3𝑅 2 + 4𝐻 2 ] 12 30 [3 𝑥 (0.5)2 + 4 𝑥 (3)2 12 = 150 𝑘𝑔. 𝑚2 QUESTION 21 The banking angle is given by: 𝑣𝑡2 tan 𝜃 = 𝑔𝑟 𝑣𝑡2 𝜃 = tan−1 𝑔𝑟 𝑘𝑚 𝑚 1ℎ𝑟 [100 𝑥 1000 𝑥 3600𝑠 ] ℎ 𝑘𝑚 tan−1 = 4.5 9.81 𝑥 1000𝑚 QUESTION 22 The velocity is: 𝑘𝑚 𝑚 1ℎ 𝑚 𝑣 = 100 𝑥 1000 𝑥 = 27.78 ℎ 𝑘𝑚 3600𝑠 𝑠 2 𝑣 𝑡 𝜃 = tan−1 𝑔𝑟 𝑚 [27.78 𝑠 ] tan−1 = 21 9.81 𝑥 200𝑚

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QUESTION 23 𝑚𝑎𝑛 = 𝜇𝑁 𝜇𝑁 𝜇𝑚𝑔 𝑎𝑛 = = = 𝜇𝑔 𝑚 𝑚 𝑚 = 0.8 𝑥 9.81 = 7.84 2 𝑚

𝑣𝑡2 𝑚 𝑎0 = 𝑜𝑟 𝑣𝑡 = √𝑎0 𝑟 = √7.84 𝑥 200 = 39.6 𝑟 𝑠 3600𝑠 1𝑘𝑚 𝑘𝑚 = 39.6 𝑥 𝑥 = 143 ℎ 1000𝑚 ℎ QUESTION 24 The centrifugal force: 𝐹𝑐 =

𝑚𝑣 2 𝑟

The frictional force 𝐹𝑓 = 𝜇𝑁 = 𝜇𝑚𝑔 𝑚𝑣 2 = 𝜇𝑚𝑔 𝑟 𝑣2 10 𝑥10 𝑟= = = 34𝑚 𝑟 0.3 𝑥 9.81 QUESTION 25 Velocity: 𝑘𝑚 𝑚 1ℎ 𝑚 𝑥 100 𝑥 = 27.78 ℎ 𝑘𝑚 3600𝑠 𝑠 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚 𝑥 𝑣 𝑚 = (1530 + 200)𝑘𝑔 𝑥 27.78 = 48,000 𝑁. 𝑠 𝑠 𝑣 = 100

QUESTION 26 The original velocity = 0.277 m/s Using the impulse-momentum principle: 𝐹Δ𝑇 = 𝑚Δ𝑣 𝐹 = 60,000 𝑥

[0.277 − 0.2] = 9300𝑁 0.5𝑠

QUESTION 27 The original velocity of car and trailer: 𝑘𝑚 𝑚 1ℎ 𝑚 𝑣 = 100 𝑥 1000 𝑥 = 27.78 ℎ 𝑘𝑚 3600𝑠 𝑠 1 Δ𝐾𝐸 = 𝑚𝑣 2 2 1 𝑚 = (1500 + 250)𝑥 (27.78 )2 = 675 𝑘𝐽 2 𝑠

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QUESTION 28 The initial velocity: 𝑣 = 65,000

𝑚 1ℎ 𝑚 𝑥 = 18.06 ℎ 3600𝑠 𝑠

The frictional force decelerating the car is: 𝐹𝑓 = 𝜇𝑁 = 𝜇𝑚𝑔 = 0.6 𝑥 3500 𝑥 9.81 = 20,600 𝑁 Using the impulse-momentum principle: 𝐹𝑓 Δ𝑇 = 𝑚Δ𝑣 20,600 𝑥 3𝑠 = 3500[18.06 − 𝑣2 ] 𝑚 𝑣2 = 0.40 𝑠

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377

Mechanics of Materials Total Questions 8–12 A. B. C. D. E. F. G. H. I. J. K.

Shear and moment diagrams Stress types (axial, bending, torsion, shear) Stress transformations Mohr’s circle Stress and strain caused by axial loads Stress and strain caused by bending loads Stress and strain caused by torsion Stress and strain caused by shear Combined loading Deformations Columns

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378

Question 1. A cylindrical vessel has an inner diameter of 4ft and a thickness of ½ inch. The maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal stress component exceeds 20,000 psi is close to: A. B. C. D.

100 200 300 400

Question 2. A force of 150 lb is applied to the edge of the member as shown in figure. The bending moment is nearly: A. 7.25 B. 11.25 C. 15.75 D. 22.50

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379

Question 3. A steel tank (specific weight = 490 lb/ft3) has an inner diameter of 24 in. and a thickness of 0.5 in. It is filled to the top with water having a specific weight of 62.4 lb/ft3. If the pressure on the tank is 1.30 psi, the circumferential stress (psi) is: A. B. C. D.

32.2 48.6 62.4 98.6

Question 4. A thin-walled box (as shown in figure) is subjected to a shear of 10,000 psi. The transverse shear flow is close to: Assume the first moment Q = 17.5 in3. A. B. C. D.

1,500 2,500 4,800 5,200

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380

Question 5. An aluminum (E = 70 GPa) rod has a circular cross section and is subjected to an axial load of 10kN. The approximate elongation (mm) of the rod when the load is applied is close to: A. B. C. D.

12 18 24 36

Question 6. A steel bar with dimensions of 1.5 m (L) x 100 mm (W) x 50 mm (thick) is subjected to a load of 80 kN. The axial elongation (μm) of the bar after applying the load is close to: A. 30 B. 60 C. 90 D. 120

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381

Question 7. An aluminum specimen has a diameter of 25mm and length of 250 mm. If a force of 165 kN elongates the length by 1.20mm, the modulus of elasticity (GPa) is nearly: A. 35 B. 70 C. 105 D. 150

Question 8. An aluminum specimen has a diameter of 25mm and length of 250 mm. A force of 165 kN elongates the length by 1.20mm. Assume the modulus of elasticity for the material is 70 GPa and shear modulus is 26 GPa. The contraction (mm) of the diameter is close to: Assume the elongation = 0.00480 mm/mm A. B. C. D.

0.042 0.064 0.088 0.098

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382

Question 9. The assembly shown in the figure consists of an aluminum tube AB having a cross-sectional area of 400 mm2. A steel rod having a diameter of 10mm is attached to a rigid collar and passes through the tube. If a load of 80 kN is applied to the rod, the deformation (m) in steel rod is close to: A. B. C. D.

0.0011 0.0022 0.0033 0.0044

Question 10. A steel bar (0.5 in) is constrained to just fit between two fixed supports when the temperature is 60 F. If the temperature is raised to 120 F, the average stress (psi) is close to: Assume temperature coefficient of expansion: 6.60 x 10-6/F and Esteel = 29 x 106 psi A. 7,500 B. 9,500 C. 11,500 D. 15,500

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383

Question 11. A solid shaft, 2-in in radius is subjected to 8,000 psi torsion stress; the resultant internal torque (kilo pounds/in2) is close to: A. B. C. D.

100 200 300 400

Question 12. A pipe has inner diameter of 80mm and an outer diameter of 100 mm. If a torque of 40 N.m is applied, the outside torsion stress (MPa) is: A. B. C. D.

0.245 0.345 0.755 0.975

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384

Question 13. A tubular shaft having an inner diameter of 30 mm and an outer diameter of 42 mm is to be used to transmit 90 kW of power. The frequency (Hz) of rotation of the shaft so that the shear stress will not exceed 50 MPa is close to: A. B. C. D.

15 27 45 76

Question 14. The gears attached to the fixed-end steel shaft are subjected to the torques as shown in figure. If the shear modulus is 80 GPa and the shaft has a diameter of 14mm, the total twist (rad) is close to: A. B. C. D.

-0.212 0.212 -0.422 0.422

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385

Question 15. The two solid steel shafts are coupled together using the meshed gears as shown in the figure. The angle of twist of end A of shaft when a torque of 45N.m is applied is nearly: Diameter of shaft: 20mm Shear modulus: 80 GPa A. B. C. D.

-0.072 +0.072 0.88 -0.88

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386

Question 16. A bronze tube with a rectangular cross-section of 60mm (3mm thick) x 40mm (5mm thick) is subjected to a torque of 35 N.m. The shear stress (MPa) along the height of the tube is close to: A. B. C. D.

0.50 1.00 1.25 1.75

Question 17. A square (3 in x 3 in) aluminum tube is subjected to a torque of 85 lb.ft. If the thickness of the tube is 0.5”, the shear stress (psi) is close to: A. 80 B. 120 C. 160 D. 220

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387

Question 18. A wooden beam (5in x 4in) is subjected to a resultant internal vertical shear force of V = 3,000 psi. The maximum shear stress in the beam is close to: A. B. C. D.

1250 2250 4500 7500

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388

Question 19. A beam is constructed from four boards is glued together is subjected to a shear of V = 850kN, the shear flow (MN/m) at B is close to: Moment of inertia: 87.52 x 10-6 m4 Q at location B: 0.271 x 10-3 m3 A. B. C. D.

1.00 1.76 2.63 3.75

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389

Question 20. A box beam is to be constructed from four boards nailed together is subjected to a vertical force of 80 lb. The shear flow is nearly: A. 2 B. 4 C. 8 D. 12

Question 21. The state plane stress at a point on a body are given as σx = -20 MPa, σy = 90 MPa. τxy = 60 MPa, the principal stresses (MPa) are: A. B. C. D.

80, 100 120, -50 100, -80 150, -100

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390

Question 22. The state of plane stress at a point on a body is represented by σx = -20 MPa, σy = 90 MPa. τxy = 60 MPa, the maximum in-plane shear stress (MPa) and the average normal stress (MPa) is close to: A. B. C. D.

81; 35 35; 88 55; 81 81; 45

Question 23. Given, σx = -12,000 psi, σy = 0 psi and τxy = 6,000 psi, the principal stresses (x1000psi)are: A. B. C. D.

1.5, 10.5 2.5, - 15 4.5, -20 -1.5, 15

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391

Question 24. Given σx = -8,000 psi, σy = 12 psi and τxy = -6,000 psi, the average stress (x1000 psi) and maximum in plane stress (x1000 psi) are close to: A. B. C. D.

1,000, 12,000 1,500, 10,000 2,000, 12,000 1,500, 15,000

Question 25. A beam (0.1075 x 0.175m x 0.015m) is subjected to a shear force of 84 kN and a moment of 30.6 KN.m, given I = 67.4 (10-6) m4, y = 0.1m; the stress (MPa) components (σ,τ) are: A. B. C. D.

-45; 35 -35; 45 55; -75 -25; 55

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392

Question 26. Due to the applied loading, the element at the point on the frame in figure is subjected to the state of plane stress as shown in the figure. The absolute maximum shear stress (psi) is close to: A. B. C. D.

20 40 60 80

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393

Question 27. The state of plane strain at a point is represented by the strain components εx = -400 (10-6), εy = 200 (10-6), γxy = 150 (10-6). The maximum in plane shear strain and the absolute maximum shear strain are close to: A. B. C. D.

100 x 10-6; 800 x 10-6 210 x 10-6; 620 x 10-6 310 x 10-6; 800 x 10-6 400 x 10-6; 900 x 10-6

Question 28. A bracket is made of steel for which E = 200 GPa, the Poisson ratio is 0.3, the shear modulus (GPa) is: A. B. C. D.

47 57 67 77

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394

Question 29. A copper bar is subjected to a uniform loading of σx = 800 MPa, σy = -500 MPa. τxy = 0 MPa and σz = 0. Take Ecu = 120 GPa and νcu = 0.34. The normal strains are: A. B. C. D.

0.008, -0.006, -0.00085 0.005, -0.006, -0.0085 0.1, -0.005, -0.009 0,0,0

Question 30. A rectangular block (4in x 3in x 2in) is subjected to a uniform pressure of 20 psi. The change in length of each side is nearly: E = 600 psi ν = 0.45 A. B. C. D.

-0.10; -0.055; -0.045 -0.055; -0.0034; -0.030 -0.0111; -0.0577; - 0.020 -0.0133; -0.00667; -0.010

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395

Question 31. A solid cast-iron shaft is subjected to a torque 400 lb.ft. The smallest radius (inches) so that it does not fail according to the maximum normal stress theory is: σult = 20,000 psi

A. B. C. D.

0.28 0.33 0.42 0.54

Question 32. A steel pipe has an inner diameter of 60mm and an outer diameter of 80mm. It is subjected to a torsional moment of 8kN and a bending moment of 3.5kN.m. If σ1 = 76.1 MPa and σ2 = -178.0 MPa, determine if these loadings cause failure as defined by the maximum-distortion theory. The yield stress for the steel found from a tension test is σY = 250 MPa. A. B. C. D.

The pipe will fail The pipe will not fail Distortion theory can’t predict failure Not enough information is provided

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396

Question 33. A steel solid shaft has a radius of 0.5in has a yielding stress of σ1 = 9,560 psi. Given σ2 = -28,660 psi and σy = 36,000 psi, determine if these loadings cause the shaft to fail according to the maximum-shear-stress theory? A. B. C. D.

The material will fail The material will not fail Can’t be predicted using this theory Not sufficient information is provided

Question 34. A 24-ft long steel having an outside diameter of 3.0 in with a thickness of 0.25in is to be used as a pin-ended column. The maximum allowable axial load (x 1000 psi) the column can support is close to: A. B. C. D.

15 35 65 88

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397

Question 35. A 12-ft long steel member is to be used as pin-connected column. The average compressive stress (psi) in the columns is: Given: Cross-section area: 9.13in2 Moment of inertia: 37.1in4 E: 29,000 psi A. B. C. D.

14.000 22.000 42.000 56.000

Question 36. A 5m long aluminum column in fixed at its bottom and is braced at its top by cables so as to prevent movement at the top along the x-axis. The largest load (kN) that can be applied is: Factor of safety: Eal: σY: A: Ix: Iy: K (x-x buckling) K (y-y buckling) A. B. C. D.

3.0 70 GPa 215 MPa 7.5 x 10-3 m2 61.3(10-6)m4 23.2(10-6)m4 2.0 0.7

100 120 140 160

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398

Question 37. A high strength steel bolt with 2.25in in length and 0.731in in diameter is chosen to support a tensile loading. The elastic strain energy (in x 1000 psi) that the bolt can absorb is: σY = 44,000 psi A. B. C. D.

0.011 0.022 0.032 0.044

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

399

SOLUTIONS QUESTION 1 The maximum stress occurs in the circumferential direction: 𝑃𝑟 𝜎𝑎 = 𝑡 𝑃(24𝑖𝑛) 20,000 = = 417 𝑝𝑠𝑖 1 𝑖𝑛 2 QUESTION 2 The maximum stress is: 𝑀𝐶 𝜎𝑚𝑎𝑥 = 𝐼 𝑀 = 150 𝑙𝑏 𝑥 5𝑖𝑛 = 750 𝑙𝑏 − 𝑖𝑛 = QUESTION 3

[750 𝑥 5𝑖𝑛] 1 3 12 (4𝑖𝑛)(10𝑖𝑛) 𝜎𝑎 =

𝑃 = 𝛾𝑧 𝑧 = 62.4 𝜎𝑎 = 1.30

= 11.25 𝑝𝑠𝑖

𝑃𝑟 𝑡

𝑙𝑏 1𝑓𝑡 𝑥 3𝑓𝑡 𝑥 = 1.30 𝑝𝑠𝑖 𝑓𝑡 3 144𝑖𝑛2

𝑙𝑏 1 𝑥 24𝑖𝑛 𝑥 = 62.4 𝑝𝑠𝑖 2 𝑖𝑛 0.5𝑖𝑛

QUESTION 4 The moment of inertia: 𝐼=

1 (6𝑖𝑛)(8𝑖𝑛)3 = 184 𝑖𝑛4 12

𝑉𝑄𝑐 𝐼 10,000 𝑝𝑠𝑖 𝑥 17.5𝑖𝑛3 /2 = = 47 60 𝑝𝑠𝑖 184 𝑖𝑛4 𝑞𝑐 =

QUESTION 5 The normal stress within each segment is: 𝑃 10(10)3 𝑁 𝜎𝐴𝐵 = = = 31.83𝑀𝑃𝑎 𝐴 𝜋 (0.01𝑚)2 𝑃 10(10)3 𝑁 𝜎𝐵𝐶 = = = 56.59 𝑀𝑃𝑎 𝐴 𝜋 (0.0075𝑚)2 31.83 𝑥 106 𝑃𝑎 𝜀𝐴𝐵 = = 0.0004547 𝑚𝑚/𝑚𝑚 70 𝑥 109 𝑃𝑎 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

400

𝜀𝐵𝐶 =

56.59 𝑥 106 𝑃𝑎 𝑚𝑚 = 0.045 9 70 𝑥 10 𝑃𝑎 𝑚𝑚

The elongation of the rod is: 𝛿 = 0.0004547(600𝑚𝑚) + 0.045(400𝑚𝑚) = 18.3𝑚𝑚 QUESTION 6 The normal stress in the bar is: 𝜎𝑡 =

𝑃 80,000𝑁 = = 16 𝑥 106 𝑃𝑎 𝐴 0.1 𝑥 0.05

E (steel) = 200 GPa (from NCEES reference handbook table) 𝜀=

𝜎 16 𝑥 106 𝑃𝑎 = = 80 𝑥 10−6 𝑚𝑚/𝑚𝑚 𝐸(𝑠𝑡𝑒𝑒𝑙) 200 𝑥 109 𝑃𝑎

The axial elongation of the bar is: 𝛿 = 𝜀𝐿 = 80 𝑥 10−6

𝑚𝑚 𝑥 1.5 𝑚 = 120 𝜇𝑚 𝑚𝑚

QUESTION 7 The normal stress is: 𝑃 165 𝑥 103 𝑁 =𝜋 = 336.1 𝑀𝑃𝑎 2 𝐴 (0.025𝑚) 4 𝜀 1.20𝑚𝑚 𝑚𝑚 𝛿= = = 0.00480 𝐿 250𝑚𝑚 𝑚𝑚

𝜎𝑡 =

The modulus of elasticity E: 𝐸=

𝜎 336.1𝑀𝑃𝑎 = = 70 𝐺𝑃𝑎 𝜀 0.00480

QUESTION 8 The Poisson’s ratio for the material is: 𝐸 2 (1 + 𝜈) 70𝐺𝑃𝑎 26 𝐺𝑃𝑎 = 𝑜𝑟 𝜈 = 0.347 2(1 + 𝜈) 𝐺=

𝜈=

0.347 =

𝜀𝑙𝑎𝑡 𝜀𝑙𝑜𝑛𝑔

𝜀𝑙𝑎𝑡 𝑚𝑚 𝑜𝑟 𝜀𝑙𝑎𝑡 = −0.00166 0.00480 𝑚𝑚

The contraction of the diameter is: 𝛿 = (0.00166)(25𝑚𝑚) = 0.0416 𝑚𝑚 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

401

QUESTION 9 The displacement of end C with respect to B is: 𝑃𝐿 80 𝑥 103 𝑁 𝑥 (0.6𝑚) 𝛿𝐶/𝐵 = = = +0.003056 𝑁 𝐴𝐸 𝜋 (0.005𝑚)2 (200 𝑥 109 2 ) 𝑚 The positive sign indicates that end C moves to the right The displacement of end B with respect to fixed A is: −80 𝑥 103 𝑁 𝑥 (0.4𝑚) = −0.001143𝑚 106 𝑚2 𝑁 2 9 (400𝑚𝑚) 𝑥 𝑥 (70 𝑥 10 2 ) 𝑚𝑚2 𝑚 The negative sign indicates that the tube shortens, so B moves to the right relative to A. Since both displacements are to the right, the resultant displacement is: 𝛿𝑐 = 0.001143 + 0.003056 = 0.0042 𝑚 𝛿𝐵 =

𝑃𝐿 = 𝐴𝐸

QUESTION 10 𝐹 𝑎 𝐹 = 𝛼Δ𝑇𝐴𝐸 𝐹 = 6.60 𝑥 10−6 𝑥 (120 − 60)(0.5𝑖𝑛)2 (29 𝑥 106 𝑃𝑠𝑖) = 2.87 𝑥 103 𝑝𝑠𝑖 2.87 𝑥 103 𝑝𝑠𝑖 𝜎= = 11,500 𝑝𝑠𝑖 (0.5𝑖𝑛)2 𝜎=

QUESTION 11 The polar moment of inertia for the cross-sectional area is: 𝜋 𝐽 = (2𝑖𝑛)4 = 25.13 𝑖𝑛4 2 𝑇𝑐 𝜏𝑚𝑎𝑥 = 𝐽 𝑇(2𝑖𝑛) 8000 = 𝑜𝑡 𝑇 = 101 𝑘𝑖𝑝. 𝑖𝑛 25.13𝑖𝑛4 QUESTION 12 The polar moment of inertia for the pipe’s cross-sectional area is: 𝜋 𝐽 = ((0.05𝑚)4 − (0.04𝑚)4 ) = 5.80 𝑥 10−6 𝑚4 2 𝑇𝑐0 40𝑁. 𝑚 𝑥 (0.05𝑚) 𝜏0 = = = 0.345𝑀𝑃𝑎 𝐽 5.80 𝑥 10−6 𝑚4 QUESTION 13 The maximum torque that can be applied to the shaft is determined from the torsional formula: 𝑇𝑐 𝜏𝑚𝑎𝑥 = 𝐽 𝑁 𝑇(0.021𝑚) 50 𝑥 106 2 = 𝜋 4 4 𝑚 2 [(0.021𝑚) − (0.015𝑚) ] FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

402

𝑇 = 538 𝑁. 𝑚 𝑃 = 2𝜋𝑓𝑇 𝑚 90 𝑥 103 𝑁. = 2𝜋𝑓(538 𝑁. 𝑚) 𝑠 𝑓 = 26.6 𝐻𝑧 QUESTION 14 Using the sign convention (right hand rule), the positive torque is directed away from the sectioned end of the shaft. 𝑇𝐴𝑐 = +150 𝑁. 𝑚; 𝑇𝐶𝐷 = −130 𝑁. 𝑚; 𝑇𝐷𝐸 = −170 𝑁. 𝑚 The polar moment of inertia: 𝜋 𝐽 = (0.075𝑚)4 = 3.77 𝑥 10−9 𝑚4 2 Angle of twist: 𝑇𝐿 𝜙= 𝐽𝐺 =

150 𝑁. 𝑚 𝑥 0.4𝑚 3.77 𝑥 10−9 𝑚4 𝑥 80 𝑥 109

𝑁 𝑚2

+

−130 𝑁. 𝑚 𝑥 0.3𝑚 3.77 𝑥 10−9 𝑚4 𝑥 80 𝑥 109

𝑁 𝑚2

−170 𝑁. 𝑚 𝑥 0.5𝑚

+

3.77 𝑥 10−9 𝑚4 𝑥 80 𝑥 109

𝑁 𝑚2

= −0.212 𝑟𝑎𝑑

QUESTION 15 The angle of twist of end A is: 𝜙=

𝑇𝐴𝐵 𝐿𝐴𝐵 𝐽𝐺

45 𝑁. 𝑚 𝑥 2𝑚 𝜋 4 9𝑁 2 (0.010𝑚) 80𝑥10 𝑚

= +0.0716 𝑟𝑎𝑑

QUESTION 16 The internal torque is 35 N.m The cross sectional area is: 𝐴𝑚 = (0.035𝑚)(0.057𝑚) = 0.00200 𝑚2 𝑇 35 𝑁. 𝑚 𝜏𝐴 = = = 1.75 𝑀𝑃𝑎 2𝑡𝐴𝑚 2 𝑥 0.005 𝑥0.00200 𝑚2 QUESTION 17 The cross-sectional area is: 𝐴𝑚 = (2.5 𝑖𝑛)(2.5𝑖𝑛) = 6.25𝑖𝑛2 𝑖𝑛 85 𝑙𝑏. 𝑓𝑡 𝑥 12 𝑇 𝑓𝑡 𝜏𝑎𝑣𝑔 = = = 163 𝑝𝑠𝑖 2𝑡𝐴𝑚 2 𝑥 0.5𝑖𝑛 𝑥 6.25𝑖𝑛2 QUESTION 18 The maximum shear stress occurs at neutral axis, since t is constant throughout the cross-section: ̅ 𝐴′ 𝑄 = 𝑦′ 2.5𝑖𝑛 = 𝑥 4𝑖𝑛 𝑥 2.5 𝑖𝑛 = 12.5 𝑖𝑛3 2 𝑉𝑄 3000 𝑝𝑠𝑖 𝑥12.5 𝑖𝑛3 𝜏𝑚𝑎𝑥 = = = 2250 𝑝𝑠𝑖 𝐼𝑡 41.7 𝑖𝑛4 𝑥 4𝑖𝑛

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403

QUESTION 19 𝑞𝐵 =

𝑉𝑄𝐵 850 𝐾𝑁 𝑥 0.271 𝑥 10−3 𝑚3 𝑀𝑁 = = 2.63 −6 3 𝐼 87.52 𝑥 10 𝑚 𝑚

QUESTION 20 The moment of inertia of the cross-sectional area about the neutral axis can be determined as: 1 1 𝐼 = (7.5𝑖𝑛)(7.5𝑖𝑛)3 − (4.5𝑖𝑛)(4.5𝑖𝑛)3 = 229.5 𝑖𝑛4 2 12 ̅ 𝐴′ = 93𝑖𝑛)(7.5𝑖𝑛)(1.5𝑖𝑛) = 33.75 𝑖𝑛3 𝑄𝐵 = 𝑦′ 𝑉𝑄𝐵 80𝑙𝑏 𝑥 33.75 𝑖𝑛3 𝑙𝑏 𝑄𝐵 = = = 11.76 𝐼 229.5 𝑖𝑛4 𝑖𝑛 QUESTION 21 Principal stresses: 𝝈𝒙 + 𝝈𝒚 𝝈𝟏,𝟐 = ± √((𝝈𝒙 − 𝝈𝒚 )/𝟐)𝟐 + 𝝉𝟐𝒙𝒚 𝟐 =

−𝟐𝟎 + 𝟗𝟎 −𝟐𝟎 − 𝟗𝟎 𝟐 ± √[ ] + (𝟔𝟎)𝟐 = 𝟑𝟓 ± 𝟖𝟏. 𝟒 𝟐 𝟐 𝝈𝟏 = 𝟏𝟏𝟔 𝑴𝑷𝒂; 𝝈𝟐 = −𝟒𝟔. 𝟒 𝑴𝑷𝒂

QUESTION 22 Maximum stress: 𝜏𝑚𝑎𝑥 = √((𝝈𝒙 − 𝝈𝒚 )/𝟐)𝟐 + 𝝉𝟐𝒙𝒚 = √[

−𝟐𝟎 − 𝟗𝟎 𝟐 ] + (𝟔𝟎)𝟐 = 𝟖𝟏. 𝟒 𝑴𝑷𝒂 𝟐

Average normal stress is: 𝜎𝑎𝑣𝑔 = QUESTION 23

−20 + 90 = 35 𝑀𝑃𝑎 2

−12 + 0 = −6𝑘𝑠𝑖; 1 𝑘𝑠𝑖 = 1000𝑝𝑠𝑖 2 𝑅 = √(12 − 6)2 − (6)2 = 8.49 𝑘𝑠𝑖

𝜎𝑎𝑣𝑔 =

𝜎1 = 8.49 − 6 = 2.49 𝑘𝑠𝑖 𝜎2 = −6 − 8.49 = −14.5 𝑘𝑠𝑖 QUESTION 24 The average stress is: −8000 + 12,000 = 2000 𝑝𝑠𝑖 2 𝑅 = √(10)2 + (6)2 = 11.66 𝑘𝑠𝑖 𝑜𝑟 11,660𝑝𝑠𝑜 𝜎𝑎𝑣𝑔 =

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QUESTION 25 𝑀𝑦 𝐼 30.6 𝑥 103 𝑥 0.100𝑚 = − = −45.4 𝑀𝑃𝑎 67.4 𝑥 10−6 𝑚4 𝑉𝑄 84 𝑥 103 𝑥 0.1075𝑚 𝑥 0.175𝑚 𝑥 0.015𝑚 𝜏= = = 35.2 𝑀𝑃𝑎 𝐼𝑡 67.4 𝑥 10−6 𝑚4 𝑥 0.010𝑚 QUESTION 26 The in-plane principal stresses can be determined from Mohr’s circle. The center of the circle is on the σ-axis at: −20 + 0 𝜎𝑎𝑣𝑔 = = −10 2 𝑅 = √(20 − 10)2 + (40)2 = 41.2 𝑝𝑠𝑖 𝜎= −

𝜎𝑚𝑎𝑥 = −10 + 41.2 = 31.2 𝑝𝑠𝑖 𝜎𝑚𝑖𝑛 = −10 − 41.2 = −51.2 𝑝𝑠𝑖 𝜏𝑚𝑎𝑥 =

𝜎𝑚𝑎𝑥 − 𝜎𝑚𝑖𝑛 31.2 − (−51.2) =( ) = 41.2 𝑝𝑠𝑖 2 2

QUESTION 27 −400 + 200 = −100𝑀𝑃𝑎 2 Since γxy/2 = 75 x 106, the reference point has the coordinates A (-400 x 106, 75 x 106). The radius of the circle is: 𝑅 = √[(400 − 100)2 + (75)2 ] 𝑥 10−6 = 309 𝑥 10−6 𝜀𝑎𝑣𝑔 =

In place principal strains we have are: 𝜀𝑚𝑎𝑥 = (−100 + 309)10−6 = 209 𝑥 10−6 𝜀𝑚𝑖𝑛 = (−100 − 309)𝑥 10−6 = −409 𝑥 10−6 The maximum in-plane shear strain is: [209 − (−409)] 𝑥 10−6 = 618 𝑥 10−6 QUESTION 28 The shear modulus is given by: 𝐸 200𝐺𝑃𝑎 𝐺= = = 76.9 𝐺𝑃𝑎 2(1 + 𝜈) 2(1 + 0.3) QUESTION 29 From Hook’s law the normal strains are: 𝜎𝑥 𝜈 𝜀𝑥 = − (𝜎𝑦 + 𝜎𝑧 ) 𝐸 𝐸 800𝑀𝑃𝑎 0.34 (−500𝑀𝑃𝑎) = 0.00808 = − 3 120 𝑥 10 𝑀𝑃𝑎 120 𝑥 103 𝑀𝑃𝑎 𝜀𝑦 = −0.00643; 𝜀𝑧 = −0.000850

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QUESTION 30 The dilation can be determined using: 1 − 2𝜈 (𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧 ) 𝐸 1 − 2𝑥0.45 𝑖𝑛3 = 𝑥 (3𝑥 (−20𝑝𝑠𝑖)) = −0.01 3 600𝑝𝑠𝑖 𝑖𝑛 The normal strain on each side can be determined from Hook’s law: 𝑒=

𝜀=

𝜎𝑥 𝜈 − (𝜎 + 𝜎𝑧 ) 𝐸 𝐸 𝑦

1 𝑖𝑛 [−20 − 0.45(−20 − 20)] = −.00333 600𝑝𝑠𝑖 𝑖𝑛 Thus the change in length of each side is: 𝛿𝑎 = −0.00333𝑥4 = −0.0133 𝑖𝑛 =

𝛿𝑏 = −0.00333 𝑥 2 = −0.00667 𝑖𝑛 𝛿𝑐 = −0.00333 𝑥 3 = −0.100 𝑖𝑛 QUESTION 31 Assuming the shaft has a radius of r, the shear stress is: 𝑇𝑐 𝜏𝑚𝑎𝑥 = 𝐽 12𝑖𝑛 400 𝑙𝑏. 𝑓𝑡 𝑥 𝑥 𝑟 3055.8 𝑙𝑏. 𝑖𝑛 𝑓𝑡 = = 𝜋 4 𝑟3 𝑥 𝑟 2 The smallest radius of the shaft is determined from: 3055.8 𝑙𝑏. 𝑖𝑛 = 20,000 𝑝𝑠𝑖 𝑟3 𝑟 = 0.535 𝑖𝑛 QUESTION 32 From the distortion theory: 𝜎12 − 𝜎1 𝜎2 + 𝜎22 ≤ 𝜎𝑦2 (76.1)2 − (76.1)(−178.0) + (−178)2 ≤ (250)2 51,000 < 62,500 Since the criterion has been met, the material within the pipe will not fail according to maximum distortion theory QUESTION 33 According to the maximum distortion theory: |𝜎1 − 𝜎2 | ≤ 𝜎𝑦 |9.56 − (−28.66)| ≤ 36 38.2 > 36 Thus shear failure of the material will occur.

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QUESTION 34 𝜋 2 𝐸𝐼 𝑃𝑐𝑟 = 𝐿2 1 1 𝜋 2 𝑥 (29 𝑥 103 𝑝𝑠𝑖)𝑥 [4 𝜋 (3)4 − 4 𝜋(2.75)4 ]𝑖𝑛4 = = 65,000 12𝑖𝑛 2 [24 𝑓𝑡 𝑥 ( )] 𝑓𝑡 QUESTION 35 𝜋 2 𝐸𝐼 𝐿2 2 (29 3 𝜋 𝑥 10 𝑝𝑠𝑖)(37.1)𝑖𝑛4 = = 512,000 𝑝𝑠𝑖 12𝑖𝑛 2 [12 𝑓𝑡 𝑥 ( )] 𝑓𝑡 When fully loaded, the average compressive strength in the column is: 𝑃𝑐𝑟 𝜎𝑐𝑟 = 𝐴 512,000 𝑝𝑠𝑖 = = 56,100 𝑝𝑠𝑖 9.13 𝑖𝑛2 QUESTION 36 𝜋 2 𝐸𝐼𝑥 𝑃𝑐𝑟 = (𝐾𝐿)2 𝑁 𝜋 2 (70𝑥109 2 ) 𝑥 (61.3 𝑥 10−6 𝑚4 ) 𝑚 = = 424 𝐾𝑁 (10𝑚)2 The allowable load is: 𝑃𝑐𝑟 424 𝐾𝑁 = = 141 𝐾𝑁 𝐹𝑆 3.0 QUESTION 37 If the bolt is subjected to its maximum tension, the maximum stress σ y = 44 ksi will occur within the 0.25in region. The tension force is: 0731 2 𝑃𝑚𝑎𝑥 = 𝜎𝑦 𝐴 = 44𝑘𝑠𝑖 (𝜋 𝑥 ) = 18.47 𝑘𝑖𝑝 2 𝑃𝑐𝑟 =

𝑁2𝐿 2𝐴𝐸 (18.47)2 (0.25𝑖𝑛) (18.47)2 (2𝑖𝑛) = + = 0.0231 𝑘𝑖𝑝. 𝑖𝑛 0.731 2 0.875 2 3 3 2 [𝜋 𝑥 2 ] 𝑥 (29 𝑥 10 ) 2 [𝜋 𝑥 2 ] 𝑥 (29 𝑥 10 ) 𝑈= ∑

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Materials Properties and Processing Total Questions 8-12 A. B. C. D. E. F. G. H. I. J. K. L.

Properties, including chemical, electrical, mechanical, physical, and thermal Stress-strain diagrams Engineered materials Ferrous metals Nonferrous metals Manufacturing processes Phase diagrams Phase transformation, equilibrium, and heat treating Materials selection Surface conditions Corrosion mechanisms and control Thermal failure

M. N. O.

Ductile or brittle behavior Fatigue Crack propagation

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QUESTION 1. The number of atoms in 100 g of silver (Ag) is close to: A. B. C. D.

3.38 x 1021 5.58 x 1023 6.75 x 1024 7.78 x 1025

QUESTION 2. The number of Fe atoms for the following data is close to: The radius of a particle: 1.5 nm Density of iron: 7.8 g/cm3 Particle shape: sphere Atomic weight of iron: 56 g/mol A. B. C. D.

1,200 2,100 2,300 2,600

QUESTION 3. The number of valance electrons in 10 cm3 of silver is close to: Density of silver: 10.49 g/cm3 Atomic weight of silver: 107.868 g/mol

A. B. C. D.

2.85 x 1023 3.85 x 1023 4.85 x 1023 5.85 x 1023

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QUESTION 4. The activation energy, Q for aluminum in a copper solvent at 575 oC is 1.6 x 108 J/kmol. The diffusion coefficient in (m2/s) if the constant proportionality D0 is 7 x 10-6 m2/s is close to: A. B. C. D.

4.04 x 10-47 2.04 x 10-20 9.75 x 10-16 2.31 x 10-5

QUESTION 5. Calculate the length after deformation of a bar, if initial length was 50 inches when a stress of 30,000 psi applied is close to: Young’s modulus: 10 x 106 psi A. B. C. D.

49.85 50.15 50.50 50.95

QUESTION 6. An aluminum alloy has an initial length of 2 inches and final length of 2.195 inches. The initial and final diameters were 0.505 and 0.398 inches, respectively. The ductility (%) is close to: A. B. C. D.

11 and 38 10 and 45 10 and 38 12 and 45

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QUESTION 7. An impact test measures: A. B. C. D.

Brittleness Hardness Toughness Ductility

QUESTION 8. A large steel plate used in a nuclear reactor has a plain strain fracture toughness of 80,000 psi √in and is exposed to a stress of 45,000 psi during service. Assuming a geometric parameter of 1.12, the crack length (inches) is close to: A. B. C. D.

0.20 0.40 0.60 0.80

QUESTION 9. A high-strength steel plate has a plane strain fracture toughness of 80 MPA √m is alternately loaded in tension to 500 MPa and in compression to 60 MPa. The crack size (mm) (assuming the geometric parameter equal to 1) is close to: A. 2 B. 4 C. 8 D. 12

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QUESTION 10. The ratio between the lateral and longitudinal strain in the elastic region is called: A. B. C. D.

Young’s modulus Poisson’s ratio Hoops’s ratio Diffusion coefficient

QUESTION 11. The heat treatment process used to eliminate some or all of the effects of cold working is known as: A. B. C. D.

Annealing Hardening Stress relief Yield strength

QUESTION 12. A mixture of ice and water is held at a constant temperature of 0oC. The degrees of freedom of the mixture is (are): A. B. C. D.

-1 0 1 2

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QUESTION 13. The degrees of freedom in a Cu-40%, Ni alloy at 1300oC using the following phase diagram is (are):

A. B. C. D.

0 1 2 3

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QUESTION 14. The composition (%) of each phase in a Cu-40%, Ni alloy at 1250oC (use the above figure) is nearly: A. B. C. D.

50 Cu and 50 Ni 45 Cu and 50 Ni 32 Cu and 40 Ni 32 Cu and 45 Ni

QUESTION 15. The amount of α and L at 1250oC in the Cu-40% Ni alloy from the above figure is: A. B. C. D.

62% and 38% 70% and 30% 65% and 35% 55% and 45%

QUESTION 16. The amount of each phase in Cu-40% Ni alloy at 1270oC is close to: A. B. C. D.

70% and 30% 77% and 23% 65% and 35% 60% and 40%

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QUESTION 17. A brass alloy is 40% zinc and 60% copper by weight. The approximate mole fraction (%) of zinc is: A. 5 B. 26 C. 39 D. 50

QUESTION 18. The composition of lead-tin alloy at 61.9% Sn using Lever rule is nearly:

A. B. C. D.

45% and 55% 40% and 60% 50% and 50% 55% and 45%

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QUESTION 19. The maximum percentage of carbon in cementite (Fe3C) is: A. B. C. D.

2.34 3.78 4.98 6.67

QUESTION 20. The percentage of carbon in pearlite is: A. B. C. D.

0.25 0.33 0.77 0.99

QUESTION 21. The hardest form of steel is: A. B. C. D.

Pearlite Ferrite Bainite Martensite

QUESTION 22. Steel is alloy of the following elements:

A. B. C. D.

Iron and silicon Iron and carbon Iron and chromium Iron and molybdenum

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QUESTION 23. The amount of Fe3C in Fe-0.60% and 0.0218% C at 726 is close to:

A. B. C. D.

91% and 9% 90% and 10% 80% and 20% 70% and 30%

QUESTION 24. Cast iron is an alloy of: A. B. C. D.

Iron, carbon and nickel Iron, carbon and zinc Iron, carbon and silica Iron, carbon and silver

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QUESTION 25. To make 1000 kg of BaTiO3 ceramic from BaCO3 and TiO2, approximately, how much (kg) barium carbonate and titanium dioxide should be mixed together? 𝐵𝑎𝐶𝑂3 + 𝑇𝑖𝑂2 → 𝐵𝑎𝑇𝑖𝑂3 + 𝐶𝑂2 A. B. C. D.

800 and 400 850 and 340 900 and 200 700 and 300

QUESTION 26. Silicon carbide particles are compacted and fired at high temperatures to produce a strong ceramic shape. The specific gravity of SiC is 3.2 g/cm3. The ceramic shape subsequently is weighed when dry (360 g), after soaking in water (385 g) and while suspended in water (22g). The porosity (%) is close to: A. B. C. D.

10 12 14 16

QUESTION 27. An example of a thermoplastic material is: A. B. C. D.

Polyurethane Polyethylene Natural rubber None of the above

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QUESTION 28 The number of monomers in 1 kg of polyethylene with an average molecular weight of 200,000 g/mol is close to: Molecular weight of ethylene: 28 g A. B. C. D.

215 x 1023 325 x 1023 450 x 1023 555 x 1023

QUESTION 29. The degree of polymerization if 6,6-nylon has a molecular weight of 120,000 g/mol is nearly: The molecular weight of 6,6-nylon is 226 g/mol A. B. C. D.

260 420 530 720

QUESTION 30. A polyethylene sample contains 4,000 chains with molecular weight between 0 and 5,000 g/mol, 8,000 chains with molecular weights between 5,000 and 10,000 g/mol, 7,000 chains with molecular weights between 10,000 and 15,000 g/mol, and 2000 chains with molecular weights between 15,000 and 20,000 g/mol, the number average molecular weight is close to: Assume the total number of chains: 21,000 A. B. C. D.

5,160 7,160 8,260 9,160

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QUESTION 31. An example of a heat stabilizer material added to thermoplastics to absorb heat and protect the polymer matrix is: A. B. C. D.

Limestone Carbon black Antimony compounds Halogen compounds

QUESTION 32. To help improve the resistance of plastics to ultraviolet degradation, the following is commonly added to the polymer: A. B. C. D.

Limestone Hydrated alumina Carbon black Talc

QUESTION 33. To retard the flammability of polymer the following is compound is additive is mixed: A. B. C. D.

Limestone Carbon black Talc Magnesium bromide

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QUESTION 34. A storage tank for liquid hydrogen will be made of metal, but must be coated with a 3-mm thickness of polymer as an intermediate layer between metal and additional insulation layers. The temperature of intermediate layer may drop to -80 C. The material suitable for this application is: A. B. C. D.

Natural rubber Thermoplastic Elastomer Carbon black

QUESTION 35. A 2% weight thallium oxide is added to nickel. Each thallium oxide particle has a diameter of 1000 Ao. The number of particles present in each cubic centimeter is close to: The density of nickel: 8.9 g/cm3 The density of thallium oxide: 9.69 g/cm3 A. B. C. D.

35 x 1012 45 x 1012 55 x 1012 75 x 1012

QUESTION 36. A cemented carbide cutting tool used for machining contains 75 wt% tungsten carbide (WC), 15 wt% titanium carbide (TiC), 5 wt% tantalum carbide (TaC) and 5 wt% of cobalt. The density of the composite (g/cm3) is close to: Density of WC: 15.77 g/cm3 Density of TaC: 14.5 g/cm3 Density of TiC: 4.94 g/cm3 Density of Co: 8.90 g/cm3 A. B. C. D.

9.50 11.50 13.50 15.50

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QUESTION 37. A silver-tungsten composite for an electrical contact is produced by first making a porous tungsten powder metallurgy compact, then infiltrating pure silver into the pores. The density of the tungsten compact before infiltration is 14.5 g/cm3. The weight percent of silver in the compact after infiltration is nearly: Density of tungsten: 19.3 g/cm3 Density of silver: 10.49 g/cm3 Fraction of pores: 0.25 A. B. C. D.

10.3 12.3 15.3 18.3

QUESTION 38. A clay-filled polyethylene composite suitable for injection molding of inexpensive components is designed. The final part must have a tensile strength of at least 3000 psi and a modulus of elasticity of at least 80,000 psi. The density of polyethylene is 0.95 g/cm3and that of clay is 2.4 g/cm3. The composite density (g/cm3) assuming the fraction of clay is 0.35 is close to: A. B. C. D.

0.45 0.90 1.45 2.45

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QUESTION 39 Boron coated with SiC reinforced aluminum containing 40 vol% fibers is an important hightemperature, lightweight composite material. The modulus of elasticity (psi) of the composite material is close to: Material

Density g/cm3

Fibers Aluminum

2.36 2.70

A. B. C. D.

Modulus of Elasticity psi 55,000,000 10,000,000

Tensile strength psi 400,000 5,000

4190 0.149 x 106 1.49 x 106 14.9 x 106

QUESTION 40. The half-life of an element is 4.3 days. The time (days) required to reduce the original amount to 1% is nearly: A. B. C. D.

4 12 19 29

QUESTION 41 A green wood has a density of 0.86 g/cm3 and contains 175% water. The density (g/cm3) of the wood after it has completely dried: A. B. C. D.

0.113 0.313 0.413 0.513

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QUESTION 42. The minimum water-cement ratio required for workability is: A. B. C. D.

0.1 0.2 0.3 0.4

QUESTION 43. The workability of a concrete structure can be measured by: A. B. C. D.

Slump test Water test Air-entrapment test Aggregate test

QUESTION 44. Calculate the total volume (ft3/sack) of concrete in 5 cubic yards of concrete bag, assuming that we want to obtain a water/cement ratio of 0.4 (by weight) and that the cement/sand/aggregate ratio is 1:2.5:4 (by weight). A normal “aggregate” will be used, containing 1% water, and the sand contains 4% water. Assume that no air is entrained into the mixture. Each sack of cement contains: 94 lbs Density of cement: 190 lb/ft3 Density of sand: 160 lb/ft3 Density of gravel: 170 lb/ft3 Density of water: 62.4 lb/ft3 A. B. C. D.

2.4 3.2 4.8 5.6

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QUESTION 45. A count of 16 grains per square inch in a photomicrograph was taken at magnification x250. The ASTM grain size number is: A. 7.64 B. 8.92 C. 10.69 D. 12.45

QUESTION 46. The electrical conductivity (ohm-1cm-1) of pure copper at -100 C is: Conductivity of pure copper: 6.98 x 10 ohm-1cm-1 Resistivity at 25C: 1.67 x 10-6 ohm.cm Temperature resistivity coefficient: 0.0068 ohm.cm/C A. B. C. D.

20 x 105 40 x 105 60 x105 80 x 105

QUESTION 47. The amount of heat that must be supplied to 250 g of tungsten to raise its temperature from 25 C to 650 C: Specific heat of tungsten: 0.032 cal/(g.K) A. B. C. D.

1,500 2,000 3,500 5,000

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QUESTION 48. The temperature of 50 g of niobium (atomic weight: 92.91 g/mol) increases 75 C when heated for a period. The heat in calories required is nearly: The heat capacity of niobium: 6 cal/(mol.C) A. B. C. D.

120 180 240 480

QUESTION 49. The dimensions (cm x cm x cm) of a pattern that will be used to produce a rectangular shaped aluminum casting having dimensions at 25C of 25 cm x 25 cm x 3 cm are: Linear coefficient of thermal expansion for aluminum: 25 x 10-6 1/C Aluminum solidification temperature: 660 C A. B. C. D.

25.1 x 25. 1 x 3.20 25.4 x 25.4 x 3.05 25.4 x 25.4 x 3.12 25.4 x 20.4 x 3.05

QUESTION 50. The chemical compound known as “rust” is: A. B. C. D.

FeO FeO3 Fe(OH)3 Fe2O3

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QUESTION 51. 1 gram of Cu2+ is dissolved in 1000 g of water to produce an electrolyte. The electrode potential (V) of the copper half-cell in this electrolyte is close to: Atomic weight of copper: 63.54 g/mol Standard potential: + 0.34V A. B. C. D.

0.15 0.30 0.45 0.60

QUESTION 52 A cold drawn steel wire is formed into a nail by additional deformation, producing the point at one end and the head at the other. The corrosion of the nail will occur at: A. B. C. D.

Head and point Shank Throughout the nail Head

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QUESTION 53. The time (years) required for a 0.1 cm nickel sheet to oxidize completely is: k for nickel: 3.9 x 10-12 cm2/s A. B. C. D.

10 15 20 25

QUESTION 54. Austenite and martensite are: A. B. C. D.

Equilibrium solutions of carbon and iron An equilibrium solution and a non-equilibrium solution, respectively of C and Fe A non-equilibrium solution and an equilibrium solution of C and Fe Both non-equilibrium solutions of C and Fe

QUESTION 55. The average molecular mass of polyvinylchloride molecules is 50,000 g. The number of mers present are: Molecular weight = 62 g/mol A. 200 B. 400 C. 600 D. 800

QUESTION 56. The linear portion of the stress-strain diagram of steel is known as: A. Elastic range B. Modulus of elasticity C. Scant modulus D. Irreversible range FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 57. All of the following procedures may reduce corrosion, except: A. B. C. D.

Avoidance of bimetallic contacts Sacrificial anodes Aeration of feed water Impressed voltages

QUESTION 58. All of the following metals will corrode if immersed in fresh water, except: A. Copper B. Magnesium C. Nickel D. Aluminum

QUESTION 59. The movement of defects through a crystal by diffusion is described by: A. Boyle’s law B. Fick’s law C. Dalton’s law D. Gibb’s law

QUESTION 60. The process of annealing can be used to achieve all of the following, except:

A. B. C. D.

Stress relief Recrystallization Grain growth Toughness

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QUESTION 61. A cylindrical test specimen with a 15 mm diameter is tested axially in tension. A 0.20 mm elongation is recorded in a length of 200 mm when the load on the specimen is 36 kN. The material behaves elastically during testing. The material of the specimen sample is: A. Aluminum B. Magnesium C. Polystyrene D. Steel

QUESTION 62. The number of atoms in a hexagonal close-packed cell is: A. B. C. D.

2 4 6 8

QUESTION 63. The ratio of stress-strain below the proportionality limit is called: A. B. C. D.

Modulus of rigidity Hooke’s constant Poisson’s ratio Young’s modulus

QUESTION 64. What does the Charpy test determine? A. B. C. D.

Endurance Yield strength Ductility Toughness

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QUESTION 65. The material property that depends only on the basic crystal structure is: A. B. C. D.

Elastic constant Fatigue strength Work hardening Fracture strength

QUESTION 66. The crystal structure of austenite is: A. B. C. D.

Body centered cubic Face centered cubic Hexagonal closed packed Body centered tetragonal

QUESTION 67. The operation in which oil is permeated into the pores of a powder metallurgy product is known as: A. B. C. D.

Mixing Sintering Impregnation Infiltration

QUESTION 68. The effective number of lattice points in the unit cell of simple cubic, body centered cubic and face centered cubic space lattices, respectively, are: A. B. C. D.

1,2,2 1,2,4 2,3,4 2,4,4

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QUESTION 69. The entrained air in concrete: A. B. C. D.

Increases workability Decreases workability Increases strength None of the above

QUESTION 70. After casting, an ordinary cement concrete on drying: A. B. C. D.

Shrinks Expands Remain unchanged None of the above

QUESTION 71. Workability of concrete can be improved by the addition of: A. B. C. D.

Iron Sodium Zinc Sulfur

QUESTION 72. Workability of concrete can be improved by: A. B. C. D.

More sand More cement More fine aggregate Fineness of coarse aggregate

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QUESTION 73. Workability of concrete can be improved by: A. B. C. D.

Increasing size of aggregate Decreasing aggregate Increasing fine aggregate Increasing flaky aggregate

QUESTION 74. Workability of concrete is inversely proportional to: A. B. C. D.

Water cement ratio Size of aggregate Time of transit None of the above

QUESTION 75. Strength of concrete is directly proportional to: A. B. C. D.

Cement water ratio Water cement ratio Water aggregate ratio Sand cement ratio

QUESTION 76. Workability of concrete mix with low water cement ration is determined by: A. B. C. D.

Slump test Tensile strength test Compaction factor test Flexural strength test

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QUESTION 77. Minimum water-cement ratio required for full hydration of cement is: A. B. C. D.

0.18 0.36 0.54 0.60

QUESTION 78 Presence of entrained air in the cement results in: A. B. C. D.

Reduced bleeding Lower density Decreased strength at all ages All of the above

QUESTION 79. For a satisfactory workable concrete with a constant W/C ratio, increase in aggregate-cement ratio: A. B. C. D.

Increases strength of concrete Decreases strength of concrete No effect on strength of concrete None of the above

QUESTION 80. Three main raw materials used in Portland cement are: A. B. C. D.

Limestone, sandstone, clay Lime, silica and clay Lime, clay and gypsum Silica, aluminum and gypsum

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QUESTION 81. When water is added to cement: A. B. C. D.

Heat is generated Heat is absorbed Chemical reaction is started Impurities are washed out

QUESTION 82. Cement is useless if it absorbs moisture in excess (%) of: A. B. C. D.

1 2 4 5

QUESTION 83. If air is entrained in cement concrete: A. B. C. D.

Workability is decreased Workability is increased Strength is increased permeability is increased

QUESTION 84. Superplasticizer in cement is added to: A. Increase extreme workability B. Reduce amount of water required C. Increase bonding properties D. All of the above FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 85. Strength of cement concrete decreases: A. If W/C is less than 0.38 B. If W/C ratio is less than 0.60 C. With entrapped air D. If cement is increased

QUESTION 86. The lower water cement ratio in concrete introduces: A. B. C. D.

Smaller creep and shrinkage Greater density and smaller permeability Improved frost resistance All of the above

QUESTION 87. Water cement ratio is: A. B. C. D.

Density of water to that of cement Weight of water to that of cement Weight of concrete to that of water Volume of concrete to that of water

QUESTION 88. Placing of concrete should preferably be done at a temperature (C) of: A. 0 B. 10 C. 20 D. 27 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 89. The factor (s) which affects workability is: A. B. C. D.

Water content and its temperature Shape and size of the aggregate Air entraining agents All of the above

QUESTION 90. Workability of concrete for given water content is good if the aggregates, are: A. B. C. D.

Rounded aggregate Irregular aggregate Angular aggregate Flaky aggregates

QUESTION 91. If the average compressive strength is 4,000 kg/cm2 and standard deviation is 500, the coefficient of variation (%) is: A. B. C. D.

10.5 12.5 15.5 18.5

QUESTION 92. Portland cement is manufactured from: A. Limestone and clay B. Gypsum and clay C. Sand, clay and limestone D. Lime, pozzolana and clay

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SOLUTIONS QUESTION 1 Number of silver atoms:

𝑎𝑡𝑜𝑚𝑠 100 𝑔 𝑥 6.023 𝑥 1023 𝑚𝑜𝑙 = 5.58 𝑥 1023 𝑔 107.868 𝑚𝑜𝑙

QUESTION 2 Volume of each iron magnetic particle: 4 𝜋(1.5𝑥10−7 𝑐𝑚) = 1.4137 𝑥10−20 𝑐𝑚3 3 Mass of each iron particle: 𝑔 7.8 3 𝑥1.4137 𝑥10−20 𝑐𝑚3 = 1.102 𝑥 10−19 𝑔 𝑐𝑚 One mole or 56 g of iron contains, 6.023 𝑥 1023 𝑎𝑡𝑜𝑚𝑠, therefore The number of atoms in one nano particle is: 1.102 𝑥10−19 𝑔 𝑎𝑡𝑜𝑚𝑠 𝑥6.023 𝑥 1023 = 1186𝑎𝑡𝑜𝑚𝑠 56𝑔 𝑚𝑜𝑙𝑒 𝑚𝑜𝑙𝑒 QUESTION 3 The valance electron of silver is one, and only one electron is expected to conduct the electricity. Mass of silver: 𝑔 10𝑐𝑚3 𝑥10.49 3 = 104.9𝑔 𝑐𝑚 Number of atoms: 𝑎𝑡𝑜𝑚𝑠 104.9 𝑔 𝑥 6.023 𝑥 1023 𝑚𝑜𝑙𝑒 = 5.85 𝑥 1023 𝑎𝑡𝑜𝑚𝑠 𝑔 107.868 𝑚𝑜𝑙𝑒 Number of valance electron: 𝑣𝑎𝑙𝑎𝑛𝑐𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 5.85 𝑥 1023 𝑎𝑡𝑜𝑚𝑠 𝑥 1 = 5.85 𝑥 1023 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑎𝑡𝑜𝑚 QUESTION 4 The diffusion coefficient is given by: 𝐷 = 𝐷0 𝑒 −𝑄/𝑅𝑇 1.6𝑥108 𝑚2 = 7 𝑥 10−6 𝑒 −8314 𝑥 848 = 9.75 𝑥 10−16 𝑠 QUESTION 5 From Hook’s law: 𝜎 30,000𝑝𝑠𝑖 𝜀= = = 0.0003 𝐸 10𝑥106 𝑙 − 𝑙0 𝑜𝑟 𝑙 = 𝑙0 + 𝜀𝑙0 𝑙0 50 + 0.0003𝑥50 = 50.15𝑖𝑛

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QUESTION 6 % 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 =

𝑙𝑓 − 𝑙0 𝑥 100 𝑙0

2.195 − 2.000 𝑥 100 = 9.75% 2.000 𝐴0 − 𝐴𝑓 % 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑎𝑟𝑒𝑎 = 𝑥 100 𝐴0 𝜋 𝜋 2 2 4 (0.505) − 4 (0.398) 𝑥 100 = 37.9% 𝜋 2 4 (0.505) QUESTION 8 𝐾𝑙𝑐 = 𝑓𝜎√𝑎𝜋 80,000 = (1.12)(45,000)√𝑎𝜋 𝑎 = 0.80 QUESTION 9 The critical crack size (ac), using the fracture toughness and the maximum stress is: 𝐾𝑓𝑐 = 𝑓𝜎√𝜋𝑎𝑐 80𝑀𝑃𝑎 √𝑚 = (1)(500𝑀𝑃𝑎)√𝜋𝑎𝑐 𝑎𝑐 = 0.0081𝑚 𝑜𝑟 8.1 𝑚𝑚 QUESTION 13 At 1300C, P = 1 since only one phase (liquid) is present; C =2 since both copper and nickel atoms are present. Therefore, 1+𝐶 =𝐹+𝑃 1 + 2 = 𝐹 + 1 𝑜𝑟 𝐹 = 2 QUESTIO 14 At 1250C, two phases are present. The tie line drawn at this temperature shows that the liquid contains 32% Ni and the solid contains 45% Ni. QUESTION 15 Using the Lever rule: (%𝑁𝑖 𝑖𝑛 𝑎𝑙𝑙𝑜𝑦) − (% 𝑁𝑖 𝑖𝑛 𝐿) (% 𝑁𝑖 𝑖𝑛 𝛼) − (%𝑁𝑖 𝑖𝑛 𝐿) 40 − 32 = 0.62 45 − 32 If we convert mass fraction to mass percent, the alloy at 1250C contains, 62%α and 38% L 𝑥=

QUESTION 16 At 1270C, 40 − 37 𝑥 100 = 23% 50 − 37 50 − 40 %𝐿 = 𝑥 100 = 77% 50 − 37

%𝛼 =

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QUESTION 17 Moles of Zn: 𝑛𝑍𝑛 =

40𝑔 𝑔 = 0.612 𝑚𝑜𝑙𝑒𝑠 65.3 𝑚𝑜𝑙

Moles of Cu: 𝑛𝐶𝑢 =

60𝑔

𝑔 = 0.944 𝑚𝑜𝑙𝑒𝑠 63.456 𝑚𝑜𝑙 Total moles: 0.612 𝑚𝑜𝑙𝑒𝑠 + 0.944 𝑚𝑜𝑙𝑒𝑠 = 1.556 Mole fraction of Zn: 0.612 𝑚𝑜𝑙𝑒𝑠 = 0.39 𝑜𝑟 39% 0.944 𝑚𝑜𝑙𝑒𝑠 QUESTION 18 97.5 − 61.9 %𝛼 (𝑃𝑏 − 19%𝑆𝑛) = 𝑥 100 = 45.35% 97.5 − 19 61.9 − 19.0 %𝛽 (𝑃𝑏 − 97.5%𝑆𝑛) = = 54.65% 97.5 − 19.0 QUESTION 25 From the reaction, one mole of BaCO3 reacts with one mole of TiO2 to form one mole of BaTiO3 and one mole of CO2. Molecular weights: BaCO3 = 197 g/mol TiO2 = 80 g/mol BaTiO3 = 233 g/mol CO2 = 44 g/mol Since 233 g of BaTiO3 is produced using 197 g of BaCO3, for 1000 kg of BaTiO3, we need 845.5 kg of BaCO3. Similarly, 233g of BaTiO3 is produced using 80 g of TiO2, for 1000 kg of BaTiO3, 343.3 kg of TiO2 is needed. QUESTION 26 The porosity: 𝑊𝑤 − 𝑊𝑑 𝑥 100 𝑊𝑤 − 𝑊𝑠 385 − 360 𝑥 100 = 15. %% 385 − 224 QUESTION 28 For 100% efficiency, we need one molecule of benzyl peroxide for each polyethylene chain. One of the free radicals would initiate one chain, the other free radical a second chain then the two chains combine into one larger one. The Molecular Weight of ethylene = 28 g/mol. The degree of polymerization: 𝑔 200,000 𝑚𝑜𝑙 = 7143 𝑔 28 𝑚𝑜𝑙

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𝑚𝑜𝑛𝑜𝑚𝑒𝑟𝑠 1000 𝑔 𝑝𝑜𝑙𝑦𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒 𝑥 6.02 𝑥 1023 𝑚𝑜𝑙 = 215 𝑥 1023 𝑚𝑜𝑛𝑜𝑚𝑒𝑟𝑠 𝑔 28 𝑚𝑜𝑙 QUESTION 29 The molecular weight of repeat unit is the sum of the molecular weights of two monomers, minus that of two water molecules that are generated. 𝑔 𝑀𝑟𝑒𝑝𝑒𝑎𝑡 = 16 + 146 − 2(18) = 226 𝑚𝑜𝑙 Degree of polymerization: 120,000 = = 531 226 QUESTION 30 Number of chains Mean M per chain (Mi) 4000 2500 8000 7500 7000 12,500 2000 17,500 ∑=21,000

xi 0.191 0.381 0.333 0.095

xiMi 477.5 2857.5 4162.5 16625 ∑=9160

Note: xi =2500/21,000 = 0.191 QUESTION 35 The volume fraction: 𝑓𝑇ℎ𝑂2 =

2 9.69

= 0.0184 2 98 + 9.69 8.9 Therefore, there is 0.0184 cm3 of ThO2 per cm3 of composite. The volume of each ThO2 sphere is: 4 4 𝑉 = 𝜋𝑟 3 = 𝜋(0.5 𝑥10−5 𝑐𝑚)3 = 0.52 𝑥 10−15 𝑐𝑚3 3 3 Concentration of ThO2 particles: 0.0184 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 = = 35.4 𝑥1012 −15 3 0.52𝑥 10 𝑐𝑚 𝑐𝑚3 QUESTION 36 Convert weight percentage to volume fractions 75/15.77 𝑓𝑤𝑐 = = 0.547 75 15 5 5 + 4.94 + + 8.9 15.77 14.5 𝑓𝑇𝑖𝐶 = 0.349 𝑓𝑇𝑎𝑐 = 0.040 𝑓𝐶𝑜 = 0.064 From the rule of mixtures, the density of the composite: 𝜌𝑐 = Σ(𝑓𝑖 𝜌𝑖 ) = (0.547)(15.77) + (0.349)(4.94) + (0.040)(14.5) + (0.064)(8.9) 𝑔 = 11.50 3 𝑐𝑚 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 37 From the rule of mixtures, 𝜌𝑐 = 𝑓𝑤 𝜌𝑤 + 𝑓𝑝𝑜𝑟𝑒 𝜌𝑝𝑜𝑟𝑒 14.5 = 𝑓𝑤 (19.3) + 𝑓𝑝𝑜𝑟𝑒 (0) 𝑓𝑤 = 0.75; 𝑜𝑟𝑓𝑝𝑜𝑟𝑒 = 1 − 0.75 = 0.25 After infiltration, the volume fraction of silver equals the volume fraction of pores 𝑓𝐴𝑔 = 𝑓𝑝𝑜𝑟𝑒 = 0.25 0.25 𝑥 10.49 𝑤𝑡% 𝐴𝑔 = = 15.3% (0.25)(10.49) + (0.75)(19.3) QUESTION 38 The composite density: 𝑔 𝜌𝑐 = (0.35)(2.4) + (1 − 0.35)(0.95) = 1.4575 3 𝑐𝑚 QUESTION 39 The modulus of elasticity perpendicular to the fibers: 1 0.6 0.4 = + = 0.06727 𝑥 10−6 6 6 𝐸𝑐 10 𝑥10 55 𝑥10 𝐸𝑐 = 14.9 𝑥 106 𝑝𝑠𝑖 QUESTION 40 0.693 0.693 𝑡1/2 = = = 0.1733𝑑𝑎𝑦 −1 𝑘 4.3𝑑𝑎𝑦𝑠 𝑁 = 𝑒 −𝑘𝑡 𝑁0 0.1733 1 − 𝑥𝑡 = 𝑒 𝑑𝑎𝑦 𝑜𝑟 𝑡 = 29 𝑑𝑎𝑦𝑠 100 QUESTION 41 A 100-cm3 sample of wood would weight 86g 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 % 𝑤𝑎𝑡𝑒𝑟 = 𝑥 100 = 175 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑑𝑟𝑦 𝑤𝑜𝑜𝑑 100𝑥𝑔𝑟𝑒𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 𝐷𝑟𝑦 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑜𝑜𝑑 = 275 100 𝑥86 = = 31.3 𝑔 275 Density of wood: 31.3𝑔 𝑔 = 0.313 3 3 100𝑐𝑚 𝑐𝑚 QUESTION 44 For each sack of cement we use the volume of materials required as: 94𝑙𝑏 𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑠𝑎𝑐 = 0.495 𝑓𝑡 3 𝑙𝑏 190 3 𝑓𝑡 2.5 𝑥 94 𝑙𝑏 𝑐𝑒𝑚𝑒𝑛𝑡 𝑠𝑎𝑛𝑑 = = 1.469𝑓𝑡 3 𝑙𝑏 160 3 𝑓𝑡

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4 𝑥 94 𝑙𝑏𝑠 = 2.212𝑓𝑡 3 𝑙𝑏 170 3 𝑓𝑡 0.4 𝑥 94 𝑙𝑏 𝑐𝑒𝑚𝑒𝑛𝑡 𝑤𝑎𝑡𝑒𝑟 = = 0.603𝑓𝑡 3 𝑙𝑏 62.4 3 𝑓𝑡 Total volume of concrete = 4.779 ft3/sack of cement. QUESTION 45 If we count 16 grains per square inch at magnification x 250 then at magnification x 100 must have: 250 2 𝑔𝑟𝑎𝑖𝑛𝑠 𝑁=( ) 𝑥 16 = 100 = 2𝑛−1 100 𝑖𝑛2 log(100) = (𝑛 − 1) log 2 𝑛 = 7.64 QUESTION 46 At -100C 𝜌 = (1.67 𝑥 10−6 )(1 + 0.0068(−100 − 25)) = 0.251 𝑥 10−6 𝑜ℎ𝑚. 𝑐𝑚 1 1 𝜎= = = 39.5 𝑥 105 (𝑜ℎ𝑚. 𝑐𝑚)−1 𝜌 0.251𝑥 10−6 QUESTION 47 Heat required: = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑥 𝑚𝑎𝑠𝑠 𝑥 ∆𝑇 𝑐𝑎𝑙 = 0.032 𝑥 250 𝑔 𝑥 (650 − 25) = 5000 𝑐𝑎𝑙 𝑔𝐾 QUESTION 48 6 𝑐𝑎𝑙 𝐶𝑝 = = 0.0646 92.91 𝑔𝐶 The total heat required is: 𝑐𝑎𝑙 = 0.0646 𝑥 50 𝑔 𝑥 75𝐶 = 242 𝑐𝑎𝑙 𝑔𝐶 QUESTION 49 The change in any dimension is given by: ∆𝑙 = 𝑙0 − 𝑙𝑓 = 𝛼𝑙0 ∆𝑇 For the 25 cm dimension, lf = 25 cm 𝑙0 − 25 = (25 𝑥10−6 )(𝑙0 )(635)𝑜𝑟 𝑙0 = 25.40 𝑐𝑚 For 3-cm dimension: 𝑙0 − 3 = (25 𝑥10−6 )(𝑙0 )(635) 𝑜𝑟 𝑙0 = 3.05 𝑐𝑚 If we design the pattern to the dimension 25.40 cm x 25.40 cm x 3.05 cm, the casting should contract to the required dimensions 𝑔𝑟𝑎𝑣𝑒𝑙 =

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QUESTION 51 The concentration of copper when 1 g is added: 1𝑔 𝐶𝑖𝑜𝑛 = = 0.0157𝑀 63.54𝑔 From the Nerst’s equation, n =2, E0 = +0.34C 0.0592 𝐸 = 𝐸0 + log(𝐶𝑖𝑜𝑛 ) 𝑛 0.0592 = 0.34 + log(0.0157) = 0.29𝑉 2 QUESTION 53 Assuming that the sheet oxidizes from both sides: 𝑦 = √𝑘𝑡 = √3.9 𝑥10−12

𝑐𝑚2 𝑥 𝑡(𝑠) 𝑠

0.1𝑐𝑚 = 0.05𝑐𝑚 2𝑠𝑖𝑑𝑒𝑠 (0.05𝑐𝑚)2 𝑡= = 20.3 𝑦𝑒𝑎𝑟𝑠 𝑐𝑚2 −12 3.9𝑥10 𝑠 QUESTION 55 The number of mers: 50,000𝑔 = 800 62 𝑔/𝑚𝑜𝑙 QUESTION 61 Stress:

Strain:

𝜎 36𝐾𝑁 =𝜋 = 203.7 𝑀𝑃𝑎 2 𝐴 (0.015𝑐𝑚) 4 𝜀=

𝛿 0.02𝑚𝑚 = = 0.001 𝐿 200𝑚𝑚

Modulus of elasticity: 𝜎 203.7𝑀𝑃𝑎 = = 203 𝐺𝑃𝑎 𝜀 0.0001 From the tables, we can find that the material is steel. 𝐸=

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

444

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445

Fluid Mechanics Total Questions 9-14 A. B. C. D. E. F. G. H. I. J.

Fluid properties Fluid statics Energy, impulse, and momentum Internal flow External flow Incompressible flow Compressible flow Power and efficiency Performance curves Scaling laws for fans, pumps, and compressors

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446

QUESTION 1. A vessel is initially connected to a reservoir open to the atmosphere. The connecting valve is closed and a vacuum of 65.6 kPa is applied to the vessel. The absolute pressure (kPa) in the vessel is nearly: Assume standard atmospheric pressure A. 36 B. 66 C. 86 D. 110

QUESTION 2. The height (m) of the capillary rise for the following conditions is: Surface tension: 0.073 N/m Contact angle: zero Diameter of the capillary: 0.1mm A. B. C. D.

0.02 0.13 0.30 0.45

QUESTION 3. 100 g of water is mixed 150 g of fluid (density = 790 kg/m3). The specific gravity of the mixture assuming complete mixing is: A. B. C. D.

0.63 0.82 0.92 0.99

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447

QUESTION 4. The film width in a surface tension (0.52 N/m) experiment is 10 cm. The maximum force (N) that can be applied is nearly: A. B. C. D.

0.052 0.104 0.208 0.412

QUESTION 5. Oil with a specific gravity of 0.72 is used as the indicating fluid in a manometer. The difference in oil levels (m) if the differential pressure is 7 kPa is close to: A. B. C. D.

0.23 0.46 0.86 1.00

QUESTION 6. A pressure of 35 kPa is measured 4 m below the surface of an unknown liquid. The specific gravity is close to: A. B. C. D.

0.09 0.89 1.00 1.25

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448

QUESTION 7. The absolute pressure (MPa) at a depth of 3000 m of sea, if the SG of sea water is 1.15 is close to: A. B. C. D.

21 34 42 59

QUESTION 8. The height (m) of a mercury column equal to a pressure of 700 kPa is close to: Density of mercury: 13,500 kg/m3 A. B. C. D.

1.6 3.4 5.3 6.9

QUESTION 9. A fluid with a vapor pressure (VP) of 0.2 Pa and a SG of 1.2 is used in a manometer. If the fluid column length is 1 m, the atmospheric pressure is close to: A. 9.8 B. 75 C. 98 D. 118

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449

QUESTION 10. The resultant force (kN) on one side of a 25 cm diameter vertical plate standing at the bottom of a 3 m pool of water is close to: A. B. C. D.

1.38 1.63 1.90 2.76

QUESTION 11. The mass flow rate (kg/s) of a liquid (density = 0.690 g/cm3) flowing through a 5-cm diameter pipe at 8.3 m/s is nearly: A. B. C. D.

11 22 69 96

QUESTION 12. Water flows through a horizontal pipe at an entrance velocity of 3.0 m/s and exits at 2.1 m/s. Neglecting frictional losses, the pressure difference (kPa) between these two points is: A. B. C. D.

0.23 2.33 4.65 9.24

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450

QUESTION 13. The headloss (m) for water passing through a horizontal pipe, if the gage pressure at the entrance is 1.03 kPa and at the exit is 1.00 kPa, assuming the velocity is constant is: A. B. C. D.

1.1 x 10-3 3.1 x 10-3 2.3 x 10-2 3.2 x 10-2

QUESTION 14. The Reynolds number for the following data is: Temperature: 40C Diameter: 47 mm Kinematic viscosity: 6.58 x 10-7 m2/s Velocity: 1.5m/s A. B. C. D.

1.02 x 103 1.02 x 104 1.02 x 105 1.02 x 106

QUESTION 15. The pump power (kW) required to pump water flowing at 10m3/s with 5m head at 70% efficiency is nearly:

A. 70 B. 350 C. 700 D. 960

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451

QUESTION 16. The velocity of water in a stream using a pitot tube is 1.2 m/s. The height (cm) of water in the pitot tube is:

A. B. C. D.

3.7 4.6 7.3 9.2

QUESTION 17. A pitot tube measured the flow of a fluid (density = 926 kg/m3) to be 2 m/s with a stagnation pressure of 14.1kPa. The static pressure (kPa) of the fluid when the measurement was taken is close to: A. B. C. D.

10.4 11.7 12.2 13.5

QUESTION 18. A sharp edged orifice with a 50-mm diameter opening in the vertical side of a large tank discharges water under a head of 5m. If the coefficient of contraction is 0.62 and the coefficient of velocity is 0.98, the discharge (m3/s) is: A. B. C. D.

0.003 0.006 0.008 0.012

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452

QUESTION 19 A 2 m tall, 0.5 m diameter tank is filled with water. A 10 cm hole is opened at 0.75 m from the bottom of the tank. The velocity (m/s) of exiting water is nearly: Assume the contraction is equal to 1. A. B. C. D.

4.75 4.85 4.95 5.25

QUESTION 20. The hydraulic diameter of a rectangular conduit 0.6m wide and 0.3m high is: A. B. C. D.

0.1 0.2 0.3 0.4

QUESTION 21. A manometer containing mercury is connected to the pitot tube indicates a height of 150 mm. If the specific weight of mercury is 133,400 N/m3, the velocity of water (m/s) is close to: A. B. C. D.

2 4 6 8

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453

QUESTION 22. The pressure (kPa) at a depth of 100 m in fresh water is close to: A. 9.81 B. 98.1 C. 981 D. 1980

QUESTION 23. An open-topped cylindrical water tank has a horizontal circular base of 3m diameter. If the water is filled to a height of 2.5 m, the force (N) on the base is close to: A. 17,000 B. 172,000 C. 510,000 D. 680,000

QUESTION 24. A floating cylinder 8 cm in diameter and weighing 9.32 N is placed in a cylindrical container that is 20 cm in diameter which is partially filled with water. The increase in depth of water (cm) when the float is placed in it is: A. B. C. D.

2 3 5 10 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

454

QUESTION 25. The velocity of dwater (m/s) in a pipe of 1 cm diameter, if the flow is 10L/min is close to: A. B. C. D.

2.13 21.3 44.4 82.6

QUESTION 26. The energy data for water flowing at two sections across the pipe is given below: _______________________________________________ Section A Section B _______________________________________________ Potential energy 20 40 Kinetic energy 15 15 Flow energy 100 75 Total 135 130 _______________________________________________ The headloss (m) in the pipe is: A. 0 B. 5 C. 130 D. 265

QUESTION 27. The power (kW) required to pump water at 400 L/min from a reservoir to a tank 120 m high is close to: A. 4 B. 8 C. 15 D. 30

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455

QUESTION 28. The head corresponding to a fluid velocity of 10m/s is nearly: A. 5 B. 10 C. 15 D. 25

QUESTION 29. The theoretical velocity (m/s) generated by a 10m hydraulic head is: A. 3 B. 9 C. 14 D. 18

QUESTION 30. A stream of fluid flowing at 30 kg/s and a velocity of 6m/s to the right has its direction reversed 180o in a “U” fitting. The force (N) exerted by the fluid on the fitting is nearest to: A. 90 B. 180 C. 270 D. 360

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456

QUESTION 31. The thrust (N) exerted by an aircraft jet engine on takeoff for each 1 kg/s of exhaust products whose velocity has been increased from 0 to 150 m/s, is close to: A. 0 B. 15 C. 75 D. 150

QUESTION 32. A 60-cm diameter pipe carries a flow of 0.1m3/s. At point A the elevation is 50 meters and the pressure is 200 kPa. At point B downstream from A, the elevation is 40 meters and the pressure is 230 kPa. The headloss between the two points A and B is: A. B. C. D.

3.92 6.94 9.87 14.7

QUESTION 33. A rectangular tank has dimensions of 3m long by 1m wide and 2m high. The tank is filled with water to the top, which is open to the atmosphere. The force (N) on the 1-m side is most nearly: A. B. C. D.

10 20 30 40

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457

QUESTION 34. The water flow rate in a 15-cm diameter pipe is measured with a differential pressure gage connected with a static pressure tap in the pipe wall and a pitot tube located in the pipe centerline. If the differential pressure is 7 kPa, the flow rate is close to: A. B. C. D.

0.005 0.066 0.50 0.65

QUESTION 35. A 0.5 m x 0.2 m flat plate is towed at 5 m/s on a 2-mm thick layer of SAE-30 oil (viscosity = 0.1) at 38C that separates it from a flat surface. The velocity distribution between the plate and the surface is assumed to be linear. The force (N) required if the plate and the surface are horizontal is close to: A. B. C. D.

150 200 225 250

QUESTION 36. A car tire is pressurized in Ohio to 250 kPa when the temperature is -15C. The car is driven to Arizona where the temperature of the tire on the asphalt reaches 65C. The gage pressure in the tire in Arizona assuming no air has leaked out and the column remains constant is close to: A. B. C. D.

475 525 575 750

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458

QUESTION 37. A pressure of 230 kPa when converted to inches of mercury is close to: A. B. C. D.

13 26 52 68

QUESTION 38. A river flowing through a campus (20C) appears quite placid. A leaf floats by and the average velocity to be about 0.2 m/s. The depth is 0.6m. The Reynolds number is close to: A. 10,000 B. 50,000 C. 100,000 D. 120,000

QUESTION 39. A flow of 20C water in an 8-mm diameter pipe is desired. A 2L container used to catch the water, which was filled in 82 s. The Reynolds number is close to: A. B. C. D.

1000 4000 6000 8000

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459

QUESTION 40. A nozzle on a hose accelerates water from a 4-cm diameter to 1-cm diameter. If the pressure is 400 kPa upstream of the nozzle, the maximum velocity (m/s) exiting the nozzle is: A. B. C. D.

10 15 25 35

QUESTION 41. Water flows in a 6-cm diameter with a flow rate of 0.02m3/s. The pipe is reduced in diameter to 2.8 cm. The mass flux (kg/s) is nearly: A. 5 B. 10 C. 15 D. 20

QUESTION 42. Water flows from a reservoir with elevation 30m out of a 5-cm diameter pipe that has 2-cm diameter nozzle attached to the end. The loss coefficient for the entire pipe is given as K =1.2, the velocity through the pipe is: Assume the nozzle is at an elevation of 10m. A. B. C. D.

10 20 40 60

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460

QUESTION 43. A farmer decides to dam up the creek next to their cabin and estimates that a head of 4m can be established above the exit to a turbine. The creek is estimated to have a flow rate of 0.8m3/s. What is the maximum power output (kW) of the turbine assuming no losses and a velocity at the turbine exit of 3.6 m/s is close to: A. B. C. D.

16 26 44 85

QUESTION 44. Air at 40C and 250 kPa is flowing in a 32-cm diameter pipe at 10 m/s. The pipe changes to 20cm and the density of the air changes to 3.5 kg/m3. The velocity in the smaller diameter pipe is nearly: A. B. C. D.

28.5 32.7 45.6 81.3

QUESTION 45. A dam is proposed on a remote stream that measures approximately 25 cm deep by 350-cm wide with an average velocity of 2.2 m/s. If the dam can be constructed so that the free subsurface above a turbine is 10m, the maximum power output (kW) of an 88 percent efficient turbine is close to: A. B. C. D.

100 170 210 330

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461

QUESTION 46. A clever design of the front of a ship is to be tested in a water basin. A drag of 12.2 KN measured on the 1:20 scale model when towed at a speed of 3.6 m/s. The corresponding drag (N) that can be expected is close to: A. B. C. D.

41,000 43,000 45,000 54,000

QUESTION 47. A pressure drop of 500 kPa is measured over 200 m of horizontal length of 8-cm diameter cast iron pipe transporting water at 20C. The flow rate (m3/s) is close to: A. B. C. D.

0.015 0.020 0.035 0.055

QUESTION 48. Water at 20C is flowing in a 2m wide rectangular, brick channel (n = 0.016) at a depth of 120 cm. The slope is 0.0012. The flow rate (m3/s) is: A. B. C. D.

1.5 2.5 3.5 4.5

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462

QUESTION 49. Water at 15C is transported in a 6-cm diameter wrought iron pipe at a flow rate of 0.004m3/s. The pressure drop (kPa) over 300m of horizontal pipe is nearly: A. 55 B. 113 C. 143 D. 245

QUESTION 50. A trapezoidal channel with bottom width of 3m and side slope of 1V:1:5H carries a discharge of 8.0m3/s with the flow depth of l.5m. The Froude number of the flow is: A. B. C. D.

0.089 0.112 0.264 0.316

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463

PART-II QUESTION 1. A venturi meter (Cv = 0.98) is to be installed in a Sch.40 4-in pipe (ID = 10.22cm) to measure the flow of water. The maximum flow rate expected is 1,400L/min at a temperature of 20C. A vertical manometer with a useful length of 100 cm, filled with mercury (sp.gr. = 13.6) is available to measure the pressure drop across the venturi. The leads to the manometer are filled with water. If the ratio of the throat area to the pipe flow area is small, the required throat diameter (cm) of the venturi for the maximum flow is most nearly: A. B. C. D.

8.2 7.7 5.2 4.4

QUESTION 2. A 12-in diameter concrete sanitary sewer (n=0.013, constant with depth) flows half-full and is constructed on a grade of 0.5%. The flow velocity (ft/s) in this sewer is most nearly: A. 1.6 B. 2.0 C. 3.2 D. 32.4

QUESTION 3. Two tanks are connected by a 500-ft length of 1-in. ID PVC pipe. The appropriate value for the Hazen-Williams coefficient is 150. Water at 60F is flowing through the pipe at a velocity of 10ft/s. The tanks are open to the atmosphere. Entrance, exit and minor losses are negligible. The difference in water surface elevation (ft) between the two tanks is most nearly: A. 81 B. 167 C. 182 D. 447

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464

QUESTION 4. Waste activated sludge can be described as a Newtonian fluid with a kinematic viscosity of 20 x 10-5 ft2/s. At the same temperature, the kinematic viscosity of water is 10-5 ft2/s. The relative roughness of the piping system is 0.001. The pressure drop for flow of water at a Reynolds number of 107 in this piping system was determined to be 1.0psi. If the waste activated sludge flows at the same velocity through the piping system, the pressure drop (psi) is most nearly: A. B. C. D.

1.0 2.0 3.0 4.0

QUESTION 5. The Reynolds number for water flowing (170 oC) at 1.5m/s inside 3-m long tubes with an ID of 2.5 cm is most nearly: A. 53,000 B. 106,000 C. 212,000 D. 424,000

QUESTION 6. An 18-in diameter pipe in a potable water distribution system with a Hazen-Williams coefficient (C) equal to 120 is flowing at 8cfs. The pressure drop (psi) in a 1,000-ft length of pipe is most nearly: A. 0.84 B. 1.89 C. 4.27 D. 10.06

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465

QUESTION 7. A sanitary sewer has the following characteristics: Manning’s roughness coefficient (n) = 0.014 Slope (s) = 0.5% Diameter = 1m The discharge (m3/s) when the sewer is flowing full is most nearly: A. 1.6 B. 2.3 C. 2.5 D. 15.7

QUESTION 8. Two open tanks are connected by a single pipe. The water surface elevations of the upstream and downstream tank are 103.0 ft and 101.o ft, respectively. The pipe is 60 ft long and 6 inches in diameter. The Darcy friction factor (f) is 0.020. The pipes are connected to the tanks with a sharp entrance and sharp exit. There are no other fittings. The expected flow rate in the pipe (cfs) is most nearly: A. B. C. D.

1.13 1.44 1.71 5.75

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466

QUESTION 9. The swale shown below has a manning’s roughness coefficient of 0.02 and a slope of 0.5%. At a flow of 30cfs, the depth (in) of water in the swale is most nearly:

A. 6 B. 9 C. 12 D. 15

QUESTION 10. A proposed storm sewer will have a slope of 0.20%. The design flow for the line has been determined to be 16 cfs. Assume steady, uniform flow and a Manning’s roughness coefficient of 0.012 that is constant for all depths of flow. The minimum circular pipe size (in) that will accommodate the design flow is most nearly: A. B. C. D.

30 24 18 12

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467

QUESTION 11. A 2,000-ft long pipeline conveys water at 70F from a reservoir with a water elevation of 100 ft to a lower reservoir. The pipe is 12-in diameter non-riveted steel pipe with a fully open globe valve and a ¼ closed gate valve. Both the pipe entrance and exit are square edged. Assuming a flow of 7cfs and a friction factor of 0.015, the elevation of the lower reservoir (ft) is most nearly: A. B. C. D.

47.4 52.6 56.2 84.4

Typical Loss Coefficients Gate Valve

Globe valve Pipe exit Pipe entrance

Fully Open ¼ Closed ½ Closed ¾ Closed Fully open Square edged Square edged

0.19 1.15 5.6 24 10 1.0 0.50

QUESTION 12. The average velocity (fps) of a steady uniform flow in a 15-in diameter sewer line with a slope of 0.35% a depth of 3in and a Manning’s roughness coefficient of 0.012 is most nearly: A. B. C. D.

0.4 2.1 3.0 6.0

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468

QUESTION 13. A long concrete-lined drainage channel has a slope of 0.001ft/ft, a bottom width of 10 ft and 2:1 (H:V) side slopes. The water depth is 5ft, and the Manning’s roughness coefficient is 0.012. The Froude number for the channel is most nearly: A. B. C. D.

0.08 0.18 0.66 0.80

QUESTION 14. The discharge (cfs) through a 6-inch diameter square edged orifice discharging freely is close to: Water surface elevation of the impoundment = 220 feet Elevation of the center of the orifice = 200 feet

A. B. C. D.

2.2 4.4 6.6 9.0

QUESTION 15. The discharge (cfs) flowing over a 60o V-notch weir if H = 4 inches is close to: Assume C = 2.5 A. B. C. D.

0.001 0.01 1.0 10.0

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469

QUESTION 16. The discharge (cfs) flowing over a rectangular broad-crested weir with a breadth 9 inches, length 4 feet and head of 1.25 feet is close to: Assume C = 3.215 A. B. C. D.

12 14 18 22

QUESTION 17. The normal depth (feet) in a 10 foot wide concrete (n = 0.013) rectangular channel having a gradient of 0.0150 ft/ft and carrying a flow of 400 cfs is close to: A. B. C. D.

2.05 2.15 3.22 4.85

QUESTION 18. The discharge of a trapezoidal channel having a brick bottom and grassy sides for the following conditions is: Depth = 6 feet Bottom width = 18 feet Top width = 18 feet Slope = 0.002 Manning’s coefficient = 0.017 (for brick) Manning’s coefficient = 0.025 (for grassy sides)

A. B. C. D.

250 450 760 990

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

470

QUESTION 19. The water flow (gpm) through a venture meter with 5-1/16 inches throat diameter when measured pressure differential is 160 inches water is nearly: Assume C =0.95 and water temperature of 50F.

A. B. C. D.

1250 1400 1750 2150

QUESTION 20. A water treatment plant pumps its raw water from a reservoir next to the plant. The intake is 12 feet below the lake surface at elevation 588. The lake water is pumped to the plant influent at elevation 611. Assuming the suction head loss for the pump is 10 feet and the loss of head in the discharge line is 7 feet. The overall efficiency of the pump is 72 percent. The plant serves 44,000 people and the water consumption is 200 gallons per capita per day. The horsepower of the pump is close to:

A. 86 B. 100 C. 134 D. 176

QUESTION 21. Water flows in a 1000-m long pipeline of diameter 200 mm at a velocity of 5m/s. the pipeline is new cast iron pipe with ε = 0.00026m. The head loss (m) in the pipe is nearly: Assume ν = 1.007 x 10-6 m2/s A. 50 B. 78 C. 135 D. 252

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

471

QUESTION 22. A pump operating at 1800 rpm delivers 180gpm at 80 feet head. If the pump is operated at 2160 rpm, head is nearly: A. B. C. D.

115 125 105 240

QUESTION 23. The head loss in 1000 ft of 6-ft diameter smooth concrete pipe carrying 80 cfs of water at 80o F is most nearly: Assume C = 0.001 ft; f = 0.014

A. B. C. D.

0.10 0.15 0.21 0.31

QUESTION 24. A turbine is to be selected for an installation where the net head is 600 ft. and the permissible flow is 200 cfs. Calculate the power (hp) required for 90% efficiency.

A. 10,500 B. 11,800 C. 12,500 D. 14,800

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

472

QUESTION 25. A pressure drop of 500 kPa is measured over 200 m of a horizontal length of 8-cm diameter cast iron pipe transporting water at 20o C. The flow rate (m3/s) is most nearly: A. B. C. D.

0.02 0.05 0.09 0.15

QUESTION 26. A 1.5-cm diameter, 20-m long plastic pipe transports water from a pressurized 400-kPa tank out a free-open end located 3 m above the water surface in the tank. There are three elbows in the water line and a square-edged inlet from the tank. The flow rate (m3/s) is most nearly: A. 8.8 x 10-3 B. 10.4 x 10-4 C. 15.8 x 10-4 D. 8.8 x 10-4

QUESTION 27. Water at 15o C is transported in a 6-cm diameter wrought iron pipe at a flow rate of 0.004 m3/s. The pressure drop (kPa) over 300 m of horizontal pipe is nearly: A. 55 B. 115 C. 230 D. 315

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473

QUESTION 28. A pressure drop of 200 kPa is measured over a 400-m length of 8-cm diameter horizontal cast iron pipe that transports water at 20o C. The flow rate (m3/s) is close to: A. B. C. D.

0.0045 0.0088 0.0105 0.0210

QUESTION 29. A cast iron pipe (L = 400 m, D = 150 mm) is carrying 0.05 m3/s of water at 15o C. The loss due to friction (m) is most nearly: A. B. C. D.

13 18 26 34

QUESTION 30. A cast iron pipe (L = 400 m, D = 150 mm) is carrying water at 15o C. The difference in piezometric head is 20 m. Assume the minor loss coefficients amount to ΣK = 2.5 and f = 0.0025. The discharge (m3/s) is close to: A. 0.015 B. 0.025 C. 0.035 D. 0.045

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

474

QUESTION 31. A small community is building a water storage standpipe. The community wants to maintain a minimum system pressure of 30 psi and not exceed a maximum system pressure of 50 psi, while providing the largest useable storage volume for the standpipe. Assume 5 psi system losses and a 10-ft tank diameter. The high water level (ft) in the standpipe and the useable storage volume (gallons) respectively, are most nearly:

A. B. C. D.

115.0; 27,000 115.0; 67,500 127.0; 27,000 127.0; 74,300

QUESTION 32. A trapezoidal channel with bottom width of 3m and side slope of 1V:1:5H carries a discharge of 8.0 m3/ sec with the flow depth of l.5m. The Froude number of the flow is: E. F. G. H.

8.24 7.82 7.64 7.38

QUESTION 33. A very tiny sphere is settling down in a viscous fluid at Reynolds number = 0.2. What is the value of its drag coefficient? A. B. C. D.

320 120 80 5

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

475

QUESTION 34. What is the velocity (mm/s) of flow of oil µ=0.5 poise, density 800 kg/m3 through a pipe of diameter 9 mm and 12 m long when the head lost is 0.75 m? A. B. C. D.

25 35 30 40

QUESTION 35. A 0.20 m diameter pipe 20 km long transports oil at n flow rate of 0.01m3/s. The power (kW) required to maintain the flow if dynamic viscosity and density of oil are 0.08P and 900 kg/ m, respectively, will be A. B. C. D.

405 808 908 3.6

QUESTION 36. Water flows in a 1000-m long pipeline of diameter 200 mm at a velocity of 5m/s. the pipeline is new cast iron pipe with ks = 0.00026m. The head loss (m) in the pipe is nearly: Assume ν = 1.007 x 10-6 m2/s A. 50 B. 78 C. 135 D. 252

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

476

QUESTION 37. The discharge (cfs) through a 6-inch diameter square edged orifice discharging freely is close to:

Water surface elevation of the impoundment = 220 feet Elevation of the center of the orifice = 200 feet

A. B. C. D.

2.2 4.4 6.6 9.0

QUESTION 38. The discharge (cfs) flowing over a rectangular broad-crested weir with a breadth 9 inches, length 4 feet and head of 1.25 feet is close to: Assume C = 3.215 A. B. C. D.

12 14 18 22

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

477

QUESTION 39. In the figure below, the pipe is steel with an internal diameter of 100 mm. Water is pumped through the system; its velocity at Point C is 2.5 m/s. The pressure at Point A is atmospheric, the gage pressure at Point B is 125 kPa, and the gage pressure at Point C is 175 kPa. The discharge at Point D is to the atmosphere.

Viscoity = 1.0 x 10-3 N.s/m2 Kinematic viscosity = 1.0 x 10-6 m2/s Density = 1,000 kg/m3

The pumping rate (m3/min) is most nearly: A. B. C. D.

1.02 1.18 1.50 4.71

QUESTION 40. The pressure drop (Pa) across each elbow is most nearly: A. B. C. D.

1,100 2,800 3,100 5,600

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478

QUESTION 41. Ideally, the work that must be supplied to the pump (J/kg) is most nearly: A. 50 B. 125 C. 175 D. 50,000

QUESTION 42. The centrifugal pump shown in the figure is to deliver 40 kg/s of water from a condenser maintained at 10 kPa to a deaerating heater maintained at 200 kPa.

Preliminary design data are as follows: Elevation head on suction, referred to pump centerline Friction head loss in suction line Diameter of suction line Diameter of pump discharge line Elevation head on discharge into heater, referred to pump centerline Friction head loss in discharge line, including valves Elevation head at pump discharge, referred to pump centerline

5.0m 0.60m 15cm 10 cm 20 m 25m 0.5 m

The total head (m) at discharge of the pump is most nearly: A. B. C. D.

1.3 21 46 66

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QUESTION 43 A mercury (γ = 13.59γwater) barometer has a column height of 728.2 mm at 20C. The barometric pressure (kPa) is close to: A. 50 B. 100 C. 150 D. 200

QUESTION 44. What is the maximum discharge in L/s of glycerin at 30C and standard atmospheric pressure from a 20 mm id (internal diameter) tube if laminar flow is maintained? (ν = 0.0005 m2/s)

A. 5 B. 16 C. 21 D. 33

QUESTION 45. A model is used in the design of a spillway for a dam. For a model-to-prototype ratio is 1:10. What is the velocity at a point in the prototype when the velocity at the corresponding point in the model is 1.4 m/s? A. B. C. D.

1.33 2.33 3.33 4.33

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QUESTION 46. A flow meter model is1/6 the size of its prototype. The model is tested with 20C water (ν = 1.006 x 10-6 m2/s) while the prototype operates at 80C (ν =0.364x10-6 m2/s). For a velocity of 3.05m/s in the 0.3m diameter throat of the prototype, what discharge (m3/s) is required in the model for similitude? A. B. C. D.

0.001 0.01 0.1 1.0

QUESTION 47. A pitot tube is used with a mercury manometer to measure the flow of an inviscid fluid with SG =1 in a 4 inch diameter pipe. The flow rate (gpm) through the duct is nearly: A. B. C. D.

250 450 650 750

QUESTION 48. Gasoline with a specific gravity SG =0.82 flows with negligible loss through a divergent section at a rate of 0.028 m3/s. What is the pressure (N/m2) at station 2 if the pressure at section 1 is 1.8 x 105 N/m2? The gasoline temperature is at 20C. The diameters at point 1 and 2 are 0.10 m and 0.18m, respectively. A. B. C. D.

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QUESTION 49. A centrifugal air compressor receives ambient air at 1 atm pressure and 300K. At a discharge pressure of 4.0 atm, a temperature of 477K and a velocity of 90 m/s, what power is required to drive the compressor for a mass flow rate of 90 kg/min? The specific heat of air is 1012.9 J/kg.K A. B. C. D.

250 370 425 575

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SOLUTIONS QUESTION 1 For vacuum pressure: 𝑃𝑎𝑏𝑠 = 𝑃𝑠𝑡𝑑 − 𝑃𝑣𝑎𝑐𝑢𝑢𝑚 101.3𝐾𝑃𝑎 − 65.5 𝐾𝑃𝑎 = 36 𝐾𝑃𝑎 QUESTION 2 The height of the capillary rise is given by: 4𝜎𝑐𝑜𝑠𝛽 𝜌𝑔𝑑 4 𝑥0.073 𝑥 𝐶𝑜𝑠 0 = = 0.30𝑚 1000 𝑥 9.81 𝑥 0.0001 ℎ=

QUESTION 3

𝑚𝑎𝑠𝑠 𝑣𝑜𝑙𝑢𝑚𝑒 100𝑔 150𝑔 3 𝑉𝑤𝑎𝑡𝑒𝑟 + 𝑉𝑓𝑙𝑢𝑖𝑑 = + 𝑔 𝑔 = 289.87𝑐𝑚 1 3 0.79 3 𝑐𝑚 𝑐𝑚 𝑚𝑤𝑎𝑡𝑒𝑟 + 𝑚𝑓𝑙𝑢𝑖𝑑 (100 + 150)𝑔 𝑔 𝜌𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = = = 0.862 3 3 𝑉𝑤𝑎𝑡𝑒𝑟 + 𝑉𝑓𝑙𝑢𝑖𝑑 289.87𝑐𝑚 𝑐𝑚 𝑔 0.862 3 𝑐𝑚 = 0.862 𝑆𝐺 = 𝑔 1 3 𝑐𝑚 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =

QUESTION 4 𝐹 𝜎= 𝐿 𝑁 10𝑐𝑚 𝐹 = 2 𝑥 0.52 𝑥 = 0.1𝑁 𝑚 100 𝑐𝑚 𝑚 Note: 2 represent the two sides of the bubble QUESTION 5 The density of oil is: 𝜌𝑜𝑖𝑙 = (𝑆𝐺)𝜌𝑤𝑎𝑡𝑒𝑟 = 0.72 𝑥 1000

𝑘𝑔 𝑘𝑔 = 720 3 3 𝑚 𝑚

∆𝑝 = 𝜌𝑔ℎ 7000𝑃𝑎 ℎ= = 1𝑚 𝑘𝑔 𝑚 720 3 𝑥 9.81 2 𝑚 𝑠 QUESTION 6

𝑃 = 𝜌𝑔ℎ = (𝑆𝐺)𝜌𝑤𝑎𝑡𝑒𝑟 𝑥 𝑔 𝑥 ℎ 35,000 𝑃𝑎 = 0.89 𝑘𝑔 𝑚 1000 3 𝑥 9.81 2 𝑥 4𝑚 𝑚 𝑠

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QUESTION 7 𝑃 = 𝑃0 + 𝛾ℎ 𝑘𝑔 𝑚 101.3 𝐾𝑃𝑎 + (1.15) (1000 3 ) (9.81 2 ) (3000𝑚) = 3.39 𝑥10 7 𝑃𝑎 𝑜𝑟 34𝑀𝑝𝑎 𝑚 𝑠 QUESTION 8 𝑃 = 𝜌𝑔ℎ 𝑃 700 𝐾𝑃𝑎 ℎ= = = 5.3𝑚 𝑘𝑔 𝑚 𝜌𝑔 13,500 3 𝑥9.81 2 𝑚 𝑠 QUESTION 9 𝑃𝑎 = 𝑃𝑣 + 𝛾ℎ 𝑘𝑔 𝑚 0.2𝑃𝑎 + (12) (1000 3 ) (9.81 2 ) 𝑥 1𝑚 = 118,000 𝑃𝑎 𝑜𝑟 118 𝐾𝑃𝑎 𝑚 𝑠 QUESTION 10 The resultant force is calculated from the average pressure on the plate which is the pressure at the plate’s centroid. 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 = 3𝑚 − 0.125𝑚 = 2.875𝑚 𝑃 = 𝜌𝑔ℎ𝑐 𝑘𝑔 𝑚 = 1000 3 𝑥9.81 2 𝑥2.875 𝑚 = 28204 𝑃𝑎 𝑚 𝑠 𝑅 = 𝑃𝑥𝐴 = 28204 𝑃𝑎 𝑥 𝜋 𝑥 (0.125𝑚)2 = 1.38𝐾𝑁 QUESTION 11 The mass flow rate is given by: 𝑚̇ = 𝜌𝐴𝑉 𝑔 𝜋 𝑐𝑚 𝑔 𝐾𝑔 0.690 3 𝑥 (5)2 𝑥 8.3 𝑚 𝑥 100 = 11,245 𝑜𝑟 11 𝑐𝑚 4 1𝑚 𝑠 𝑠 QUESTION 12 From Bernoulli’s equation: 𝑃2 𝑣22 𝑃1 𝑣12 + + 𝑧2 + ℎ𝑓 = + + 𝑧1 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔 Since the pipe is horizontal, z1 and z2 are equal. [𝑣12 − 𝑣22 ] 𝜌 [𝑣12 − 𝑣22 ] 𝑃2 − 𝑃1 = 𝜌𝑔 = [ ] 2𝑔 2 2 𝑘𝑔 [32 − 2.12 ] 1000 3 = 2300 𝑃𝑎 𝑜𝑟 2.3𝐾𝑃𝑎 𝑚 2 QUESTION 13 Using Bernoulli’s’ equation: 𝑃2 𝑣22 𝑃1 𝑣12 + + 𝑧2 = + + 𝑧1 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔 Since, z1 and z2 are equal, v1 and v2 are equal. 𝑃1 − 𝑃2 [1003 − 1000]𝑃𝑎 ℎ𝑓 = = = 3.1 𝑥10−3 𝑚 𝑘𝑔 𝑚 𝜌𝑔 1000 3 𝑥9.81 2 𝑚 𝑠

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QUESTION 14 𝐷𝑉𝜌 𝐷𝑉 𝑜𝑟 𝜇 𝜗 𝑚 1.5 𝑠 𝑥 0.0447𝑚 = 1.02 𝑥105 𝑚2 −7 6.58 𝑥 10 2 𝑅𝑒 =

QUESTION 15 The pump power is given by: 𝑘𝑔 𝑚3 𝑚 10 𝑄𝜌𝑔ℎ 𝑠 𝑥1000 𝑚3 𝑥9.81 𝑠 2 𝑥 5𝑚 𝑤= = = 700 𝐾𝑊 𝑒 0.70 QUESTION 16 The equation for pitot tube is: 2(𝑃0 − 𝑃𝑠 ) 2𝜌𝑔ℎ 𝑣= √ =√ = √2𝑔ℎ 𝜌 𝜌 ℎ=

𝑣2 (1.2𝑚)2 = 𝑚 = 7.3 𝑐𝑚 2𝑔 2 𝑥9.81 2 𝑠

QUESTION 17 The equation for pitot tube is: 𝑣= √

2(𝑃0 − 𝑃𝑠 ) 𝜌

𝑘𝑔 𝑚 2 926 3 𝑥 (2 𝑠 ) 𝜌𝑣 2 𝑚 𝑃𝑠 = 𝑃0 − = 14.1 − = 12.2 𝐾𝑃𝑎 𝑃𝑎 2 2 𝑥 1000 𝐾𝑃𝑎 QUESTION 18 𝑄 = 𝐶𝐴√2𝑔ℎ 𝜋 𝐴 = (0.05)2 = 0.00196 𝑚2 4 𝐶 = 𝐶𝑐 𝐶𝑣 = 0.62 𝑥 0.98 = 0.6076 𝑄 = 0.6076 𝑥 0.00196√2𝑥9.81𝑥5 = 0.012

𝑚3 𝑠

QUESTION 19 The hydraulic head at the hole is: 2𝑚 − 0.75𝑚 = 1.25𝑚 𝑣 = √2𝑔ℎ = √2𝑥9.81 𝑥 1.25 = 4.95

𝑚 𝑠

QUESTION 20 𝐴 = 0.6 𝑚 𝑥 0.3𝑚 = 0.18 𝑚2 𝑊𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 2[0.6 + 0.3] = 1.8𝑚 𝐴 0.18𝑚2 𝐷𝐻 = 4 = = 0.4𝑚 𝑊𝑃 1.8𝑚

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QUESTION 21 𝛾𝑚 133,400 𝑚 − 1] = √2𝑥9.81 𝑥0.15 𝑥[ − 1] = 6.09 𝛾 9810 𝑠

𝑣 = √2𝑔ℎ[

QUESTION 22 𝑃 = 𝜌𝑔ℎ = 𝛾ℎ 9810 𝑥 100𝑚 = 981,000 𝑃𝑎 𝑜𝑟 981 𝐾𝑃𝑎 QUESTION 23 The pressure at the base of the tank:

Force on the tank base:

𝑁 𝑃 = 𝛾ℎ = 9810 𝑥 2.5𝑚 = 24325 2 𝑚 𝜋 2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 𝑏𝑎𝑠𝑒 = 𝑥(3𝑚) = 7.07𝑚2 4 𝐹 = 𝑃𝐴 = 24325

𝑁 𝑥 7.07𝑚2 = 171,978 𝑁 𝑚2

QUESTION 24 𝑊 9.32𝑁 = 0.00095 𝑚3 = 950 𝑐𝑚3 𝛾 9810 𝑁/𝑚3 The change in total volume ∆𝑉 beneath the water surface equals the area of the cylindrical container A x the change in water level dh or 𝑑𝑉 𝑑𝑉 = 𝐴 𝑥 𝑑ℎ; 𝑑ℎ = 𝐴 950 𝑐𝑚3 𝑑ℎ = 𝜋 = 3.02 𝑐𝑚 2 𝑥 20 𝑐𝑚 4 𝑉𝑏 =

QUESTION 25 𝑄 = 𝐴𝑣 𝐿 1 𝑚3 1𝑚𝑖𝑛 𝑚3 −6 𝑄 = 10 𝑥 𝑥 = 167 𝑥 10 𝑚𝑖𝑛 100𝐿 60𝑠 𝑠 𝜋 2 −6 2 𝐴 = 𝑥(0.01𝑚) = 78.5 𝑥 10 𝑚 4 3 −6 𝑚 𝑄 167 𝑥 10 𝑠 = 2.13 𝑚 𝑣= = −6 𝐴 78.5 𝑥 10 𝑚2 𝑠 QUESTION 26 Applying the energy balance: 𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡 + 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑠 = 𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑝𝑢𝑡𝑠 135 = 130 + ℎ𝐿 − 0 𝑜𝑟 ℎ𝐿 = 5𝑚

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QUESTION 27 Power required: 𝑄𝜌𝑔ℎ 𝑒 𝐿 1 𝑚3 1𝑚𝑖𝑛 𝑚3 𝑄 = 400 𝑥 𝑥 = 6.67 𝑥10−3 𝑚𝑖𝑛 100𝐿 60𝑠 𝑠 3 𝑚 𝑤 = 9810 𝑥 6.67 𝑥10−3 𝑥 120 𝑚 = 7.85𝐾𝑊 𝑠 𝑤=

QUESTION 28 The head is: 𝑣 2 (10𝑚)2 ℎ= = = 5.10𝑚 2𝑔 2𝑥 9.81 QUESTION 29 𝑣2 𝑜𝑟 𝑣 = √2𝑔ℎ 2𝑔 𝑚 𝑣 = √2𝑥9.81𝑥10 = 14 𝑠 ℎ=

QUESTION 30 The impulse-momentum equation: 𝐹 = 𝑚[𝑣2 − 𝑣1 ] If the pressure is zero at each end of the tube, then 𝑘𝑔 𝑚 𝑘𝑔. 𝑚 [−6 − 6] = 360 𝐹 = 30 = 360𝑁 𝑠 𝑠 𝑠2 QUESTION 31 The impulse momentum equation is: 𝐹 = 𝑚[𝑣2 − 𝑣1 ] 𝑘𝑔 𝑚 [150 − 0] = 150 𝑁 𝐹=1 𝑠 𝑠 QUESTION 32 Using the Bernoulli’s equation: 𝑃2 𝑣22 𝑃1 𝑣12 + + 𝑧2 + ℎ𝑓 = + + 𝑧1 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔 The pipe diameter is unchanged; continuity equation requires that velocity be the same at both points. The equation reduces to: 𝑃2 𝑃1 + 𝑧2 + ℎ𝑓 = + 𝑧1 𝜌𝑔 𝜌𝑔 [200 − 230] ℎ𝑓 = + 50 − 40 = 6.94 𝑚 9.81 QUESTION 33 𝐹 = 𝛾ℎ𝐴 𝐾𝑁 9.81 3 𝑥1𝑚 (1𝑚𝑥2𝑚) = 19.6 𝐾𝑁 𝑚

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QUESTION 34 The equation for pitot tube is: 2𝑔∆𝑃 7000 𝑚 𝑉= √ = √2𝑥9.81 𝑥 = 3.74 𝛾 9810 𝑠 𝑄 = 𝐴𝑣 =

𝜋 𝑚 𝑚3 (0.15𝑚)2 𝑥 3.74 = 0.0661 4 𝑠 𝑠

QUESTION 35 The velocity gradient is calculated as: 𝑑𝑣 5−0 = = 2500 𝑑𝑦 0.002 The force is the stress multiplied by the area: 𝑑𝑣 𝐹 = 𝜏𝑥𝐴= 𝜇 𝑥𝐴 𝑑𝑦 = 0.1 𝑥 2500 𝑥 0.5 𝑥 2 = 250 𝑁 QUESTION 36 Assuming the volume does not change: 𝑃2 𝑇2 = 𝑃1 𝑇1 423 𝑃2 = (250 + 100) = 574 𝐾𝑃𝑎 258 QUESTION 37 𝑃 = 𝛾ℎ 230,000 = (13.6 𝑥 9800)ℎ ℎ = 1.726 𝑚 𝑓𝑡 𝑖𝑛 = 1.726 𝑥 3.281 𝑥 12 = 68 𝑖𝑛𝑐ℎ𝑒𝑠 𝑜𝑓 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑚 𝑓𝑡 QUESTION 38 𝑅𝑒 =

𝑣ℎ 0.2 𝑥 0.6 = = 120,000 𝜗 10−6

QUESTION 39 𝑄 2𝑥10−3 = = 0.485 𝐴 𝜋(0.004)2 𝑥 82 0.485 𝑥 0.008 𝑥 0.5 𝑅𝑒 = = 3880 10−6 𝑣=

QUESTION 40 The continuity equation relates the velocities as: 𝐴1 𝑣1 = 𝐴2 𝑣2 𝜋(2)2 𝑣1 = 𝜋(0.5)2 𝑣2 𝑣2 = 16𝑣1 Using the Bernoulli’s equation: 𝑣12 400,000 256𝑣12 100,000 + + 𝑔ℎ1 = + + 𝑔ℎ2 2 1000 2 1000

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Solving for v1: 𝑣1 = 1.534

𝑚 𝑚 𝑎𝑛𝑑 𝑣2 = 24.5 𝑠 𝑠

QUESTION 41 The mass flux is: 𝑚 = 𝜌𝑄 = 0.02

𝑚3 𝑘𝑔 𝑘𝑔 𝑥 1000 3 = 20 𝑠 𝑚 𝑠

QUESTION 42 The energy equation is written in the form: 𝑊 𝑣22 𝑃2 𝑣12 𝑃1 𝑣2 = + 𝑧2 + − − 𝑧1 − + 𝑘 𝑚𝑔 2𝑔 𝛾2 2𝑔 𝛾1 2𝑔 Where the pressure is zero at the surface. The velocities are related as: 𝐴2 𝑑22 4 𝑣= 𝑣2 𝑜𝑟 2 𝑣2 = 𝑣 𝐴 𝑑 25 2 𝑣22 4 2 𝑣22 0= + 10 − 30 + 1.2 ( ) 2𝑔 25 2𝑔 𝑚 𝑣2 = 19.5 𝑠 QUESTION 43 The mass flow rate: 𝑘𝑔 𝑚 = 𝜌𝑄 = 1000 𝑥 0.8 = 800 𝑠 𝑚𝑣12 𝑤 = 𝑚𝑔𝑧 − 2 (3.6)2 = 800 𝑥 9.81 𝑥 4 − 800 = 26.2 𝐾𝑊 2 QUESTION 44 The continuity equation 𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2 𝑃1 𝜋 2 𝜋 𝑥 𝑑1 𝑣1 = 𝜌2 𝑑22 𝑣2 𝑅𝑇1 4 4 Calculating for v2: (0.32)2 𝑥350 𝑚 𝑥 10 = 28.5 3.5 𝑥 (0.2)2 𝑥 0.287 𝑥 313 𝑠 QUESTION 45 The flow rate of water passing through the turbine: 𝑚3 𝑄 = 𝐴𝑣 𝑜𝑟 0.25 𝑥 3.5 2.2 = 1.925 𝑠 The energy equation is applied between the surface of the reservoir behind the dam, where P1 =0, v1 = 0; z1 = 0, z1 = 10m, v2 = 0, P2 = 0, and z2 = 0 𝑤 = 𝑚𝑔ℎ = 1000 𝑥 1.925 𝑥 9.981 𝑥 10 = 189,000 𝑊 The maximum turbine output is: 𝑤 = 0.88 𝑥 189 𝐾𝑊 = 166 𝐾𝑊

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QUESTION 46 The Froude number guides the model study of ship, since gravity effects are more significant than viscous effects. 𝑣𝑝 𝑣𝑚 = √𝑙𝑝 𝑔𝑝 √𝑙𝑚 𝑔𝑚 Where, p = prototype and m = model 𝑙𝑝 𝑚 𝑣𝑝 = 𝑣𝑚 √ = 3.6 𝑥 √20 = 16.1 𝑙𝑚 𝑠 QUESTION 47 The relative roughness (from Moody’s diagram) 𝑒 0.26 = = 0.00325 𝐷 80 From Moody’s friction diagram: f = 0.026 The headloss is given by: ∆𝑃 500,000 ℎ𝑓 = = = 51𝑚 𝛾 9810 The average velocity is: 2𝑔𝐷ℎ𝑓 2𝑥9.81𝑥0.08𝑥51 𝑚 𝑣= √ = √ = 3.92 𝑓𝐿 0.026𝑥200 𝑠 The volumetric flow rate: 𝑚 𝑚3 𝑄 = 𝐴 𝑣 = 𝜋(0.04) 𝑥3.92 = 0.0197 𝑠 𝑠 2

QUESTION 48 The hydraulic radius is: 𝑅=

𝐴 𝑏𝑦 2𝑥1.2 = = = 0.545𝑚 𝑊𝑃 𝑏 + 2𝑦 2 + 2𝑥1.2

Using the Manning’s equation: 𝑄= =

1 2/3 1/2 𝐴𝑅 𝑆 𝑛

1 𝑚3 𝑥(2𝑥1.2)𝑥(0.545)2/3 𝑥(0.0012)1/2 = 3.47 0.016 𝑠

QUESTION 49 The average velocity and Reynolds number are: 𝑄 0.004 𝑚 𝑣= = = 1.415 2 𝐴 𝜋(0.03) 𝑠 𝑚 1.415 𝑠 𝑥 0.06 𝑅𝑒 = = 7.44 𝑥 104 1.14 𝑥10−6 From Moody’s diagram: 𝑒 0.046 = = 0.00077 𝐷 60 The friction factor f = 0.0225 The pressure drop:

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𝐿 𝑣2 ∆𝑃 = 𝛾ℎ𝐿 = 𝜌𝑓 𝐷 2 300 (1.415)2 = 1000 𝑥 0.0225 𝑥 𝑥 = 113,000 𝑃𝑎 𝑜𝑟 113 𝐾𝑃𝑎 0.6 2 QUESTION 50 Q = 8 m3/s 1 [3 + 7.5]𝑥 1.5 = 7.8752 𝑚2 2 𝑄 8 𝑚 𝑣= = = 1.015 𝐴 7.8752 𝑠 1 1.5 = 𝑜𝑟 𝑥 = 2.25 𝑚 1.5 𝑥 𝑇 = 3 + 2 𝑥 2.25 = 7.5 𝑚 𝐴 7.875 𝐷= = = 1.05 𝑇 7.5 𝑣 1.05 𝐹𝑟 = = = 0.316 √𝑔𝐷 √9.81 𝑥 1.05 𝐴=

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PART-II QUESTION 1. For a venturi, the relationship between flow and pressure drop is given under Venturi Meters in the Fluid Mechanics section of the FE Reference Handbook.

𝑄=

𝑃 𝑃 𝐶𝑣 𝐴2 √2𝑔 ( 𝛾1 + 𝑍1 − 𝛾2 − 𝑍2 ) 2 √1 − (𝐴2 ) 𝐴 1

Where, Point 2 is at the throat of the venturi. 𝐴

2

Assume √1 − (𝐴2 ) is approximately unity. 1

Z2 = Z1 𝜋

A2 = 4 𝐷22 𝑄=

𝜋 𝐶𝑣 𝑥 4 𝐷22 √2(𝑃1 − 𝑃2 )/𝛾 4

For a manometer measuring the pressure difference between Points 1 and Points 2, the equation from Manometry in the Fluid mechanics section of the FE Reference Handbook can be used. 𝑃1 − 𝑃2 = ℎ(𝛾2 − 𝛾1 ) = ℎ𝛾1 (𝑆𝐺𝑚𝑓 − 1) 𝑃1 − 𝑃2 980.7𝑐𝑚 = 𝑔ℎ((𝑆𝐺𝑚𝑓 − 1) = ( ) (100𝑐𝑚)(13.6 − 1) = 1.236 𝑥 106 𝑐𝑚2 /𝑠 2 𝛾 𝑠2 Solving the venturi equation for D22 gives: 𝐷22 =

4 𝑥 1,400 𝐿/𝑚𝑖𝑛 𝜋(0.98)√(2)(1.236 𝑥 106

𝑐𝑚2 ) 𝑠2

(

1,000𝑐𝑚3 1 𝑚𝑖𝑛 )( ) = 19.3 𝑐𝑚2 𝐿 60𝑠

𝐷2 = 4.4 𝑐𝑚

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QUESTION 2. 𝑉𝑓𝑢𝑙𝑙 =

1.486 2/3 1/2 𝑅 𝑆 0.013

Where, 𝑅 =

𝐷 4

=

1.0 4

= 0.25 𝑓𝑡 𝑉𝑓𝑢𝑙𝑙 =

2 1 1.486 𝑓𝑡 (0.25)3 (0.005)2 = 3.208 0.013 𝑠

From the Hydraulic-elements graph for Circular sewers in the Civil Engineering section of the FE Supplied-Reference Handbook, the velocity in a pipe flowing half-full is the same as for a pipe flowing full.

QUESTION 3. Use Hazen-Williams equation. 𝑉 = (1.318)𝐶𝑅 0.63 𝑆 0.54 𝑆=[

1 𝑉 ]0.54 0.63 1.318𝐶𝑅

= 𝑆𝐿 = (500 𝑓𝑡)[

10

1

0.54 0.63 ]

1 12 1.318(150) ( 4 )

= 182 𝑓𝑡 QUESTION 4. 𝐿 𝑉2 𝐻𝐿 = 𝑓 ( )( ) 𝐷 2𝑔 For the piping system and velocity, HL is proportional to f. For this case, the Reynolds number is very large, corresponding to fully turbulent flow conditions. Referring to the Moody’s Stanton diagram in the Fluid Mechanics section of the FE Supplied-Reference Handbook, it is apparent that f depends on the pipe roughness but not Reynolds number for e/D = 0.001, and Reynolds number >106. Thus, f = 0.002 for both applications, and pressure drops are equal. FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 5. The Reynolds number is found in the Fluid Mechanics section of the FE Supplied-Reference Handbook. 𝑅𝑒 =

𝐷𝑣𝜌 𝜇

V = 1.5 m/s D = 2.5 x10-2 m ρ = 898 kg/m3 ν = 1.59 x 10-4 N.s/m2 𝑘𝑔 1.5𝑚 ( 𝑠 ) (2.5 𝑥 10−2 𝑚) (898 3 ) 𝑚 𝑅𝑒 = = 211,792 𝑁. 𝑠 −4 1.59 𝑥 10 ( 2 ) 𝑚

QUESTION 6. 𝑉 = 𝐾1 𝐶𝑅 0.63 𝑆 0.54 𝑉=

𝑄 = 𝐴

8𝑐𝑓𝑠 = 4.527 𝑓𝑝𝑠 18𝑖𝑛 2 ( ) 𝜋 12𝑖𝑛 4

C = 120 K1 = 1.318 (English units) R=

𝐷 4

=

18𝑖𝑛 ) 12𝑖𝑛

(

4

= 0.375 𝑓𝑡 𝑆=

𝐻𝐿 𝐻𝐿 = 𝐿 1000

𝐻𝐿 = 1,000[ = 1,000[

1 𝑉 0.54 ] 𝐾1 𝐶𝑅 0.63

1 4.527 0.54 = 4.355 𝑓𝑡 ] (1.318)(120)(0.375)0.63

62.4𝑙𝑏 ) 𝑓𝑡 3 = 1.89 𝑝𝑠𝑖 144 𝑖𝑛2 /𝑓𝑡 2

(4.355𝑓𝑡)( ∆𝑃 = 𝐻𝐿 𝛾 =

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QUESTION 7. 𝑄=

𝑘 𝐴𝑅 2/3 𝑆 1/2 𝑛

k = 1 (metric units) n = 0.014 𝜋𝐷 2

𝜋(1𝑚)2

= 4 = 0.785𝑚2 4 𝐷 1𝑚 𝑅= = = 0.25𝑚 4 4 S = 0.5% = 0.005 A=

𝑄=

2 1 1 (0.785)(0.25)3 (0.005)2 = 1.57𝑚3 /𝑠 0.014

QUESTION 8. Energy equation: 𝑃1 𝑉12 𝑃2 𝑉22 + 𝑧1 + + ℎ𝑚 = + 𝑧2 + + ℎ𝐿 𝛾 2𝑔 𝛾 2𝑔 𝑃1 𝑃2 𝑉12 𝑉22 = ; = = 0, ℎ𝑚 = 0 (𝑛𝑜 𝑝𝑢𝑚𝑝) 𝛾 𝛾 2𝑔 2𝑔 𝐿 𝑉2 𝑉2 ℎ𝐿 = 𝑓 + ∑𝑘 𝐷 2𝑔 2𝑔 Therefore, 𝑧1 − 𝑧2 = [𝑓

𝐿 𝑉2 + ∑ 𝑘] 𝐷 2𝑔

Entrance k = 0.5 Exit k = 1.0 f =0.020 L = 60 ft D = 0.5 ft g = 32.2 ft/s2 z1 = 103.0ft z2 = 101.0 ft

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1 𝑧1 − 𝑧2 𝑉 = [( ) 2𝑔]2 𝐿 𝑓𝐷 +∑𝑘 1 2

= [(

103.0 − 101.0 𝑓𝑡 ) 2(32.2)] = 5.75 60 𝑠 (0.02) ( ) + 1.5 0.5

𝑄 = 𝑉𝐴 = 5.75

𝑓𝑡 (0.5𝑓𝑡)2 𝑥𝜋𝑥 = 1.128𝑐𝑓𝑠 𝑠 4

QUESTION 9. The depth may be found using a trial and error method to solve Manning’s equation (below) for the depth y. 𝑄=

1.486 2/3 1/2 𝐴𝑅 𝑆 𝑛

Where, A = area R = hydraulic radius S = slope = 0.5% = 0.005 n = 0.02 𝐴 = (𝑏 + 𝑧𝑦)𝑦 Where, b = the bottom width and z = 3 Since the side slopes are 3:1 (H:V) 𝑅=

𝑄=

(𝑏 + 𝑧𝑦)𝑦 𝑏 + 2𝑦√1 + 𝑧 2

1.486 𝐴 𝑅 2/3 (0.005)1/2 = 30 0.02

Trial Depth (in) Depth(ft) R (ft) A (ft2) Q (cfs) 1 12 1.00 0.601 5.00 18.7 2 15 1.25 0.726 7.188 30.5 The depth is most nearly 15 inches.

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QUESTION 10. Solve Manning’s equation for the diameter D. 2.159 𝑥 𝑄 𝑥 𝑛 3/8 𝐷= [ ] 𝑠1/2 Q = 16cfs n = 0.012 s = 0.20% = 0.002 𝐷=[

2.159 𝑥 16 𝑥 0.012 1 (0.002)2

3 8

] = 2.30 𝑓𝑡 𝑥

12𝑖𝑛 = 27.65𝑖𝑛 𝑓𝑡

QUESTION 11. Determine the combined minor loss coefficient for the gate valve, the globe valve, the pipe entrance and the pipe exit. 𝐾 = 1.15 + 10 + 1.0 + 0.5 = 12.65 𝑉2 ℎ𝐿 = 𝐾 2𝑔 Where, hL = headloss K = minor loss coefficient V = velocity, Q/A g = acceleration due to gravity Q = flow rate A = area 𝑄 2 (𝐴 ) 7 1 ℎ𝐿 = 12.65 𝑥 = 12.65 𝑥( )2 = 15.6𝑓𝑡 2 2𝑥32.2 3.1𝑥0.5 2𝑥32.2 Find the headloss in the pipe using Darcy-Weisbach equation. 𝐿 𝑉2 ℎ𝐿 = 𝑓 𝐷 2𝑔 2,000𝑥79.516 ℎ𝐿 = 0.015 𝑥 = 37𝑓𝑡 (2𝑥1𝑥32.2) Total headloss = 37 + 15.6 = 2.6𝑓𝑡 The water elevation of the lower reservoir is equal to the water elevation of the upper reservoir minus the total headloss which is: 100 ft-52.6ft =47.4 ft

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QUESTION 12. 𝑉=

1.486 2/3 1/2 𝑅 𝑆 𝑛

Where, n = 0.012 S = 0.35% = 0.0035 ft/ft R = 0.151 ft (hydraulic radius) 1.486 𝑉= (0.151)2/3 (0.0035)1/2 = 2.08𝑓𝑝𝑠 0.012

QUESTION 13. Use Manning’s equation to calculate the velocity of flow, and then calculate the Froude number. Manning’s equation: 𝑉=

1.486 2/3 1/2 𝑅 𝑆 𝑛

By geometry of trapezoid:

1 𝐴𝑟𝑒𝑎 = 𝐴 = (𝑏𝑜𝑡𝑡𝑜𝑚 𝑤𝑖𝑑𝑡ℎ + 𝑡𝑜𝑝 𝑤𝑖𝑑𝑡ℎ)𝑥 ℎ𝑒𝑖𝑔ℎ𝑡 2 =

1 (10 + 30)𝑥5 = 100 𝑓𝑡 2 2

𝑊𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 𝑃 = 11.18 + 10 + 11.18 = 32.36 𝑓𝑡 𝑅=

𝐴 100 = = 3.09𝑓𝑡 𝑃 32.36

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𝑉=

2 1 1.486 (3.09)3 (0.001)2 = 8.31 𝑓𝑝𝑠 0.012

Hydraulic depth: 𝐷ℎ =

𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 100𝑓𝑡 2 = = 3.33𝑓𝑡 𝑡𝑜𝑝 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑓𝑙𝑜𝑤 30𝑓𝑡

Froude number: 𝐹𝑟 =

𝑉 √𝑔𝐷ℎ

Where, V = velocity g = acceleration due to gravity Dh = hydraulic depth

𝐹𝑟 =

8.31𝑓𝑝𝑠 √32.2 𝑓𝑡2 𝑥 3.33 𝑓𝑡 𝑠

= 0.80

QUESTION 14. The cross sectional area: 𝐴 = 𝜋𝑟 2 = 0.196 𝑓𝑡 2 The total head is calculated as the difference between elevations: ℎ = 220 𝑓𝑡 − 200 𝑓𝑡 = 20 𝑓𝑡 The discharge: 𝑄 = 𝐶𝐴√2ℎ𝑔 = (0.62)(0.196𝑓𝑡 2 )√2 𝑥 32.2

𝑓𝑡 𝑥 20𝑓𝑡 = 4.4 𝑐𝑓𝑠 𝑠2

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QUESTION 15. The angle: 𝜃=

60 = 30 2

The discharge: 5

𝑄 = 𝐶(𝑡𝑎𝑛𝜃)𝐻 2 5

4𝑖𝑛 2 = 2.5 (tan 30) ( ) = 0.090 𝑐𝑓𝑠 12𝑖𝑛

QUESTION 16. The discharge: 3

3

𝑄 = 𝐶𝐿𝐻 2 = 3.215 𝑥 4 𝑥 (1.25)2 = 18 𝑐𝑓𝑠

QUESTION 17. In open channel problems, when Q is given D is found by trial and error. From the answer choices given, please select B or C. Select Depth = 2.15 ft (from the choics) Hydraulic radius: 𝐴𝑟𝑒𝑎 𝐴 = 21.5 𝑓𝑡 2 𝑊𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 14.3 𝑓𝑡 𝑅= 𝑄=

21.5𝑓𝑡 = 1.50𝑓𝑡 14.3𝑓𝑡

2 1 1.49 𝑥 21.5𝑓𝑡 2 (1.5)3 (0.015)2 = 396 𝑐𝑓𝑠 0.013

This is close to 400cfs given in the problem statement. Therefore the normal depth = 2.15 ft

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QUESTION 18. The discharge portion of flow in the rectangular section: 𝑅 (ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑟𝑎𝑑𝑖𝑢𝑠) = 𝑄=

6𝑓𝑡 𝑥 12 𝑓𝑡 = 6𝑓𝑡 12 𝑓𝑡

2 1 1.49 𝑥 72 𝑓𝑡 2 𝑥 (6𝑓𝑡)3 𝑥 (0.002)2 = 930.75 𝑐𝑓𝑠 0.017

The discharge portion in the grassy subsection of the channel: 1 𝐴 = ( ) (3𝑓𝑡)(6𝑓𝑡) = 9𝑓𝑡 2 2 𝑅=

9𝑓𝑡 2 = 1.35 𝑓𝑡 4𝑓𝑡

For both sides of flow: 2 1 1.49 𝑥 9𝑓𝑡 2 𝑥 (1.35𝑓𝑡)3 𝑥(0.002)2 ] = 58.59 𝑐𝑓𝑠 0.025

𝑄 = 2[

Total discharge from the channel is: 930.75 𝑐𝑓𝑠 + 58.59 𝑐𝑓𝑠 = 989.34 𝑐𝑓𝑠

QUESTION 19. Flow in a venture can be calculated as: 𝑃1 − 𝑃2 0.5 𝑄 = 𝐶𝐴 [2𝑔 ] 𝛾 (0.422)2 160 𝑥 62.4 0.5 = 0.95 𝑥 𝜋𝑥 [2 𝑥 32.2𝑥 ] = 3.90 𝑐𝑓𝑠 𝑜𝑟 1750 𝑔𝑝𝑚 4 12 𝑥 62.4

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QUESTION 20. The discharge Q: 44,000 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 𝑥 200 𝑔𝑝𝑐𝑑 = 8.8 𝑀𝐺𝐷 = 13.6 𝑐𝑓𝑠 The effective head: 𝐻 = (611 − 588)𝑓𝑡 + 10 𝑓𝑡 + 7𝑓𝑡 = 40 𝑓𝑡 The overall horsepower: 62.4𝑙𝑏 𝑄𝛾𝐻 13.6 𝑐𝑓𝑠 𝑥 𝑓𝑡 3 𝑥 40 𝑓𝑡 ℎ𝑝 = = = 85.7 ℎ𝑝 550𝑓𝑡 550𝑒 𝑥 0.72 𝑙𝑏. ℎ𝑝 QUESTION 21. The headloss is computed b Darcy-Weisbach equation: ℎ𝑙 = 𝑓 𝑥

𝐿 𝑣2 𝑥 𝐷 2𝑔

To determine the friction factor from Moody’s diagram, the Reynolds number must be computed: 𝐷𝑉𝜌 𝐷𝑉 𝑅𝑒 = = = 𝜇 𝜈

200𝑚𝑚 5 𝑥 1000𝑚𝑚 𝑚2 1.007 𝑥 10−6 𝑠

= 9.93 𝑥 105

The relative roughness: 𝑒 0.00026 = = 0.0013 𝐷 0.20 From the Moody’s diagram, f = 0.021 5𝑚 2 ( 1000 𝑠 ) = 133.8 𝑚 ℎ𝑙 = 0.021 𝑥 𝑥 𝑚 200 2𝑥9.81 2 𝑠 1000

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QUESTION 22. 𝐻1 𝑛12 = 2 𝐻2 𝑛2 𝐻2160 = 𝐻1800 [

2160 2 ] = 115𝑓𝑡 1800

QUESTION 23.

𝑉=

𝑄 80𝑥4 𝑓𝑡 = = 2.83 𝐴 𝜋 𝑥 6 𝑥6 𝑠

𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 =

𝐷𝑉 6 𝑥 2.83 = = 1.20 𝑥 106 −5 𝜈 1.41 𝑥 10

Given C = 0.001 ft; 𝑒 0.001 = = 0.00017 𝐷 6 Given f = 0.014 Using head-loss equation:

𝐿 𝑉2 1000 𝑥 2.832 ℎ𝐿 = 𝑓 = 0.014 𝑥 = 0.29 𝑓𝑡 𝐷 2𝑔 6 𝑥 64.4

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QUESTION 24. The turbine power can be calculated using: 𝑃=

𝛾𝑄ℎ𝜂 550

Where, P = power in hp γ = Specific weight of water Q = volumetric flow of fluid η = efficiency With a 90% efficiency, the available power is: =

62.4 𝑥 200 𝑥 600 𝑥0.9 = 12,300 ℎ𝑝 550

QUESTION 25.

Step 1: The relative roughness is: 𝑒 0.25 = = 0.00325 𝐷 80 Assuming a completely turbulent flow, the friction factor from Moody’s diagram is f =0.025.

Step 2: The head loss is: ℎ𝐿 =

∆𝑃 500,000 = = 51𝑚 𝛾 9800

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Step 3: The average velocity is: 𝑉=√

2𝑔𝐷ℎ𝐿 2𝑥9.8𝑥0.08𝑥51 = √ = 3.92 𝑚/𝑠 𝑓𝐿 0.025 𝑥 200

(Note: make sure the Reynolds number is in the turbulent region) Step 4: The flow rate is: Q

=

=

VA

=

π x (0.04)2 x 3.92 0.02 m3/s

QUESTION 26. The energy equation is applied between the tank and the faucet exit: 0=

𝑉22 −𝑉12 2𝑔

+

𝑝2 −𝑝1 𝛾

+ 𝑧2 − 𝑧1 + ℎ𝐿

Where, ℎ𝐿 = [

𝑓𝐿 𝐷

+ 3𝐾𝑒𝑙𝑏𝑜𝑤 + 𝐾𝑒𝑛𝑡𝑟𝑎𝑛𝑐𝑒 ]

𝑉2 2𝑔

Assuming that the pipe has e/D = 0, and that Re = 2x105, so that the Moody diagram yields f =0.016. The energy equation yields: 𝑉2 400,000 0.016𝑥20 𝑉2 0= − +3+[ + 3𝑥1.6 + 0.5] 2𝑥9.8 9800 0.015 2𝑥9.8 V = 5.18 m/s (By trial and error method) The Reynolds number is then: 𝑅𝑒 =

5.18𝑥0.15 10−6

= 7.8 x104

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Try f =0.018, then 𝑉2 400,000 0.018𝑥20 𝑉2 0= − +3+[ + 3𝑥1.6 + 0.5] 2𝑥9.8 9800 0.015 2𝑥9.8 V = 4.95 m/s (By trial and error method) Thus the Reynolds number equals to: 𝑅𝑒 =

4.95 𝑥 0.15 = 7.4𝑥104 −6 10

This is close enough so use V = 5.0 m/s, the flow rate is: π x 0.00752 x 5 = 8.8x10-4 m3/s

Q = VA =

QUESTION 27.

The average velocity is: 𝑉=

𝑄 𝐴

=

0.004 𝜋𝑥0.032

= 1.415 𝑚/𝑠

The Reynolds number is: 𝑅𝑒 =

𝑣𝐷 𝜈

=

1.415𝑥0.06 1.14𝑥10−6

= 7.44x104

The value of e is found on the Moody diagram so that 𝑒 0.046 = = 0.00077 𝐷 60 The friction factor is found from the Moody diagram to be: f = 0.0225 The pressure drop in the pipe is: 𝐿 𝑉2 300 1.4153 Δ𝑝 = 𝛾ℎ𝐿 = 𝜌𝑓 = 1000𝑥0.0225𝑥 𝑥 = 113.000 𝑃𝑎 𝐷 2 0.06 2 =113 kPa

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QUESTION 28. The relative roughness is: 𝑒 0.26 = = 0.00325 𝐷 80 The head loss is: ℎ𝐿 =

Δ𝑝 200 000 = = 20.41 𝑚 𝛾 9800

Assuming completely turbulent flow, Moody’s diagram yields: f = 0.026

The average velocity in the pipe is: 2𝐷𝑔ℎ𝐿 2𝑥20.41𝑥0.08𝑥9.81 𝑉=√ = √ = 1.76 𝑚/𝑠 𝑓𝐿 0.026𝑥400

The Reynolds number is: 𝑅𝑒 =

𝑣𝐷 1.76 𝑥 0.08 = = 1.4𝑥105 𝜐 10−6

At this Reynolds number and e/D = 0.0325, Moody’s diagram provides f = 0.026, so the friction factor does not have to be adjusted. The flow rate is then expected to be: Q =VA = π x 0.042 x 1.76 = 0.0088 m3/s

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QUESTION 29. First determine the friction factor and the Hazen-Williams coefficient 𝑉=

𝑄 0.05𝑥4 𝑚 = = 2.83 𝐴 𝜋 𝑥 0.152 𝑠 ν = 1.141x10-6 m2/s

𝑅𝑒 = 𝑒 𝐷

𝑉𝐷 = 3.72 𝑥105 𝜈

=

0.26 150

=0.00173

From Moody’s diagram, f = 0.024, C=100 Using Hazen-Williams formula: ℎ𝐿 =

10.59𝑥400 𝑥0.051.85 = 𝟑𝟒. 𝟏 𝒎 1001.85 𝑥0.154.87

QUESTION 30. First compute the resistance coefficient: 𝑅=

1 𝐿 [𝑓 + ∑ 𝐾] 2𝑔𝐴2 𝐷

1 400 𝑠2 4 𝑅= + 2.5] = 1.13𝑥10 5 [0.025𝑥 2𝑥9.81𝑥𝜋𝑥0.15𝑥0.15 0.15 𝑚 4 The discharge is: 𝑄=√

𝐻𝐴 −𝐻𝐵 𝑅

= √

20 1.13𝑥104

= 0.042 m3/s

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QUESTION 31. Change the pressure from psi to ft as: 30 𝑝𝑠𝑖 =

2.31𝑓𝑡 = 69.3 𝑓𝑡 𝑝𝑠𝑖

50 𝑝𝑠𝑖 =

2.31 𝑓𝑡 = 115.5 𝑓𝑡 𝑝𝑠𝑖

5 𝑝𝑠𝑖 =

2.31 𝑓𝑡 = 11.55 𝑓𝑡 𝑝𝑠𝑖

Maximum height = 115.5 ft + 11.55 ft = 127.05 ft 𝑉𝑜𝑙𝑢𝑚𝑒 =

𝜋𝑥102 4

𝑥 (115.5 − 69.3) = 3.628.5 𝑓𝑡 3

3528.5𝑓𝑡 3 𝑥7.48

𝑔𝑎𝑙 𝑓𝑡 3

= 27,141 gal

QUESTION 32. Area: =

1 (3 + 7.5)𝑥 1.5 = 7.852 𝑚2 2

Velocity: 𝑄 8 𝑚 = = 1.015 𝐴 7.852 𝑠 Since the ration of horizontal to vertical is 1:1.5 1 1.5 = ; 𝑥 = 2.25 1.5 𝑥 Wetted perimeter: = 3 + 2𝑥2.25 = 7.5𝑚 Hydraulic diameter: 7.875 = 1.05 7.5 Froude number: =

=

𝑣 √𝑔𝐷

=

1.05 √9.81 𝑥 1.05

= 0.316

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QUESTION 33. For Re < 0.2, 𝐶𝐷 =

24 24 = = 120 𝑅𝑒 0.2

QUESTION 34. Given, µ = 0.5 poise = 0.05 Pa 𝑅𝑒 =

𝐷𝑣𝜌 800 𝑥 𝑉 𝑥 0.009 = = 144𝑉 𝜇 0.05 𝑓=

64 64 = = 0.444𝑉 𝑅𝑒 144𝑉 ℎ𝑓 = 𝑓

𝐿 𝑉2 𝐷 2𝑔

0.444 𝑉2 1 0.0248𝑚 𝑚𝑚 0.75 = 12 𝑥 𝑥 = 𝑜𝑟 24.8 𝑉 2 𝑥 9.81 9𝑥10−3 𝑠 𝑠 QUESTION 35. Velocity: 𝑉=

𝑄 0.01 1 =𝜋 = 𝑚/𝑠 2 𝐴 𝜋 4 (0.2)

Reynolds number: 1 𝐷𝑣𝜌 900 𝑥 𝜋 𝑥 0.02 𝑅𝑒 = = = 716 < 2100 𝜇 0.08 The flow is laminar. Loss of head for laminar flow: ℎ𝑓 =

32𝜇𝐿𝑉 32 𝑥 0.8𝑥(20𝑥1000)𝑥 1/𝜋 = = 4527 𝑚 2 𝑤𝑑 900 𝑥 (0.2)2

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Power required to maintain the flow: 𝑃=

𝑊𝑄ℎ𝑓 900𝑥0.01𝑥4527 = = 542 𝐻𝑃 𝑜𝑟 405 𝑘𝑊 75 75

QUESTION 36. The headloss is computed by Darcy-Weisbach equation: ℎ𝑓 = 𝑓

𝐿 𝑉2 𝐷 2𝑔

To determine the friction factor from Moody’s diagram, the Reynolds number must be calculated. 200 5𝑥(1000) 𝐷𝑣𝜌 𝑅𝑒 = = = 9.93𝑥105 𝜇 1.007𝑥10−6 The relative roughness is: 𝑒 0.00026 = = 0.0013 𝐷 0.2 From Moody’s diagram, the friction factor, f = 0.021 ℎ𝑓 = 0.02 𝑥

1000 52 𝑥 = 133.8𝑚 200 2 𝑥 9.81 1000

QUESTION 37. The cross-sectional area: 𝐴 = 𝜋𝑟 2 = 0.196𝑓𝑡 2 The total head as the difference between elevations: ℎ = 220 − 200 = 20𝑓𝑡 𝑄 = 𝐶𝐴√2𝑔ℎ = (0.62)(0.196)√2𝑥32.2 𝑥 20 = 4.4 𝑐𝑓𝑠

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QUESTION 38. 𝑄 = 𝐶𝐿𝐻 3/2 = 3.215𝑥4𝑥(1.25)3/2 = 18𝑐𝑓𝑠

QUESTION 39. The cross sectional area of the pipe is: 𝜋 𝜋 𝐴 = 𝐷2 = (0.10)2 = 0.007854𝑚2 4 4 The flow rate is: 𝑄 − 𝐴𝑉 = 0.007854(2.5)(60) = 1.178 𝑚3 /𝑚𝑖𝑛 QUESTION 40. ∆𝑃𝑒𝑙𝑏𝑜𝑤

𝜌𝑉 2 1,000 (2.5)2 (0.9) = 2,812.5 𝑃𝑎 = 𝐶= 2 2

QUESTION 41. 𝑊𝑖𝑑𝑒𝑎𝑙 =

∆𝑃 𝑃𝑐 − 𝑃𝐵 175 − 125 𝑘𝐽 𝐽 = = = 0.050 = 50 𝜌 𝜌 1,000 𝑘𝑔 𝑘𝑔

QUESTION 42. The cross sectional area of the pipe is: 𝐴=

𝜋 2 𝜋 𝐷 = (0.10)2 = 0.007854𝑚2 4 4

The velocity of flow in the discharge pipe is: 𝑉=

𝑚 40 𝑚 = = 5.093 𝜌𝐴 1000(0.007854) 𝑠

The head at the discharge into the heater is: (5.903)2 𝑃 𝑉2 200,000 𝐻ℎ = + + 𝑧2 = + + 20 = 41.7𝑚 𝜌𝑔 2𝑔 1000 (9.807) 2(9.807) The head at the pump discharge is: 𝐻𝑑 = 41.7 − 0.5 + 25 = 66.2 𝑚

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QUESTION 43. A force balance on the fluid column gives: 𝑃𝑏 = 𝑃𝑎𝑡𝑚 = 𝑃𝑣 + (𝛾. ℎ) The vapor pressure of mercury is negligible. Therefore, 𝑃𝑏 = 𝛾𝐻𝑔 ℎ 𝑚 𝑘𝑔 𝑘𝑔 𝛾𝐻𝑔 = (9.81 2 ) (13579 3 ) = 133165 2 2 𝑠 𝑚 𝑚 𝑠 𝑘𝑔 𝑘𝑔 𝑃𝑏 = 133165 2 2 𝑥 0.728 𝑚 = 96,944 𝑚 𝑠 𝑚. 𝑠 2 QUESTION 44. To assure laminar flow the Reynolds number should not exceed 2000. Therefore 𝐷𝑉 2000𝜈 𝑅𝑒 = ≤ 2000 𝑜𝑟 𝑉 = 𝜈 𝐷 2000 𝑥 0.0005 𝑚 𝑉≤ = 50 20 𝑥 10−3 𝑠 And the maximum discharge is: 𝑚 𝜋 1000𝐿 𝐿 𝑄 = 𝑉𝐴 = 50 𝑥 𝑥(0.02𝑚)2 ( ) = 15.7 3 𝑠 4 𝑚 𝑠 QUESTION 45. Assuming geometric similarity of model and prototype, equal Froude numbers assure dynamic similarity. 𝑉𝑝 𝑉𝑚 = √𝑔𝑚 𝑙𝑚 √𝑔𝑝 𝑙𝑝 Since gm = gp, 𝑉𝑝 = 𝑉𝑚 √

𝑙𝑝 𝑚 𝑚 = 1.4 √10 = 4.43 𝑙𝑚 𝑠 𝑠

QUESTION 46. For incompressible flow, identical Reynolds numbers will assure similarity, assuming geometrical similarity. Therefore: 𝑅𝑒𝑚 = 𝑅𝑒𝑝 𝑉𝑝 𝐷𝑝 𝑉𝑚 𝐷𝑚 = 𝜈𝑚 𝜈𝑝 1.006 𝑥10−6 𝑚 𝑚 𝑉𝑚 = 6 𝑥 𝑥 3.05 = 50.6 −6 0.364 𝑥 10 𝑠 𝑠

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This velocity results in a discharge volumetric flow rate of: 𝑚 𝜋 0.3 2 𝑚3 𝑄𝑚 = 𝑉𝑚 𝐴𝑚 = 50.6 𝑥 𝑥 ( ) = 0.1 𝑠 4 6 𝑠 QUESTION 47. Assume that the flow is steady, incompressible and frictionless, making Bernoulli’s equation valid: 𝑃1 𝑉12 𝑃2 𝑉22 + 𝑧1 + = + 𝑧2 + 𝛾 2𝑔 𝛾 2𝑔 Station 2 is a stagnation point, where the flow is completely stopped (V2 =0) and the two stations are at the same elevation (z1=z2), therefore 𝑉12 𝑃2 − 𝑃1 = 2𝑔 𝛾 The pressure differential can be determined by using the hydrostatic equation P=γh, along the path 1-3-4-2 through the manometer. 𝑃1 + 𝛾ℎ + 𝛾𝐻𝑔 ℎ0 − 𝛾(ℎ + ℎ0 ) = 𝑃2 𝑃2 − 𝑃1 𝛾𝐻𝑔 = ℎ − ℎ0 = ℎ0 (𝑆𝐺𝐻𝑔 − 1) 𝛾 𝛾 0 Substituting for

𝑃2 −𝑃1 𝛾

𝑉1 = √2𝑔ℎ0 (𝑆𝐺𝐻𝑔 − 1) = √2 𝑥 32.2 𝑄 = 𝐴𝑉 =

𝑓𝑡 2 𝑓𝑡 ( 𝑓𝑡)(13.6 − 1) = 11.63 2 𝑠 12 𝑠

𝜋 𝑓𝑡 𝑔𝑎𝑙 60𝑠 (4/12𝑓𝑡) 2 (11.63 ) (7.48 3 ) ( ) = 455.49 𝑔𝑎𝑙/𝑚𝑖𝑛 4 𝑠 𝑓𝑡 min

QUESTION 48. The volumetric flow Q is constant and 𝑄 = 𝐴1 𝑉1 = 𝐴2 𝑉2 𝑚3 0.028 𝑠 𝑚 𝑚 𝑉1 = 𝜋 = 3.56 ; 𝑉2 = 1.10 2 𝑠 𝑠 4 (0.18) From Bernoulli equation: 𝑃2 = 𝑃1 +

𝛾(𝑉12 − 𝑉22 ) 2

Substituting the values gives: 𝑃2 = 1.8 𝑥105 + 820

𝑘𝑔 3.562 − 1.102 𝑁 = 1.847 𝑥105 2 3 𝑚 2 𝑚

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515

QUESTION 49. Assume the inlet velocity so low that its effect is negligible. Take the flow as steady, assume the process is adiabatic (no heat transfer) and neglect elevation changes. The energy equation reduces to: 𝑉22 𝑊 = ℎ1 − ℎ2 − 2 The corresponding rate equation is assuming a constant specific heat: 𝑉22 𝑊 = 𝑚[𝑐𝑝 (𝑇1 − 𝑇2 ) − 2 𝑘𝑔 (90𝑚)2 𝑊 = 1.5 [1012.9(300 − 477) − = −369 ℎ𝑝 𝑠 2

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516

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PART-III QUESTION 1 Methane at 22C flows through a pipeline at a velocity of 416.4 m/s. The flow is: Assume: k = 1.32; R = 52.9 A. B. C. D.

Subsonic Supersonic Hypersonic None of the above

QUESTION 2 Chlorine gas at 51F flows through a pipeline. The velocity at which the flow will be sonic is: Assume: k = 1.34; R = 21.8 A. B. C. D.

500 600 700 800

QUESTION 3 Air flows past an object at 500 fps. The stagnation temperature (F) in the standard atmosphere at elevation of 5000 ft is: Assume: Cp = 6000 lb-ft/slug R R = 1716 lb-ft/slug R Standard temperature = 60F

A. B. C. D.

-27 50 62 79

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518

QUESTION 4 Air flows past an object at 180 m/s at an elevation of 2km. The stagnation temperature in the standard atmosphere at the above elevation is close to: Assume: R = 287 Cp = 1003 Temperature at 2km = 2C A. B. C. D.

10 18 24 36

QUESTION 5 Air at 250 psia is moving at 550 fps in a high-pressure wind tunnel at a temperature of 100F. The stagnation pressure is close to: Assume k = 1.40 A. B. C. D.

100 200 300 400

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519

QUESTION 6 A cylinder containing 2kg nitrogen at 0.14MPa absolute and 5C compressed isentropically to 0.30 MPa abs. The final temperature (C) is: A. B. C. D.

23 33 53 73

QUESTION 7 The speed of sound in dry air when temperature is -67F is close to: Assume R = 1716 A. B. C. D.

670 970 1017 1210

QUESTION 8 In an isentropic process, 1 kg of oxygen at 17C has its absolute pressure doubled. The final temperature (C) is: A. B. C. D.

24 64 84 99

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520

QUESTION 9 A projectile moves through water at 80F at 1800 fps. The Mach number is: A. B. C. D.

0.167 .0254 0.367 0.555

QUESTION 10 The speed of sound through hydrogen at 75F is close to: A. B. C. D.

2300 4300 6925 7891

QUESTION 11 The speed of sound in carbon monoxide at 200C is: Assume R = 0.297 kJ/kg.K A. 443 B. 520 C. 625 D. 877

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QUESTION 12 Steam at 450F and 80 psia is compressed isentropically to 100 psia. The new temperature (F) is: Assume an ideal gas with k = 1.33 A. B. C. D.

200 300 400 500

QUESTION 13 An airplane flies at 360 m/s through air at -10C, and 40kPa. The airplane is: A. B. C. D.

Subsonic Supersonic Hyper sonic None of the above

QUESTION 14 Isentropic flow of air occurs at a section of a pipe where p = 50 psia, T = 100F and v = 540 ft/s. An object is immersed in the flow, which brings the velocity to zero. The temperature (F) and pressure (psia) at stagnation point are: A. B. C. D.

100; 60 125; 58 100; 58 58;124

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522

QUESTION 15 Air flows from a reservoir at 80C and 6atm. The temperature (C) at a section where M = 0.65 is: A. B. C. D.

25 52 65 75

QUESTION 16 Air flows from a reservoir at 80C and 6atm. The pressure at a section where M = 0.65 is: A. B. C. D.

258 358 458 678

QUESTION 17 A flow of air (Cp = 1005 J/kg.K, k = 1.40) has V = 850 m/s,p = 130kPa and T= 180C. The pressure (kPa) and temperature (C) of the air if brought isentropically to rest are: A. B. C. D.

1000; 540 540; 1000 500; 500 1000; 1000

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523

QUESTION 18 Compute the frontal stagnation temperature (C) of an airplane flying at a Mach of 2.0 at 8000 m altitude. Assume the temperature to be 236.1K A. B. C. D.

125 150 175 200

QUESTION 19 At what Mach number a plane would fly to have a frontal stagnation temperature of 380C? Assume the plane altitude is 800m (T = 236.1K) A. B. C. D.

1 2 3 5

QUESTION 20 Air flows adiabatically through a duct. At one section, v1 = 420 fps, T1 =210F, and p1 =35psia. While farther downstream, v2 = 1000 fps, and p2 = 18psia. The ratio of p01/p02 is: A. B. C. D.

0.25 0.50 0.75 1.00

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524

QUESTION 21 Air flows isentropically from a reservoir, where p = 320 kPa and T = 490K, to section 1 in a duct, where A1 = 0.2 m2 and v1 = 600 m/s. Compute the temperature (T1): A. B. C. D.

100K 200K 310K 420K

QUESTION 22 Air flows isentropically from a reservoir, where p = 320 kPa and T = 490K, to section 1 in a duct, where A1 = 0.2 m2 and v1 = 600 m/s. Compute p1 (kPa) A. B. C. D.

25 45 58 65

QUESTION 23 A perfect gas at 305 kPa expands isentropically through a supersonic nozzle with an exit area 5 times its throat area. If the exit Mach number is 3.8, the specific heat ratio of the gas is: A. B. C. D.

1.1 1.3 1.5 1.7

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525

QUESTION 24 A perfect gas at 305 kPa expands isentropically through a supersonic nozzle with an exit area 5 times its throat area. If the exit Mach number is 3.8, and the specific heat ratio of the gas is 1.67, the exit pressure (kPa) of the gas is: A. B. C. D.

1.72 3.72 8.98 9.75

QUESTION 25 The pressure, velocity and temperature just upstream of a normal shock wave in air are 12 psia, 2300 fps and 24F. The pressure (psia) just downstream of the wave is nearly: Assume k = 1.4 and R = 1715 A. 30 B. 60 C. 90 D. 120

QUESTION 26 Consider a supersonic flow of air through a stationary duct wherein a stationary shock is present. The Mach ahead of the shock is 2.1 and the pressure and temperature are 101.3 kPa absolute and 37C. What is the velocity (m/s) of propagation of the shock relative to the air ahead of the shock? A. B. C. D.

371 522 741 981

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526

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SOLUTIONS QUESTION 1 𝑁𝑀 =

𝑣 √𝑘𝑔𝑅𝑇

416.4 √(1.32)(9.807)(52.9)(22 + 273) Since NM 1.0, it is supersonic.

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529

QUESTION 14 𝑉2 𝑘𝑅 =[ ] (𝑇 − 𝑇) 2 𝑘−1 0 (540)2 [1.40)(1716) = [𝑇0 − (100 + 460)] 2 1.40 − 1 𝑇0 = 584𝑅 𝑜𝑟 124𝐹 𝑝2 𝑇2 = 𝑇1 [ ](𝑘−1)/𝑘 𝑝1 584 = (𝑝2 /50)(1.40−1)/1.40 𝑜𝑟 𝑝2 = 57.9 𝑝𝑠𝑖𝑎 100 + 460 QUESTION 15 𝑇0 𝑘−1 2 = 1+[ ]𝑁𝑀 𝑇 2 273 + 80 1.4 − 1 =1+[ ] (0.65)2 𝑇 2 𝑇 = 325 𝐾 𝑜𝑟 52𝐶 QUESTION 16 𝑝0 𝑘−1 2 𝑘 = { 1 + ([ ] 𝑁𝑀 }𝑘−1 𝑝 2 6 𝑥 101.310 1.4 − 1 = {1 + [ ] (0.65)2 }1.40/(1.40−1) = 458 𝑘𝑃𝑎 𝑝 2 QUESTION 17 (850)2 𝑉2 (180 𝑇0 = 𝑇 + = + 273) + = 812𝐾 𝑜𝑟 539𝐶 2𝑐𝑝 2𝑥1005 1.67 𝑇0 𝑘 1150 𝑝0 = 𝑝 ( )𝑘−1 = (130)[ ]1.67−1 = 1326 𝑘𝑃𝑎 𝑇 180 + 273 QUESTION 18 𝑇0 𝑘−1 2 = 1+[ ]𝑁𝑀 𝑇 2 1.40 − 1 𝑇0 = 236.1 [1 + ] (2)2 = 425𝐾 𝑜𝑟 152 𝐶 2 QUESTION 19 𝑇0 𝑘−1 2 = 1+[ ]𝑁𝑀 𝑇 2 (380 + 273) = 236.1{1 + (1.40 − 1)/2 }𝑁𝑀2 𝑁𝑀 = 2.97

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530

QUESTION 20 𝑝0 = (𝑝){[1 + 𝑁𝑀1 =

𝑉1

= 0.331 √(1.40)(1716)(210 + 460) 1.40 [1.40 − 1) (𝑝0 )1 = 35{1 + ](0.331)2 }1.40−1 = 37.76 𝑝𝑠𝑖𝑎 2 1.40 [1.40 − 1) (𝑝0 )2 = 18{1 + ](0.831)2 }1.40−1 = 28.31 𝑝𝑠𝑖𝑎 2 (𝑝0 )1 28.31 = = 0.75 (𝑝0 )2 37.76 √𝑘𝑅𝑇1

=

𝑘−1 2 𝑘 ]𝑁𝑀 }𝑘−1 2 420

QUESTION 21 𝑇1 = 𝑇0 −

(600)2 𝑉12 = 490 − = 311𝐾 (2)(1005) 2𝑐9

QUESTION 22 𝑝1 = (𝑝0 )/{[1 + 320

𝑘−1 2 𝑘 ]𝑁𝑀1 }𝑘−1 2 1.40

= 64.8𝑘𝑃𝑎

1.40−1 1.40 − 1 (1.70)2 } {1 + [ ] 2

QUESTION 23 [2 + (𝑘 − 1)(𝑀)2𝑀𝑒 ] (𝑘+1)/2(𝑘−1) 𝐴𝑒 1 = [ ] { } 𝐴∗ 𝑁𝑀𝑒 𝑘+1 Substituting the values, (𝑘 − 1)(3.8)2 ] 𝑘+1 1 5 = ( ) {[2 + }2(𝑘−1) 3.8 𝑘+1 By trial and error, k = 1.67 QUESTION 24 𝑝𝑒 = (𝑝0 )/{[1 + =

𝑘−1 2 𝑘 ]𝑁𝑀𝑒 }𝑘−1 2

305 1.66 = 3.72 𝑘𝑃𝑎 1.66 − 1 2 }1.66−1 {1 + [ ](3.8) 2

QUESTION 25 𝑘−1 } 𝑘+1 𝑐 = √𝑘𝑅𝑇 = √(1.4)(1715)(484) = 1078 𝑓𝑝𝑠 𝑉 2300 𝑁𝑀 = = = 2.13 𝑐 1078 [2(1.4)(2.13)2 − 0.4] (12) 𝑝2 = { } = 61.5 𝑝𝑠𝑖𝑎 2.4 𝑝2 = 𝑝1 {[2𝑘(𝑁𝑀 )12 −

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531

QUESTION 26 𝑁𝑀1 = 2.1 =

𝑉1 𝑉1 = 𝑐1 √𝑘𝑅𝑇1 𝑉1

√(1.40)((287)(37 + 273)

= 741

𝑚 𝑠

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532

Thermodynamics Total Questions 13-20 A. B. C. D. E. F. G. H. I. J. K.

Properties of ideal gases and pure substances Energy transfers Laws of thermodynamics Processes Performance of components Power cycles, thermal efficiency, and enhancements Refrigeration and heat pump cycles and coefficients of performance Nonreacting mixtures of gases Psychrometrics Heating, ventilating, and air-conditioning (HVAC) processes Combustion and combustion products

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533

PART-I QUESTION 1. 2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increases by 10 kJ and 140 kJ of heat is transferred to the surroundings. The work (kJ) done by the gas during the process is: A. B. C. D.

-150 -130 -85 -45

QUESTION 2. During a process, 30 J of work are done by a closed stationary system on its surroundings. The internal energy of the system decreases by 40 J. The heat transfer is close to: A. B. C. D.

10 J released to surroundings 10 J absorbed by the system 70 J released to the surroundings 70 J absorbed by the system

QUESTION 3. 55 m3 of water passes through a heat exchanger and absorbs 2,800,000 kJ. The exit temperature is 95C, the entrance temperature (C) is nearly: A. B. C. D.

49 56 68 83

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534

QUESTION 4. The mass flow of a Freon refrigerant through a heat exchanger is 5 kg/min. The enthalpy of entry Freon is 238kJ/kg, and the enthalpy of exit Freon is 60.6kJ/kg. Water coolant is allowed to rise 6C. The water flow rate in kg/min is: A. B. C. D.

24 35 83 99

QUESTION 5. A 35 m x 15 m swimming pool is filled with water to depth of 3m. The amount of heat (GJ) required to raise the temperature of the water in the pool from 10C to 25C is nearly: A. 99 B. 230 C. 990 D. 1,900

QUESTION 6. The change in specific internal energy (kJ/kg) of air cooled from 550C to 100C is: A. B. C. D.

320 390 450 550

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535

QUESTION 7. Gas initially at 1 MPa and 150C receives 7.2MJ of work while 1.5 kW of heat removed from the system. The internal energy (MJ) change for the system over a period of an hour is: A. B. C. D.

-5.7 1.8 8.7 13

QUESTION 8. A fluid has a mass of 5 kg and occupies a volume of 1 m3 at a pressure of 150 kPa. If the internal energy is 2,500 kJ/kg, the total enthalpy is: A. 2.5 B. 9.8 C. 12.7 D. 20.7

QUESTION 9. An ideal gas at a gage pressure of 0.3 MPa and 25C is heated in a closed container to 75C. The final pressure (MPa) is close to: A. B. C. D.

0.35 0.47 0.90 1.20

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536

QUESTION 10. Steam exists at a pressure of 120 Pa at a temperature of 250K. The number of molecules present in 2 cm3 at these conditions is: A. B. C. D.

5 x 1012 5 x 1016 7 x 1016 8 x 1018

QUESTION 11. A 2 m3 tank contains 40 kg of oxygen at 40C. The gage pressure (MPa) in the tank is: Specific gas constant: 0.259 kJ/kg.K A. B. C. D.

0.61 1.10 1.30 1.53

QUESTION 12. A total of 3 kg of air is in rigid container at 250 kPa and 50C. The volume (m3) of the container, if the gas constant is 0.287 kJ/kg.K is close to: A. B. C. D.

1.1 2.2 2.8 3.1

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537

QUESTION 13. The density (kg/m3) of air 1 atm and 600 C is close to: (Molecular Weight of air: 28.967g) A. B. C. D.

0.12 0.40 0.59 0.68

QUESTION 14. A boy on the beach holds a spherical balloon filled with air. At 10AM, the temperature on the beach is 20C and the balloon has a diameter of 30 cm. Two hours later, the balloon diameter is 30.5cm. Assuming that the air is an ideal gas and that no air was lost or added, the temperature (C) at noon is: A. B. C. D.

21 25 32 35

QUESTION 15. A mixture at 100 kPa and 20C consists of 30% by weight of CO2 (MW=44) and 70% by weight of nitrogen (MW=28). The partial pressure of CO2 (kPa) is close to: A. B. C. D.

21.4 31.5 68.3 78.6

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538

QUESTION 16. Dry air has an average molecular weight of 28.79 consists of 21 mole-percent oxygen and 78 mole-percent of nitrogen and 1 mole-percent of argon. The weight percent of oxygen is nearly: A. B. C. D.

21.0 22.4 23.2 24.6

QUESTION 17. A 0.75 m3 tank contains nitrogen at 150 kPa and 40C. 2.0 kg of oxygen are added to the tank. The final volumetric percentage of nitrogen is: A. B. C. D.

14 17 38 41

QUESTION 18. By weight, atmospheric air is approximately 23.15% oxygen and 76.85% nitrogen. The partial pressure (kPa) of oxygen in the air at standard temperature and pressure is: A. B. C. D.

21 23 26 30

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539

QUESTION 19. A Carnot machine operates between a reservoir at 200C and a cold reservoir at 20C. When operated as an engine, it receives 1000 kJ/kg. The work output (kJ/kg) is close to: A. B. C. D.

181 281 381 481

QUESTION 20. The maximum thermal efficiency that can be obtained in an ideal reversible heat engine operating between 833 C and 170 C is close to: A. B. C. D.

60 70 80 90

QUESTION 21. A 2.2 kW refrigerator or heat pump operates between -17C and 38C. The maximum theoretical heat (kW) can be transferred from the cold reservoir is nearest to: A. 4.7 B. 7.6 C. 10.2 D. 15.6

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540

QUESTION 22. At the start of compression in an air-standard Otto cycle with a compression ratio 10, air is at 100 kPa and 40C. A heat addition of 2,800 kJ/kg is made. The thermal efficiency is close to: A. B. C. D.

52 60 64 70

QUESTION 23. A Carnot refrigeration system receives heat from a cold reservoir at 0C. The power input is 1,750 W per ton of refrigeration. The COP is nearly: A. B. C. D.

1.4 1.6 1.8 2.0

QUESTION 24. A Carnot engine receives 100 kJ of heat from a hot reservoir at 370 C and rejects 37 kJ of heat. The temperature (C) of the cold reservoir is: A. B. C. D.

- 35 100 130 230

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541

QUESTION 25. A refrigeration cycle has a coefficient of performance 80% of the value of a Carnot refrigerator operating between the temperature limits of 50C and -5C for a 3 kW of cooling, the power input (KW) required is: A. B. C. D.

0.53 0.62 0.77 0.89

QUESTION 26. A heat pump takes from groundwater at 7C and maintains a room at 21C. The COP possible for this heat pump: A. B. C. D.

1.4 2.8 5.6 21

QUESTION 27. A Carnot cycle refrigerator operates between -11C and 22C. The COP is:

A. B. C. D.

0.50 1.1 4.3 7.9

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542

QUESTION 28. A reversible thermodynamic system is made to follow the Carnot cycle between the temperature limits of 300C and 75C. 300 kW of heat are supplied per cycle to the system. The change in entropy (kW/K) during the heat addition is close to:

A. B. C. D.

0.52 0.86 1.0 4.0

QUESTION 29. A plane wall is 2m high 3m wide and is 20 cm thick. It is made of material that has thermal conductivity of 0.5 W/(m.K). A temperature difference of 60C is imposed on the two large faces. The heat flow (W) is: A. B. C. D.

450 600 750 900

QUESTION 30. A steam pipe with a surface area of 5 m2 and a surface temperature of 600C radiates into a large room. The surface of which is at 25C. The emissivity is 0.6. The heat flux (W/m2) in the room: A. B. C. D.

1.95 x 104 2.50 x 105 3.50 x 106 4.50 x 107

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543

QUESTION 31. Water at an average temperature of 20C flows through a 5-cm diameter pipe is 2m long. The pipe is heated by steam and is held at 100C. The convective heat transfer coefficient is 2.2 x 10 4 W/m2.K. The heat flux (MW/m2) is close to: A. B. C. D.

1.26 1.76 2.76 4.34

QUESTION 32. A well-insulated copper wire 1m long with a cross-sectional area of 0.084 m2 connects two reservoirs, one with boiling water at 100C and the other with ice at 0C. The time it takes for 100 cal energy to move through the wire is nearly: Assume K for copper = 388 W/m.K A. 7 B. 14 C. 21 D. 35

QUESTION 33. In 1 hr, the amount (kJ) of black-body radiation escapes a 1cm x 2cm rectangular opening in a kiln operating at 980C is: A. 20 B. 100 C. 150 D. 200

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544

QUESTION 34. A 3m x 3m plate at 500C is suspended vertically in a very large room. The plate has an emissivity of 0.13. The net heat transfer (kW) in the room with an initial temperature of 25C is nearly: A. B. C. D.

8.3 24 46 86

QUESTION 35. On a hot summer day, a stone will reach a maximum temperature of 50C. The stone has an emissivity of 0.95. The radiant heat energy transfer (W/m2) from the wall per square meter of this wall at that temperature is close to: A. 190 B. 590 C. 1100 D. 3800

QUESTION 36. Hot air at an average temperature of 100C flows through a 3m long tube with an inside diameter of 60mm. The temperature of the tube is 20C along its entire length. The convective film coefficient 20.1W/m2K. The rate of convective heat transfer from the air to the tube: A. 520 B. 850 C. 910 D. 1070

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545

QUESTION 37. The isentropic compression of 1m3 of air from 20 kPa to 100 kPa results in a final volume of: A. B. C. D.

0.16 0.20 0.32 0.40

QUESTION 38. The work (kJ) of a polytropic (n= 1.21) compression of air from 15 kPa with a1m 3 volume to 150 kPa and 0.15 m3 volume is nearly: A. B. C. D.

-35.7 - 324 1,080 5,000

QUESTION 39. Air is compressed isentropically such that its pressure is increased by 50%.The initial temperature is 70C and the final temperature (C) is close to: A. 80 B. 110 C. 140 D. 240

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546

QUESTION 40. A gas (MW=55) is initially at 2MPa, 200C and 0.5 m3 expands with the relation of PV1.35 = C to 0.2 MPa. The work for the expansion process is close to: A. B. C. D.

1.3 2.3 4.1 5.9

QUESTION 41. Air is compressed in a piston-cylinder arrangement to 1/10 of its initial volume. If the initial temperature is 35C and the process is frictionless and adiabatic, the final temperature (K) is: A. B. C. D.

350 460 620 770

QUESTION 42. The state of an ideal gas in an open system is changed in a steady-state process from 400 kPa and 1.2 m3 to 300 kPa and 1.5 m3. The relationship between pressure and volume during the process is PV1.3 = constant. The work (kJ) performed is close to: A. 130 B. 930 C. 1210 D. 2500

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547

QUESTION 43. Argon (MW = 40) is compressed in a closed system from 100 kPa and 30C to 500 kPa and 170C. The specific entropy (kJ/kg.k) change is close to: A. B. C. D.

-0.010 -0.14 0.37 0.56

QUESTION 44. The volume occupied by 10 kg of water at 200C and 2 MPa is close to: A. B. C. D.

0.099 m3 0.012 m3 9.4 L 11.8 L

QUESTION 45. The temperature (K) of 2kg of air contained in a 40L volume at 2MPa is close to: A. B. C. D.

120 140 160 180

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548

QUESTION 46. A 200-mm diameter piston is lowered by increasing the pressure from 100 kPa to 800 kPa such that the P-V relationship PV2 = constant. The work done (kJ) on the system if V1 =0.1 m3 is nearly: A. B. C. D.

-18.3 -24.2 -31.6 -42.9

QUESTION 47. Ten kilograms of air at 800 kPa is heated at constant pressure from 170C to 400 C. The work (kJ) required is: A. 115 B. 660 C. 960 D. 1150

QUESTION 48. Ten kilograms of saturated steam at 800 kPa is heated at constant pressure to 400C. The work (kJ) required is: A. 115 B. 660 C. 960 D. 1150

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549

QUESTION 49. The work (kJ) needed to compress 2 kg of air in an insulated cylinder from 100 kPa to 600 kPa if T1 = -20C is close to: A. B. C. D.

-469 -390 -280 -220

QUESTION 50. Air is compressed adiabatically from 100 kPa and 20C to 800 kPa. The new temperature (C) is close to: A. B. C. D.

260 290 360 440

QUESTION 51. The work (kJ) required to compress 2kg of an air insulated cylinder from 100C and 100 kPa to 600 kPa is: A. B. C. D.

220 280 360 460

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

550

QUESTION 52. Methane is heated at constant pressure of 200 kPa from 0C to 300C. The amount of heat (kJ/kg) needed is: A. B. C. D.

623 676 692 731

QUESTION 53. The equilibrium temperature if 10 kg of ice at 0C is mixed with 60kg of water at 20C is close to: A. B. C. D.

1.1 2.1 5.8 12

QUESTION 54. One kilogram of air is heated in a rigid temperature from 20C to 300C. The entropy change (kJ/K) is: A. B. C. D.

0.34 0.48 0.54 0.64

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551

QUESTION 55. The work (kJ) required to isentropically compress 2 kg of air in a cylinder at 400 kPa and 400C to 2MPa is: A. 560 B. 780 C. 940 D. 1,020

QUESTION 56. 200 kg of a solid at 0C is added to 42 kg of water at 80C. The final temperature (C) is: A. B. C. D.

42 48 52 58

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552

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

553

PART-II QUESTION 1. A low-energy gas is burned in a furnace with excess air. The feed rate is 2,500 m 3/h of gas at 0C and 1 atm.

Low energy gas CO CO2 H2 CH4

Mol % 16.0 12.6 19.0 52.4

Gas Composition MW Flue gas (dry basis) 28 CO2 44 N2 2 O2 16

Mol% 13.5 84.4 2.1

The amount of air supplied per 100 kmol of dry gas is almost nearly: A. 10 B. 17 C. 110 D. 400

QUESTION 2. It is desired to burn liquid propane, C3H8 with excess air, both supplied at 25C. Assume that air is 21vol% oxygen, and the balance is nitrogen. If 1.0 mol of propane is burned with x moles of air, how many moles of oxygen appear in the combustion products? A. B. C. D.

3.0 + 0.21x 4.0 + 0.79x 0.79x 0.21x -5.0

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554

QUESTION 3. A 30-L cylinder is filled with air to a pressure of 10 MPa. The valve to the cylinder is then closed. Thermodynamic data for dry air are given in the table below.

Assuming that the temperature in the cylinder immediately after filling is 600K and that no air has been removed from the cylinder, the pressure (MPa) in the cylinder a week after filling (when the temperature is 300K) is most nearly: A. B. C. D.

4.17 4.94 5.06 5.20

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555

QUESTION 4. The experimental heat capacity of a particular ideal gas is tabulated below: Temperature (C) 20 40 60 80 100

Cp 22 22.5 23.4 23.8 24.4

𝑇

Enthalpy change is: ∆ℎ = ∫𝑇 2 𝐶𝑝 (𝑇)𝑑𝑇 1

If this gas is heated from 20 to 100 C, the enthalpy change for 1 mole of gas can be calculated using numerical integration. Using Simpson’s rule, the result is most nearly: A. B. C. D.

∆h = 20(22+22.5+23.4+23.8) ∆h = 20(22.5+23.4+23.8+24.4) ∆h = 20/3(22+4(22.5) + 2(23.4)+ 4(23.8) +24.4) ∆h = 20/2 (22+ 2(22.5 + 23.4+23.8) + 24.4)

QUESTION 5. An insulated, steam-heated single-stage laboratory evaporator is used to determine data for the design of large units. Saturated steam at 150C is used as the heating medium. The steam condensate is all liquid at 150C. The total heat duty is 225 kW. The required steam rate (kg/s) is most nearly: A. B. C. D.

9.40 8.56 0.11 0.08

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556

QUESTION 6. One mole of an ideal gas with Cp = 3.5 𝑅̅ and Cv = 2.5 𝑅̅ , where 𝑅̅ is the universal gas constant expands from P1 = 103 kPa and V1 = 0.001m3 to P2 = 100kPa. If the gas follows a path such that the volume of the gas is constant, then the amount of work (kJ) done for the process is most nearly: A. 0.0 B. 1.4 C. 20.8 D. 29.0

QUESTION 7-8. The pump shown in the figure is used to pump 50,000 kg of water per hour into a boiler. Pump suction conditions are 40C and 100kPa. Pump discharge conditions are 40C and 14.0 MPa. Boiler outlet conditions are 500C and 14.0 MPa. The boiler efficiency is 88%.

P T Condition H (MPa) C (kJ/kg) 14.0 40 Compressed liquid 167.6 14.0 336.75 Saturated liquid 1,571.1 14.0 336.75 Saturated valor 2,637.6 14.0 500 Superheated vapor 3,322

If the pump efficiency is 80%, the pump power requirement (kW) is most nearly: A. B. C. D.

150 190 240 300

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557

QUESTION 8 (DATA FROM ABOVE). The total heat transfer (MW) to the working fluid that occurs in the boiler is most nearly: A. B. C. D.

10 15 29 44

QUESTION 9. Air is to be considered as an ideal gas with the following properties:

Cp = 1.0 kJ/(kg.K) Cv = 0.718 kJ/(kg.K) k = 1.4 R = 0.287 kJ/(kg.K) One kilogram of air at 172 kPa and 100C is heated reversibly at constant volume until the pressure is 344 kPa. The specific volume of the air (kJ/kg) at State 1 is most nearly: A. B. C. D.

0.17 0.62 0.93 1.28

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558

QUESTION 10 (DATA FROM ABOVE QUESTION). The change in entropy [kJ(kg.K)] between States 1 and 2 (s2-s1) is most nearly: A. B. C. D.

-0.498 0 0.498 0.693

QUESTION 11-13. A vapor-compression refrigeration cycle using HFC-134a as the refrigerant has the pressureenthalpy diagram shown below. The evaporator temperature is 0C, and the condenser temperature is 40C.

Assume the compression process is reversible and adiabatic. If the vapor entering the compressor is saturated, the work done on the compressor (kJ/kg) is most nearly: A. B. C. D.

15 25 35 42

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559

QUESTION 12. The cooling produced by the evaporator (kJ/kg) is most nearly: A. 28 B. 143 C. 169 D. 210

QUESTION 13. The process 3-4 is: A. B. C. D.

Constant entropy Constant enthalpy Reversible Both constant entropy and enthalpy

QUESTION 14. A 200kg automobile travelling at 90km/h hits the rear of a stationary, 1000-kg automobile. After the collision that large automobile slows to 50 km/h, and the smaller vehicle has a speed of 88km/h. The increase in internal energy (kJ) taking both vehicles as the system is nearly: A. B. C. D.

100 159 176 210

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560

QUESTION 15. Four kg of water is heated at a pressure of 220 kPa to produce a mixture with quality x =0.8. The final volume (m3) occupied by the mixture is: A. B. C. D.

2.7 3.7 4.7 5.7

QUESTION 16. An automobile tire with a volume of 0.6 m3 is inflated to a gage pressure of 200 kPa. The mass of air in the tire if the temperature is 20C is nearly: A. B. C. D.

1.14 2.14 3.14 4.14

QUESTION 17. Ten pounds of steam is contained in a volume of 50 ft3. The quality of the steam if the temperature is 263 F is nearly: A. B. C. D.

0.22 0.33 0.44 0.55

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561

QUESTION 18. Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 2MPa. If the pressure is reduced to 400 kPa by cooling the vessel, the mass (kg) of vapor is: A. 6.69 B. 8.89 C. 10.69 D. 12.78

QUESTION 19. A pressurized can contains air at a gage pressure of 40 Psi when the temperature is 70F. The can will burst when the gage pressure reaches 200 psi. At what temperature (F) the can will burst? A. B. C. D.

1000 1250 1420 1620

QUESTION 20. One kg of steam with a quality of 20 percent is heated at a constant pressure of 200 kPa until the temperature reaches 400C. The work (kJ) done by the steam is close to: A. B. C. D.

125 175 250 275

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

562

QUESTION 21. Energy is added to a piston-cylinder arrangement and the piston is withdrawn in such a way that the quantity PV remains constant. The initial pressure and volume are 200 kPa and 2m3, respectively. If the final pressure is 100 kPa, the work done by the gas on the piston is nearly: A. B. C. D.

107 157 177 280

QUESTION 22. The horse power required to overcome the wind drag on a modern car travelling 25m/s if the drag coefficient is 0.2 is nearly: The drag formula is given by: FD = ½ ρV2ACD The density of air = 1.23 kg/m3 Area = 2.3 m2 A. B. C. D.

2 4 6 8

QUESTION 23. A 10-m long by 3 m high wall is composed of an insulation layer with R = 2 m 2.K/W and a wood layer with R = 0.5 m2.K/W. The heat transfer rate through the wall if the temperature difference is 40C is nearly: A. B. C. D.

240 480 670 980

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

563

QUESTION 24. The heat transfer from a 2-m diameter sphere to a 25C air stream over a time interval of one hour is 3000kJ. The surface temperature of the sphere if the heat transfer coefficient is 10 W/m2.K- is: A. B. C. D.

16 32 64 88

QUESTION 25. Air is compressed in a cylinder such that the volume changes from 100 to 10 m 3. The initial pressure is 50 psia and the temperature is held constant at 100F. The work done is close to: A. B. C. D.

250 -250 960 -960

QUESTION 26. A 5-hp fan is used in a large room to provide for air circulation. Assuming a well-insulated sealed room, the internal energy (J) increase after 1 h of operation is: A. B. C. D.

1.3 x 104 1.3 x 105 1.3 x 106 1.3 x 107

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

564

QUESTION 27. The specific heat of superheated steam at approximately 150 kPa can be determined by the equation: 𝑇 − 400 𝑘𝐽 𝐶𝑝 = 2.07 + 1480 𝑘𝑔. 𝐶 The enthalpy change between 300 C and 700 C for 3 kg of steam is nearly: A. B. C. D.

2100 2250 2350 2550

QUESTION 28. A piston-cylinder arrangement contains 0.02 m3 of air at 50C and 400kPa. Heat is added in the amount of 50kJ and the work is done by the paddle wheel until the temperature reaches 700C. if the pressure is held constant, how much paddle-wheel work (kJ) must be added to the air? A. B. C. D.

-1.5 -2.5 -5.1 -6.1

QUESTION 29. Steam enters a turbine at 4000 kPa and 500C and leaves as shown. For an inlet velocity of 200 m/s, the turbine power (MW) output is close to: A. B. C. D.

1.5 2.5 3.5 4.5

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

565

QUESTION 30. A liquid flowing at 100 kg/s, enters a heat exchanger at 450C and exists at 350C. The specific heat of the liquid is 1.25kJ/kg.C. Water enters at 5000 kPa and 20c. The minimum mass flux of the water so that the water does not completely vaporize is nearly: A. B. C. D.

1.6 2.6 3.6 4.6

QUESTION 31. A refrigerator is situated in an insulated room, it has a 2-hp motor that drives a compressor. Over a 30-minute period of time it provides 5300 kJ of cooling to the refrigerated space and 8000 kJ of heating from the coils on the back of the refrigerator. The increase in the internal energy (kJ) in the room is close to: A. B. C. D.

1200 1800 2600 4400

QUESTION 32. A 5kg block of copper at 300C is submerged in 20 liters of water at 0C contained in an insulated tank. The final equilibrium temperature is: A. B. C. D.

2.8 4.8 6.8 9.8

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

566

QUESTION 33. Helium is contained in a 2 m3 rigid volume at 50C and 200 kPa. The heat transfer (kJ) needed to increase the pressure to 800 kPa is close to: A. B. C. D.

1200 1400 1600 1800

QUESTION 34. The air in the cylinder of an air compressor is compressed from 100 kPa to 10Mpa. The work (kJ/kg) required if the air is initially at 100C is nearly: A. B. C. D.

139 259 739 980

QUESTION 35. Steam with a mass flux of 600 lbm/min exits a turbine as saturated steam at 2 psia and passes through a condenser (a heat exchanger). The mass flux of cooling water is needed if the steam is to exit the condenser as saturated liquid and the cooling water is allowed a 15F temperature rise is: A. B. C. D.

21,000 41,000 62,000 88,000

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

567

QUESTION 36. A refrigeration unit is cooling a space to -5C by rejecting energy to the atmosphere at 20C. It is desired to reduce the temperature in the refrigerated space to -25C. The minimum percentage in work required assuming a Carnot refrigerator is close to: A. B. C. D.

74 84 94 99

QUESTION 37. A power utility company desires to use the got groundwater from a hot spring to power a heat engine. If the groundwater is at 95C, the maximum power (kW) output if a mass flux of 0.2 kg/s is close to: Assume the atmosphere at 20C. A. 7 B. 13 C. 19 D. 21

QUESTION 38. Two Carnot engines operate in series between two reservoirs maintained at 600F and 100F, respectively. The energy rejected by the first engine is input into the second engine. If the first engine’s efficiency is 20 percent greater than the second engine’s efficiency, the intermediate temperature (F) is nearly: A. B. C. D.

285 326 421 555

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

568

QUESTION 39. A heat engine operates on a Carnot cycle with an efficiency of 75%. What COP would a refrigerator operating on the same cycle have? The low temperature is 0C. A. B. C. D.

0.111 0.222 0.333 0.444

QUESTION 40. Two Carnot refrigerators operate in series between two reservoirs maintained at 20C and 200C, respectively. The energy output by the first refrigerator is used as the heat energy input to the second refrigerator. If the COPs of the two refrigerators are the same, the intermediate temperature (C) is close to: A. 50 B. 100 C. 150 D. 200

QUESTION 41. Air is contained in an insulated, rigid volume at 20C and 200 kPa. A paddle-wheel inserted in the volume does 720 kJ of work on the air. If the volume is 2 m3, the entropy increase (kJ/K) assuming constant specific heats is close to: A. B. C. D.

1.1 1.3 1.6 1.9

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

569

QUESTION 42. After a combustion process in a cylinder the pressure is 1200 kPa and the temperature is 350C the gases are expanded to 140kPa with a reversible adiabatic process. The work done by the gas (kJ/kg) is close to: A. B. C. D.

100 200 300 400

QUESTION 43. Two kg of superheated steam at 400C and 600 kPa is cooled at constant pressure by transferring heat from a cylinder until the steam is completely condensed. The surroundings are at 25C. The entropy change (kJ/K) of the universe during the process is: A. 3 B. 6 C. 9 D. 12

QUESTION 44. A Carnot engine delivers 100 kW of power by operating between temperature reservoirs at 100 C and 1000C. The net entropy change (kJ/kg) of the two reservoirs after 20 min operation is nearly: A. B. C. D.

0 1 2 3

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570

QUESTION 45. Two kilograms of air is heated at constant pressure of 200 kPa to 500C. The entropy change (kJ/kg) if the initial volume is 0.8 m3 is:

A. B. C. D.

1 2 3 4

QUESTION 46. Air expands from 200 to 1000 cm3 in a cylinder while the pressure s held constant at 600 kPa. If the initial temperature is 20C, the heat transfer is nearly: A. B. C. D.

1.1 1.3 1.7 2.5

QUESTION 47. The steam in a Carnot engine is compressed adiabatically from 10 kPa to 6 MPa with saturated liquid occurring at the end of the process. If the work output is 500 kJ/kg, the quality at the end of the isothermal expansion is: A. B. C. D.

0.36 0.56 0.76 1.00

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

571

QUESTION 48. Steam expands isentropically through a turbine from 6MPa to 10 kPa. The power output (kW) if the mass flux is 2 kg/s is nearly: A. B. C. D.

2770 2915 3254 4851

QUESTION 49. An adiabatic compressor receives 20m3/min of air from the atmosphere at 20C and compresses it to 10MPa. The minimum power requirement is: A. B. C. D.

217 317 525 874

QUESTION 50. A six-cylinder engine with a compression ratio of 8 and a total volume of 600 mL intakes atmospheric air at 20C. The maximum temperature during a cycle is 1500C. Assuming an Otto cycle, the heat supplied per cycle is nearly: A. B. C. D.

1.4 2.4 3.4 4.5

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

572

QUESTION 51. Air enters the compressor of a simple gas refrigeration cycle at -10C and 100 kPa. For a compression ratio of 10 and a turbine inlet of 30C, the COP is close to:

A. B. C. D.

0.70 1.07 2.10 3.52

QUESTION 52. A rigid tank contains 2kg of nitrogen and 4kg of CO2 at a temperature of 25C and 2 MPa. The partial pressure of nitrogen is nearly:

A. B. C. D.

0.44 0.66 0.88 1.00

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

573

QUESTION 53. Gravimetric analysis of a mixture of three gases indicates 20% nitrogen, 40% CO2 and 40% oxygen. The heat transfer needed to increase the temperature of 20 lbm of the mixture from 80F to 300 F in a rigid tank is close to:

A. B. C. D.

500 700 800 900

QUESTION 57 Determine the steady state rate of heat transfer per unit area through a 4.0 cm thick homogeneous slab with its two faces maintained at uniform temperatures of 38 C and 21C. The thermal conductivity of the material is 0.19W/m.K A. B. C. D.

20 40 60 80

QUESTION 58 The forced convective heat transfer coefficient for a hot fluid flowing over a cool surface is 225 W/m2 C for a particular question. The fluid upstream of the cool surface is 120C and the surface is held at 10C. The heat transfer rate (W/m2) per unit area from the fluid to the surface is: A. B. C. D.

12,500 25,000 50,000 75,000

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

574

QUESTION 59 The air inside an electronics package housing has a temperature of 50C.A chip in this house has internal thermal power generation rate of 3x10-3W. This chip is subjected to an airflow resulting in a convective coefficient of 9W/m2 over its two main surfaces which are 0.5cm x 1.0 cm. The chip surface temperature (C) neglecting radiation and heat transfer from the edges is close to: A. B. C. D.

23 33 53 63

QUESTION.60 A laboratory furnace wall is constructed of 0.2m thick fireclay brick having Ka = 1.0 W/m.K. This is covered on the outer surface with a 0.03 m thick layer of insulating material having Kb = 0.07 W/m.K. The furnace inner brick surface is at 1250K and the outer surface of the insulation material is 310K. The interfacial temperature T2 (K) between the brick and the insulation is: A. B. C. D.

278 678 951 998

QUESTION 61 A composite three-layered wall is formed of a 0.5 cm thick aluminum plate, a 0.25 cm thick layer of sheet asbestos, and a 2cm thick layer of rock wool; the asbestos is the central layer. The outer aluminum surface is at 400C and the outer rock wool surface is at 50C. The heat flow (W) per unit area is: Kal = 249 W/m.K Kasb = 0.166 W/m.K Krw = 0.0548 W/m.K A. B. C. D.

250 400 600 920

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

575

QUESTION 62 A steel pipe (K = 45 W/m.K) having a 5 cm OD is covered with a 4.2 cm thick layer of magnesia (K = 0.07 W/m.K) which is in turn covered with a 2.4 cm layer of fiberglass insulation (K = 0.048 W/m.K). The pipe wall outside temperature is 370K and the outside surface temperature of the fiberglass is 305K. The interfacial (K) temperature between the magnesia and the fiberglass is: A. B. C. D.

130 230 330 425

QUESTION 63 A thick walled copper cylinder has an inside radius of 1cm and an outside radius of 1.8cm. The inner and outer surface temperatures are held at 305C and 295C, respectively. Assume K =367 W/m.K, the heat loss (kW) per unit length is nearly: A. B. C. D.

39 49 59 69

QUESTION 64 A 6inch concrete wall, having a thermal conductivity of 0.50 Btu/h.ft.F is exposed to air at 70F on one side and air at 20F on the opposite side. The heat transfer coefficients are 2.0 Btu/h.ft2.F on the 70F side and 10 Btu/h.ft2. F on the 20F side. The heat transfer rate (Btu/h.ft2) is: A. B. C. D.

10 20 35 45

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

576

QUESTION 65 In a food processing plant a brine solution is heated from 6C to 12C in a double-pipe heat exchanger by water entering at 50C and leaving at 40C at the rate of 0.166 kg/s. If the overall heat transfer coefficient is 859 W/m2, the heat exchanger area (m2) is required for a parallel flow is: A. 0.15 B. 0.23 C. 0.58 D. 0.75

QUESTION 66 In a double-pipe counter flow heat exchanger, water at the rate of 60 lbm/min is heated from 65F to 95F by oil having a specific heat of 0.36 Btu/lbm.F. The oil enters the exchanger at 200F and leaves at 140F. The heat exchanger area (ft2) for an overall heat transfer coefficient of 50 Btu/h.ft2.F is nearly: A. B. C. D.

12 25 37 43

QUESTION 67 When new, a heat exchanger transfers 10% more heat than it does after being in service for six months. Assuming that it operates between the same temperature differentials and that there is insufficient scale buildup to change the effective surface area, the effective fouling factor in terms of its clean (new) overall heat transfer coefficient is: A. B. C. D.

0.10 0.15 0.20 0.30 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

577

QUESTION 68 Water enters a counterflow, double-pipe heat exchanger at 38C, flowing at the rate of 0.75 kg/s. It is heated by oil (Cp =1.884 kJ/kg.K) flowing at the rate of 1.5 kg/s from an inlet temperature of 116C. For an area of 13 m2 and an overall heat transfer coefficient of 340 W/m2.K, the number of transfer units are: A. B. C. D.

0.57 1.57 2.57 3.33

QUESTION 69 The total incident radiant energy upon a body which partially reflects, absorbs and transmits radiant energy is 2200 W/m2. Of this amount, 450 W/m2 is reflected and 900 W/m2 is absorbed by the body. The transmissivity is close to: A. B. C. D.

0.186 0.386 0.486 0.666

QUESTION 70 The total emissivity power of a black body at 1000C is nearly: A. B. C. D.

100 150 200 350

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

578

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

579

SOLUTIONS

PART-I QUESTION 1 Applying the first law of thermodynamics: 𝑊 = 𝑄 − ∆𝑈 When heat is transferred to the surroundings, it is negative. 𝑊 = −140 𝑘𝐽 − 10 𝑘𝐽 = −150𝑘𝐽 QUESTION 2 From first law of thermodynamics 𝑄 = 𝑊 + ∆𝑈 𝑄 = −40𝐽 + 30𝐽 = −10𝐽 Heat is released to the surroundings. QUESTION 3 𝑄 = 𝑚𝐶𝑝 ∆𝑇 𝑘𝑔 𝐾𝐽 (95 − 𝑇1 ) 2,800,000 = 55𝑚3 𝑥1000 3 𝑥 4.18 𝑚 𝐾. 𝑘𝑔 𝑇 = 82.8𝐶 QUESTION 4 Over 1min period, the heat is given by water equals the heat loss by Freon 𝑚𝑐𝑤 𝐶𝑝 ∆𝑇 = 𝑚𝐹 [ℎ1 − ℎ2 ] 𝐾𝐽 (6𝐾) = 5[238 − 60.6] 𝑚𝑐𝑤 𝑥4.18 𝐾. 𝐾𝑔 𝑚𝑐𝑤 = 35.2 𝑘𝑔/𝑚𝑖𝑛 QUESTION 5 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝐴𝑟𝑒𝑎 𝑥 𝑑𝑒𝑝𝑡ℎ = 35 𝑚 𝑥 15𝑚 𝑥 3𝑚 = 1575𝑚3 𝑘𝑔 𝑚𝑎𝑠𝑠 = 𝑑𝑒𝑛𝑖𝑠𝑡𝑦 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒; 1000 3 𝑥1575𝑚3 = 1.575 𝑥 106 𝑘𝑔 𝑚 𝑄 = 𝑚𝐶𝑝 ∆𝑇 𝐾𝐽 = 1.575 𝑥 106 𝑘𝑔 𝑥 4.18 𝑥 (25 − 10) = 99 𝐺𝐽 𝑘𝑔. 𝐾 QUESTION 6 ∆𝑈 = 𝐶𝑣 ∆𝑇 𝐾𝐽 𝐶𝑣 = 0.718 𝐾𝑔. 𝐾 𝐾𝐽 ∆𝑈 = 0.718 𝑥 (550 − 100) = 323 𝐾𝑔 QUESTION 7 For a closed system and from first law of thermodynamics, ∆𝑈 = 𝑄 − 𝑊 𝑠 = −1500𝑊 𝑥 3600 (1ℎ) + 7.2 𝑥106 𝐽 = 1.8 𝑥 106 𝐽 = 1.8𝑀𝐽 ℎ

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

580

QUESTION 8 𝑉 1𝑚3 𝑚3 = = 0.2 𝑚 5𝑘𝑔 𝑘𝑔 ℎ = 𝑈 + 𝑃𝜗 𝐾𝐽 𝐾𝑁 𝑚3 𝐾𝐽 = 2500 + (150 2 ) (0.2 ) = 2530 𝐾𝑔 𝑚 𝑘𝑔 𝐾𝑔 𝐾𝐽 𝐻 = 𝑚𝑥ℎ = 5𝑘𝑔 𝑥 2530 = 12,650 𝐾𝐽 𝑜𝑟 12.7𝑀𝐽 𝐾𝑔 𝜗=

QUESTION 9 From ideal gas equation: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 Since volumes are equal, 𝑃1 𝑇2 (0.3𝑀𝑃𝑎 + 0.101 𝑀𝑃𝑎)(75 + 273) 𝑃2 = = = 0.468 𝑀𝑃𝑎 𝑇1 25 + 273 QUESTION 10 𝑃𝑉 = 𝑛𝑅̅ 𝑇 𝑃𝑉 120𝑃𝑎 𝑥 2 𝑥 10−6 𝑚3 𝑛= = = 1.155 𝑥 10−10 𝐾𝑚𝑜𝑙 𝑅̅ 𝑇 8314 𝑚3 . 𝑃𝑎 𝑥250𝐾 𝐾. 𝐾𝑚𝑜𝑙 Number of molecules: 𝑘𝑔 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 1.155 𝑥 10−10 𝐾𝑚𝑜𝑙 𝑥 100 𝑥 6.023 𝑥 1023 = 7𝑥1016 𝑚𝑜𝑙𝑢𝑐𝑢𝑙𝑒𝑠 𝐾𝑚𝑜𝑙 𝑚𝑜𝑙𝑒 QUESTION 11 𝑃𝑉 = 𝑚𝑅𝑇 𝐾𝐽 𝑚𝑅𝑇 40𝑘𝑔 𝑥 0.2598 𝐾𝑔. 𝐾 𝑥 (40 + 273)𝐾 𝑃= = = 1627 𝐾𝑝𝑎 𝑉 2 𝑚3 𝑃(𝑔𝑎𝑔𝑒) = 𝑃𝑎𝑏𝑠 − 𝑃𝑎𝑡𝑚 = 1627 𝐾𝑃𝑎 − 101.3 𝐾𝑝𝑎 = 1530 𝐾𝑃𝑎 𝑜𝑟 1.53 𝑀𝑃𝑎 QUESTION 12 𝑃𝑉 = 𝑚𝑅̅ 𝑇 250 𝑥 𝑉 = 3 𝑥 0.287 𝑥 323 𝑜𝑟 𝑉 = 1.11𝑚3 QUESTION 13 𝑃𝑉 = 𝑚𝑅̅ 𝑇 𝑚 𝑃 =𝜌 = 𝑉 𝑅̅ 𝑇 𝐾𝑃𝑎 1 𝑎𝑡𝑚𝑥 101.3 𝑎𝑡𝑚 𝐾𝑔 = 0.404 3 𝐾𝐽 𝑚 0.2870 𝐾𝑔. 𝐾 𝑥 (600 + 273)

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

581

QUESTION 14 Volume of balloon at 20C: 4 4 0.3𝑚 3 𝑉 = 𝜋𝑟 3 = 𝜋 ( ) = 0.01414𝑚3 3 3 2 Volume of balloon at T: =

4 0.305𝑚 3 𝜋( ) = 0.01486𝑚3 3 2

Since pressure is constant 𝑇2 =

𝑉2 0.01486𝑥(20 + 273) 𝑇1 = = 308𝐾 𝑉1 0.01414 𝑇2 = 308 𝐾 − 273𝐾 = 35𝐶

QUESTION 15 Moles of carbon dioxide: 0.30 = 0.00682 44 Moles of nitrogen: 0.70 = 0.25 28 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 = 0.03182 Mole fraction of carbon dioxide: =

0.00682 𝑥 100 𝐾𝑃𝑎 = 21.4 𝐾𝑃𝑎 0.03182

QUESTION 16 Component MW Mole Fraction Weight (Kg) Wt (%) Oxygen 32 0.21 6.72 23.2 Nitrogen 28 0.78 21.80 75.4 Argon 40 0.01 0.42 1.4 QUESTION 17 For nitrogen: 𝑃𝑉 = 𝑛𝑅̅ 𝑇 150 𝐾𝑃𝑎 𝑥 0.75𝑚3 𝑛= = 0.0432 𝐾𝑚𝑜𝑙 𝐾𝐽 8.3134 𝑥(40 + 273)𝐾 𝐾𝑚𝑜𝑙. 𝐾 For oxygen: 2𝑘𝑔 = 0.0625 𝐾𝑚𝑜𝑙 32𝑘𝑔 Mole fraction of nitrogen: 0.0432 𝑥100 = 41% 0.0432 + 0.0625 QUESTION 18 Mole fraction of oxygen: 0.2315 32 = = 0.209 0.2315 0.7685 32 + 28 Partial pressure: FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

582

= 𝑥𝑖 𝑃 = 0.209 𝑥 101 𝐾𝑃𝑎 = 21.1𝐾𝑃𝑎 QUESTION 19 𝐶3 𝐻8 + 5𝑂2 = 3𝐶𝑂2 + 4𝐻2 𝑂 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 3𝑥44 + 4𝑥18 𝑥 = = 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 3𝑥12 + 8𝑥1 11𝑔 X = 51g

QUESTION 20 𝑇𝐻 − 𝑇𝐿 200 − 20 = = 0.381 𝑇𝐻 473 𝐾𝐽 𝐾𝐽 𝑊 = 𝜂𝑡ℎ 𝑄𝑖𝑛 = 0.381 𝑥 1000 = 381 𝑘𝑔 𝐾𝑔 𝜂𝑡ℎ =

QUESTION 21 The maximum efficiency is achieved with Carnot engine. 𝑇𝐿 = 170 + 273 = 443𝐾 𝑎𝑛𝑑 𝑇𝐻 = 833 + 273 = 1106𝐾 𝑇𝐻 − 𝑇𝐿 1106 − 443 𝜂𝑡ℎ = = = 60% 𝑇𝐻 1106 QUESTION 22 𝑇𝐿 256 𝐶𝑂𝑃 = = = 4.56 𝑇𝐻 − 𝑇𝐿 311 − 256 𝑞𝐿 𝐶𝑂𝑃 = ; 𝑞𝐿 = 4.56 𝑥 2.2 = 10.2 𝐾𝑊 𝑊 QUESTION 23 The efficiency of Otto cycle is: 𝜂 = 1 − 𝑟 1−𝑘 = 1 − 101−1.4 = 0.602 𝑜𝑟 60% QUESTION 24 The COP is the ratio of heat transfer to work input. In SI system, one tone of refrigeration corresponds to 3516 W 𝑊 𝑞𝐿 3516 𝑡𝑜𝑛 𝐶𝑂𝑃 = = = 2.0 𝑊 1750 𝑊 𝑡𝑜𝑛 QUESTION 25 𝑄𝑛𝑒𝑡 100 𝐾𝐽 − 37 𝐾𝐽 𝜂= = = 63% 𝑄𝑖𝑛 100 𝐾𝐽 For Carnot cycle, thermal efficiency is: (370 + 273) − (𝑇𝐿 + 273) 𝑇𝐻 − 𝑇𝐿 𝜂𝑡ℎ = = 0.63 = 𝑇𝐻 (370 + 273) 𝑇𝐿 = −35𝐶

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QUESTION 26 𝐶𝑂𝑃𝑖𝑑𝑒𝑎𝑙 =

𝑇𝐿 −5𝐶 + 273 = = 4.87 (50 + 273) − (−5 + 273) 𝑇𝐻 − 𝑇𝐿 𝑞𝐿 𝑞𝐿 𝐶𝑂𝑃𝑎𝑐𝑡𝑢𝑎𝑙 = ; 𝑊 = 𝑊 𝐶𝑂𝑃𝑎𝑐𝑡𝑢𝑎𝑙 3 𝐾𝑊 = = 0.77𝐾𝑊 0.8𝑥4.87

QUESTION 27 The upper limit for the COP of a heat pump is set by the COP of a Carnot heat pump. 𝑇𝐻 21 + 273 𝐶𝑂𝑃 = = = 21 (21 + 273) − (7 + 273) 𝑇𝐻 − 𝑇𝐿 QUESTION 28 The COP of a refrigeration cycle is: 𝑇𝐿 −11 + 273 𝐶𝑂𝑃 = = = 7.9 (22 + 273) − (−11 + 273) 𝑇𝐻 − 𝑇𝐿 QUESTION 29 The heat addition occurs during the isothermal process at high temperatures 𝑄 300𝐾𝑊 𝐾𝑊 Δ𝑆 = = = 0.524 𝑇0 (300 + 273)𝐾 𝐾 QUESTION 30 𝑇𝐻 − 𝑇𝐶 0.5𝑥(3𝑥2)(60) 𝑄 = 𝐾𝐴 = = 900𝑊 𝑥 0.2 QUESTION 31 𝑄 = 𝜎𝐴𝜀[𝑇14 − 𝑇24 ] = 5.67 𝑥10−8 𝑥5𝑚2 𝑥0.6[(600 + 273)4 − (25 + 273)4 ] = 9.75 𝑥104 𝑊 Heat flux: 𝑄 9.75𝑥104 𝑊 𝑊 = = 1.95 𝑥104 2 2 𝐴 5𝑚 𝑚 QUESTION 32 𝑄 = ℎ𝐴[𝑇𝐻 − 𝑇𝑐 ] = 2.2 𝑥 104 𝑥(𝜋𝑥0.05𝑥2)(100 − 20) = 5.53 𝑥105 𝑊 𝑄 5.53 𝑥105 𝑊 𝑀𝑊 = = 1.76 2 (𝜋𝑥0.05𝑥2) 𝐴 𝑚 QUESTION 33 𝑇1 − 𝑇2 𝑇1 − 𝑇2 𝑄 = 𝐾𝐴 𝑜𝑟 𝑄𝑡 = 𝐾𝐴𝑡 𝐿 𝐿 4.18𝐽 100 𝑐𝑎𝑙 𝑥 𝑥 1𝑚 𝑐𝑎𝑙 𝑡= = 21 𝑚𝑖𝑛 1𝑚 2 𝑊 (100𝑐𝑚 ) (388 𝑚𝐾 ) (0.0839)𝑥 60 𝑠/ min 𝑥 100

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QUESTION 34 𝑄 = 𝜎𝐴𝜀𝑇 4 1𝑚 2 −8 2 = (1)( 5.67 𝑥10 )(2 𝑐𝑚 ) ( ) (900 + 273)4 = 28𝑊 100𝑐𝑚 𝑠 𝑄 = 𝑄𝑡 = 28𝑊 𝑥 1ℎ 𝑥 3600 = 100 𝐾𝐽 ℎ QUESTION 35 The plate has two surfaces which can radiate. 𝑄 = 𝜎𝐴𝜀[𝑇14 − 𝑇24 ] −8 = 5.67 𝑥10 𝑥 2𝑥(3𝑚 𝑥 3𝑚)𝑥0.13[(500 + 273)4 − (25 + 273)4 ] = 46𝐾𝑊 QUESTION 36 𝑄 = 𝜎𝐴𝜀𝑇 4 𝑄 𝑊 = 𝜎𝜀𝑇 4 = (0.95)(5.67 𝑥10−8 )(50 + 273)4 = 586 2 𝐴 𝑚 QUESTION 37 The heat transfer area: 𝜋𝑥60𝑚𝑚 𝐴 = 𝜋𝑥𝑑𝑥𝐿 = 𝑥30𝑚 = 0.565 𝑚2 1000𝑚𝑚/𝑚 𝑄 = ℎ𝐴 [𝑇𝑎𝑖𝑟 − 𝑇𝑤𝑎𝑙𝑙 ] = 20.1𝑥0.565[100 − 20] = 910𝑊 QUESTION 38 𝐶𝑝 𝑃1 𝑉1𝑘 = 𝑃2 𝑉2𝑘 ; 𝑤ℎ𝑒𝑟𝑒 𝑘 = 𝐶𝑣 𝑘 1.4 20𝑥(1) = 100𝑥(𝑉2 ) 𝑉2 = 0.317 𝑚3 QUESTION 39 𝑃2 𝑉2 − 𝑃1 𝑉1 𝑤= 1−𝜂 [150𝑥0.15 − 15𝑥1.0] = = −35.7 𝐾𝐽 1 − 1.21 Which is work done on the gas QUESTION 40 For a closed isentropic system: 𝑇1 𝑃2 1−𝑘 =[ ] 𝑘 𝑇2 𝑃1 1 1−1.4 𝑇2 = 343 ( ) 1.4 = 385.1 𝑜𝑟 112𝐶 1.5 QUESTION 41 𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛 𝑉21.35 =

2𝑀𝑃𝑎 (0.5𝑚3 )1.35 = 2.752 𝑚3 0.2𝑀𝑃𝑎

For a polytropic closed system: (0.2𝑥2.751) − (2𝑥0.5) 𝑃2 𝑉2 − 𝑃1 𝑉1 𝑤= = = 1.285𝑀𝐽 1−𝑛 1 − 1.35 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

585

QUESTION 42 𝑇2 𝑉1 = [ ]𝑘−1 𝑇1 𝑉2 10 1.4−1 𝑇2 = (35 + 273) ( ) = 773𝐾 1 QUESTION 43 The work for an open polytropic process: 𝜂[𝑃2 𝑉2 − 𝑃1 𝑉1 ] 1.3[(300𝑥1.5) − (400𝑥1.2) 𝑤= = = 130𝐾𝐽 1−𝜂 1 − 1.3 QUESTION 44 𝑇2 𝑉2 Δ𝑆 = 𝐶𝑣 ln ( ) + 𝑅 ln( ) 𝑇1 𝑉1 ̅ 𝑇2 𝑅 𝑃1 𝑇2 = 𝐶𝑣 ln ( ) + ln[ ] 𝑇1 𝑀𝑊 𝑃2 𝑇1 170 + 273 8.314 ln[100𝑥(170 + 273)] 𝐾𝐽 = 0.312𝑥𝑙𝑛 [ ]+( ) = −0.14 [500𝑥(30 + 273)] 30 + 273 40 𝐾𝑔𝐾

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PART-II QUESTION 1. A nitrogen balance ties the flue gas to the air: N2 in the flue gas: = (100𝑘𝑀𝑜𝑙 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠)(

0.844𝑘𝑚𝑜𝑙 𝑁2 ) = 84.4 𝑘𝑚𝑜𝑙 𝑁2 𝑘𝑚𝑜𝑙 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠

N2 in air: = (𝑌 𝑘𝑚𝑜𝑙 𝑎𝑖𝑟)(0.70 𝑘𝑚𝑜𝑙

𝑌=

𝑁2 = 𝑁2 𝑖𝑛 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠 𝑘𝑚𝑜𝑙 𝑖𝑛 𝑎𝑖𝑟

84.4 𝑘𝑚𝑜𝑙 𝑁2 = 107 𝑘𝑚𝑜𝑙 𝑎𝑖𝑟 0.79

QUESTION 2. Propane is burned with oxygen according to the following raction: 𝐶3 𝐻8 + 5𝑂2 → 3𝐶𝑂2 + 4 𝐻2 𝑂 Since there are x moles of air furnished, if 1 mole of propane is burned, the O2 remaining will be: 0.21x -5 QUESTION 3. Given the following: 1. Rigid walled cylinder, 30L volume 2. Initial state of air in the cylinder: 10 MPa, 600K 3. Final state of air in the cylinder: ?MPa, 300K Volume of gas and number of moles of gas are fixed. Initially at 10MPa, 600K, ν = 0.018 m3/kg Final state: ν = 0.018 m3/kg, T = 300K Interpolate between 4 and 6 MPa in the table. P is proportional to 1/ν 𝑃 =4+

0.018 − 0.0214 = 4.94 𝑀𝑃𝑎 0.0142 + 0.0214

QUESTION 4. FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

587

Simpson’s rule is under Numerical methods in the Mathematics section of the FE Reference Handbook. Options A and B are rectangular integration. Option D is the Trapezoidal rule. Therefore, the correct answer is Choice B.

QUESTION 5. Referring to the Thermodynamics section of the FE Reference Handbook, ∆Hvap of steam at 150C is 2,114.3 kJ/kg. 225𝑘𝑊 𝑥

𝑘𝐽 1𝑘𝑔 𝑥 = 0.106 𝑘𝑔/𝑠 𝑘𝑊. 𝑠 2,114.3 𝑘𝐽

QUESTION 6. Given: 1. Closed system 2. Only boundary expansion work 𝑊𝑜𝑟𝑘 = ∫ 𝑃 𝑑𝑣 With a constant volume path, dv = 0 So the work done is zero. QUESTION 7. ν1 = 0.001 m3/kg The pump power is: 𝑊𝑝 =

𝑚𝜈1 (𝑃2 − 𝑃1 ) 𝜂𝑝

50,000𝑘𝑔 0.001𝑚3 (14,0000100)𝑘𝑁/𝑚2 ) 3600𝑠 ( 𝑘𝑔 ) = = 241.3 𝑘𝑊 0.80 QUESTION 8. 𝑄 = 𝑚(ℎ3 − ℎ2 ) =

50,000𝑘𝑔 (3,322 − 167.6)𝑘𝐽 = 43,811𝑘𝑊 = 44𝑀𝑊 3600𝑠 𝑘𝑔

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QUESTION 9. The specific volume is found using the ideal gas equation of state. 𝑝𝜈 = 𝑅𝑇 0.287𝑘𝐽 𝑅𝑇 ( 𝑘𝑔. 𝐾 ) (273 + 100)𝐾 𝑚3 𝜈= = = 0.622 𝑝 172 𝑘𝑃𝑎 𝑘𝑔 QUESTION 10. The change in Entropy is found from the equations in the Thermodynamics section of the FE Reference Handbook. 𝑇2 𝑃2 Δ𝑠 = 𝑐𝑝 ln ( ) − 𝑅 ln ( ) 𝑇1 𝑃1 For a constant volume process, 𝑃1 𝑃2 = 𝑇1 𝑇2 Therefore, 𝑃2 𝑇2 = 𝑃1 𝑇1 𝑃2 𝑃2 Δ𝑠 = 𝑐𝑝 ln ( ) − 𝑅 ln ( ) 𝑃1 𝑃1 𝑃2 𝑃2 = (𝑐𝑝 − 𝑅) ln ( ) = 𝑐𝑣 ln ( ) 𝑃1 𝑃1 𝑘𝐽 = 0.718 ln(2) = 0.498 𝑘𝑔. 𝐾 QUESTION 11. From the P-h Diagram for Refrigerant HFC-134a given in the Thermodynamics section of the FE Supplied-Reference Handbook: h1 = 400 kJ/kg h2 = 425 kJ/kg h3 = h4 = 257 kJ/kg 𝑊 25𝑘𝐽 = (ℎ2 − ℎ1 ) = (425 − 400) = 𝑚 𝑘𝑔 QUESTION 12. Evaporator cooling, 𝑄𝑒𝑣𝑎𝑝 𝑘𝐽 = (ℎ1 − ℎ4 ) = (400 − 257) = 143 𝑚 𝑘𝑔 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

589

QUESTION 13. The process from 3 to 4 is a throttling process, which involves both a drop in pressure and temperature. It is an irreversible process, generally considered a constant enthalpy. QUESTION 14. Change all speeds to m/s. The kinetic energy before the collision is: 1 1 𝐾𝐸1 = 𝑚𝑎 𝑣𝑎2 = 𝑥 2200 𝑥252 = 687,500 𝐽 2 2 After the collision, the kinetic energy is: 1 1 1 1 𝐾𝐸2 = 𝑚𝑎 𝑣𝑎2 + 𝑚𝑏 𝑣𝑏2 = 𝑥 2200 𝑥 13.892 + 𝑥 1000 𝑥 24.442 = 510,900 𝐽 2 2 2 2 The conservation of energy requires that 𝐾𝐸1 + 𝑈1 = 𝐾𝐸2 + 𝑈2 𝑈2 − 𝑈1 = 687500 − 510 900 = 176, 600 𝑜𝑟 176.6 𝑘𝐽 QUESTION 15. To determine the steam volume, we linearly interpolate between 0.2 and 0.3 MPa. This provides, at 220 kPa, 220 − 200 𝑚3 𝑚3 𝑣𝑔 = ( ) (0.6508 − 0.8857) + 0.8857 = 0.8297 , 𝑣𝑓 = 0.0011 300 − 200 𝑘𝑔 𝑘𝑔 𝑚3 𝑣 = 𝑣𝑓 + 𝑥( 𝑣𝑔 − 𝑣𝑓 ) = 0.0011 + 0.8((0.8927 − 0.0011) = 0.6640 𝑘𝑔 The total volume occupied by 4 kg is 𝑚3 𝑉 = 𝑚𝑣 = 4 𝑘𝑔 𝑥 0.6640 = 2.656𝑚3 𝑘𝑔 QUESTION 16. In the ideal gas equation, we use absolute pressure and absolute temperature. P = 200 +103 = 303 kPa T = 20 +273 =293 K The mass is then calculated as: 300,000𝑁 𝑥 0.6𝑚3 𝑃𝑉 2 𝑚 𝑚= = = 2.14 𝑘𝑔 𝑅𝑇 (287𝑁. 𝑚 . 𝐾) (293𝐾) 𝑘𝑔

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QUESTION 17. We interpolate between temperatures of 260F and 270F to find: 3 𝑣𝑔 = ( ) (10.066 − 11.768) + 11.768 = 11.257, 𝑣𝑓 = 0.017 10 From the given information we calculate: 𝑉 50 𝑓𝑡 3 𝑣= = = 5.0 𝑚 10 𝑙𝑏𝑚 The quality is found as: 5 = 0.017 + 𝑥(11.257 − 0.017) = 0.4433 QUESTION 18. The initial specific volume is found as: 𝑚3 𝑘𝑔 Since the vessel is rigid, the specific volume does not change. Hence the specific volume at a pressure of 400 kPa is also 0.07994. We can then find the quality as follows: 𝑣 = 𝑣𝑓 + 𝑥(𝑣𝑔 − 𝑣𝑓 ) = 0.00118 + 0.8(0.09963 − 0.00118) = 0.07994

0.07994 = 0.0011 + 𝑥(0.4625 − 0.0011) = 0.1709 The total mass of water is: 𝑚 =

𝑉 𝑣

5

= 0.07994 = 62.55𝑘𝑔

The mass of vapor: 𝑚𝑔 = 𝑥𝑚 = (0.1709)(62.55) = 10.69 𝑘𝑔

QUESTION 19. We will assume the volume to remain constant a the temperature increases. 𝑇1 𝑇2 = 𝑃1 𝑃2 𝑇2 = 𝑇1

(200 + 14.7)(144) 𝑃2 = (70 + 460) = 2080 𝑅 𝑜𝑟 1620 𝐹 (40 + 14.7)(144) 𝑃1

QUESTION 20. The work is given by: 𝑊 = ∫ 𝑃 𝑑𝑉 = 𝑃(𝑉2 − 𝑉1 ) = 𝑚𝑃(𝑣2 − 𝑣1 ) To evaluate work we must know, v1 and v2. 𝑚3 𝑘𝑔 From the superheated table we locate state 2 at T2 = 400C and P2 = 0.2 MPa; v2 = 1.549 m3/kg. 𝑣1 = 𝑣𝑓 + 𝑥(𝑣𝑔 − 𝑣𝑓 ) = 0.001061 + 0.2(0.8857 − 0.001061) = 0.1780

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The work is then: 𝑊 = (1)(200)(1.549 − 0.1780) = 274.2 𝑘𝐽 QUESTION 21. The work is found from: 𝑉2

𝑉2

𝐶 𝑑𝑉 2 2 𝑉 Where, we have used PV =C. To calculate the work we must find C and V2. The constant C is found from 𝐶 = 𝑃1 𝑉1 = (200)(2) = 400 𝑘𝐽 To find V2 we use, P2V2 = P1V1, the equation that would result from an isothermal process involving an ideal gas. This can be written as: (200)(2) 𝑉2 = = 4𝑚3 100 The work done is: 4 400 ln 4 𝑊= ∫ 𝑑𝑉 = 400 = 277 𝑘𝐽 2 2 𝑉 QUESTION 22. The drag force is: 1 = 𝑥 1.23 𝑥 252 𝑥2.3 𝑥 0.2 = 177𝑁 2 To move this drag force at 25 m/s, the engine must do the work at the rate: 𝑊 = 𝐹𝐷 𝑉 = 177 𝑥 25 = 4425 𝑊 The horsepower is: 4425𝑊 = 5.93ℎ𝑝 746 𝑊/ℎ𝑝 𝑊 = ∫ 𝑃𝑑𝑉 = ∫

QUESTION 23. The total resistance to heat flow through the wall is: 𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 + 𝑅𝑤𝑜𝑜𝑑 = 2 + 0.5 = 2.5 𝑚2 . 𝐾/𝑊 The heat transfer rate is then: 𝑄=

𝐴 𝑅𝑡𝑜𝑡𝑎𝑙

∆𝑇 =

10𝑥3 𝑥 40 = 480𝑊 2.5

QUESTION 24. The heat transfer is: 𝑄 = ℎ𝑐 𝐴(𝑇𝑠 − 𝑇∞ )∆𝑡 3 𝑥 106 = 10 𝑥 4𝜋𝑥(1)2 (𝑇𝑠 − 25)𝑥 3600 = 31.6𝐶 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 25. The work is given by: 𝑊 = ∫ 𝑃𝑑𝑉 For the isothermal process, the equation of state allows us to write 𝑃𝑉 = 𝑚𝑅𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since the mass m, the gas constant R and the temperature T are all constant. Therefore, 𝑉2

𝑊 = 𝑃1 𝑉1 ∫ 𝑉1

𝑑𝑉 100 10 = (50)(144) ( ) ln ( ) = −959𝑓𝑡 − 𝑙𝑏𝑓 𝑉 1728 100

QUESTION 26. By assumption, Q = 0, with ∆PE = ∆KE =0, the first law becomes, -W = ∆U The work input is: 𝑊 𝑠 𝑊 = (−5ℎ𝑝)(1ℎ) (746 ) (3600 ) = −1.343 𝑥 107 𝐽 ℎ𝑝 ℎ The negative sign results because the work is input to the system. Finally the internal energy increase is: ∆𝑈 = −(− − 1.343 𝑥 107 𝐽) = 1.343 𝑥 107 𝐽 QUESTION 27. The enthalpy change is found to be: 𝑇2

700

∆𝐻 = 𝑚 ∫ 𝐶𝑝 𝑑𝑇 = 3 ∫ 𝑇1

300

[2.07 +

𝑇 − 400 ] 𝑑𝑇 = 2565𝑘𝐽 1480

QUESTION 28. The first law may be written as: 𝑄 − 𝑊𝑝𝑎𝑑𝑑𝑙𝑒 = 𝑚(ℎ2 − ℎ1 ) = 𝑚𝐶𝑝 (𝑇2 − 𝑇2 ) To find m we use the ideal gas equation. It gives us: 𝑃𝑉 400,000𝑥0.02 𝑚= = = 0.0863 𝑘𝑔 𝑅𝑇 (287)(273 + 50) From the first law the paddle-wheel work is found to be: 𝑊𝑝𝑎𝑑𝑑𝑙𝑒 = 𝑄 − 𝑚𝐶𝑝 (𝑇2 − 𝑇2 ) = 50 − (0.0863)(1.00)(700 − 50) = −6.095𝑘𝐽

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QUESTION 29. The energy equation in the form: 𝑊𝑇 − (ℎ2 − ℎ1 )𝑚 200 𝜋(0.025)2 (0.08643) 1 𝑘𝑔 𝑚 = 𝜌1 𝐴1 𝑉1 = 𝐴1 𝑉1 = 4.544 𝑣1 0.08643 𝑠 The enthalpies are found from the table: 𝑘𝐽 𝑘𝐽 ℎ1 = 3445.2 , ℎ2 = 2665.7 𝑘𝑔 𝑘𝑔 The maximum power output is then: 𝑘𝐽 −(2665.7 − 3445.2)(4.544) = 3542 , 3.542 𝑀𝑊 𝑠 QUESTION 30. The energy equation is: 𝑚𝑠 𝐶𝑝 (𝑇𝑠1 − 𝑇𝑠2 ) = 𝑚𝑤 (ℎ𝑤1 − ℎ𝑤2 ) Using the given values, and using steam tables: 100(1.25)(450 − 350) = 𝑚𝑤 (2792.8 − 88.7) 𝑚𝑤 = 4.623

𝑘𝑔 𝑠

QUESTION 31. For an insulated room (Q=0), the first law provides: 𝑄 − 𝑊 = ∆𝑈 𝑘𝑊 ∆𝑈 = −(−2ℎ𝑝) (0.746 ) (1800 𝑠) = 2686 𝑘𝐽 ℎ𝑝 QUESTION 32. Conservation of energy requires that the energy lost by the copper block is gained by the water. This is expressed as: 𝑚𝑐 𝐶𝑝,𝑐𝑜𝑝𝑝𝑒𝑟 (∆𝑇)𝑐 = 𝑚𝑤 𝐶𝑝,𝑤𝑎𝑡𝑒𝑟 (∆𝑇)𝑤 Using the values from the tables in FE-Reference handbook (5)(0.30)(300 − 𝑇2 ) = (0.02)(1000)(4.18)(𝑇2 − 0) Solving for T2: 𝑇2 = 6.84𝐶

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QUESTION 33. The work is zero for this constant volume process. The first law gives: 𝑄 = 𝑚∆𝑢 = 𝑚𝐶𝑣 ∆𝑇 =

𝑃𝑉 𝐶 (𝑇 − 𝑇1 ) 𝑅𝑇 𝑣 2

The ideal gas PV = mRT allows us to write: 𝑃1 𝑃2 200 800 = ; = ; 𝑇2 = 1292𝐾 𝑇1 𝑇2 323 𝑇2 The heat transfer is given by: (200)(2) (3.116)(1292 − 323) = 1800 𝑘𝐽 𝑄= (2.077)(323) QUESTION 34. The final temperature in the adiabatic process is: 1.4−1

𝑃2 𝑘−1 10,000 1.4 𝑇2 = 𝑇1 [ ] 𝑘 = (373) ( ) = 1390𝐾 𝑃1 100 The work is found by using the first law with Q = 0 𝑊 = −∆𝑢 = −𝑐𝑣 (𝑇2 − 𝑇1 ) = −(0.717)(1390 − 373) = −729 𝑘𝐽/𝑘𝑔 QUESTION 35. The heat transfer rate for the steam is assuming no pressure drop through the condenser is: 𝑄 = 𝑚𝑠 (ℎ𝑠2 − ℎ𝑠1 ) = 600(94.02 − 1116.1) = −613,200 𝐵𝑡𝑢/𝑚𝑖𝑛 The energy is gained by the water. Hence: 𝑄𝑤 = 𝑚𝑤 𝐶𝑝 (𝑇𝑤2 − 𝑇𝑤1 ) = 613,200 = 𝑚𝑤 (1.00); 𝑚𝑤 = 40,880 𝑙𝑏𝑚/𝑚𝑖𝑛 QUESTION 36. For a Carnot refrigerator: 𝑄𝐿 1 𝐶𝑂𝑃 = = 𝑇𝐻 𝑊 −1 𝑇𝐿 For the first situation, we have 𝑇𝐻 293 𝑊1 = 𝑄𝐿 ( − 1 ) = 𝑄𝐿 ( − 1) = 0.0933𝑄𝐿 𝑇𝐿 268 For the second situation: 293 𝑊2 = 𝑄𝐿 ( − 1) = 0.181𝑄𝐿 248

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The percentage increase in work is then: 𝑊2 − 𝑊1 0.181𝑄𝐿 − 0.0933𝑄𝐿 = 𝑥 100 = 94% 𝑊1 0.0933𝑄𝐿 QUESTION 37. The maximum possible efficiency is: 𝑇𝐿 293 =1− = 0.2038 𝑇𝐻 368 Assuming the water is rejected at atmospheric temperature. The rate of heat transfer from the energy source is: 𝑄 = 𝑚𝐶𝑝 Δ𝑇 = (0.2)(4.18)(95 − 20) = 62.7 𝑘𝑊 The maximum power output is: 𝑊 = 𝜂𝑄 = (0.2038)(62.7) = 12.8 𝑘𝑊 𝜂 =1−

QUESTION 38. The efficiencies of the two engines are: 𝑇 560 ; 𝜂2 = 1 − 1060 𝑇 Where T is the unknown intermediate temperature in R. It is given that: 𝜂1 = 1 −

𝜂1 = 𝜂2 + 0.2𝜂2 Substituting for 𝜂1 𝑎𝑛𝑑 𝜂2 we have: 𝑇 560 = 1.2[1 − ] 1060 𝑇 𝑇 2 + 212𝑇 − 712,320 = 0; 𝑇 = 744.6𝑅 𝑜𝑟 285𝐹 1−

QUESTION 39. The efficiency of a heat engine is given by: 𝑇𝐿 𝑇𝐻 𝑇𝐿 273 𝑇𝐻 = = = 1092𝐾 1 − 𝜂 1 − 0.75 𝜂 = 1−

The COP for the refrigerator: 𝐶𝑂𝑃𝑅 =

𝑇𝐿 273 = = 0.333 𝑇𝐻 − 𝑇𝐿 1092 − 273

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QUESTION 40. The COP for a refrigerator is given by: 𝐶𝑂𝑃 =

𝑇𝐿 𝑇𝐻 − 𝑇𝐿

Requiring that the two COPs be equal gives: 293 𝑇 = 𝑇 − 293 473 − 𝑇 𝑇 2 = 138589; 𝑇 = 100𝐶 QUESTION 41. To determine the final state of the process we use the energy equation, assuming zero heat transfer. We have –W = ∆U = mCp∆T. The mass m is found from the deal-gas equation as: (200)(2) 𝑃𝑉 𝑚= = = 4.76𝑘𝑔 𝑅𝑇 (0.287)(293) From the first law, taking the paddle wheel work as negative is then: 720 = (4.76)(0.717)(𝑇2 − 293), 𝑇2 = 504 𝐾 For the constant volume process: 𝑇2 504 Δ𝑠 = 𝑚𝐶𝑣 ln = (4.76)(0.717) ln ( ) = 1.851 𝑘𝐽/𝑘𝑔 𝑇1 293 QUESTION 42. The first law can be applied with zero heat transfer to give −𝑊 = Δ𝑢 = 𝐶𝑣 (𝑇2 − 𝑇1 ) The temperature T2 is found from: 1.4−1 1.4

𝑃2 𝑘−1 140 𝑇2 = 𝑇1 [ ] 𝑘 = 623 ( ) 𝑃1 1200

= 337𝐾

The work is: (0.717(623 − 337) = 205 𝑘𝐽/𝑘𝑔

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QUESTION 43. The entropy of the steam which defines our system decreases since the heat is transferred from the systems to the surroundings. From the steam tables this change is found to be: 𝑘𝐽 𝑘𝑔 The heat transfer to the surrounds occurs at constant temperature. Hence the entropy change of the surrounds is: 𝛿𝑄 𝑄 Δ𝑠𝑠𝑢𝑟𝑟 = ∫ = 𝑇 𝑇 The heat transfer for the constant-pressure process is: 𝑄 = 𝑚Δℎ = 2(3270.2 − 670.6) = 5199 𝑘𝐽 Δ𝑠𝑠𝑦𝑠 = 𝑚(𝑠2 − 𝑠1 ) = (2)(1.9316 − 7.7086) = −11.55

Giving Δ𝑠𝑠𝑢𝑟𝑟 =

5199 298

= 17.45

𝑘𝐽 𝑘

Δ𝑠𝑢𝑛𝑖𝑣 = Δ𝑠𝑠𝑢𝑟𝑟 + Δ𝑠𝑠𝑦𝑠 = 17.45 − 11.55 = 5.90

𝑘𝐽 𝑘

QUESTION 44. The efficiency of the engine is given by: 𝑇𝐿 373 𝜂 = 1− =1− = 0.7070 𝑇𝐻 1273 The high-temperature heat transfer is then: 𝑊 100 𝑄𝐻 = = = 141.4 𝑘𝑊 𝜂 0.7070 The low-temperature heat transfer is: 𝑄𝐿 = 𝑄𝐻 − 𝑊 = 141.4 − 100 = 41.4 𝑘𝑊 The entropy change of reservoirs are then (141.4)(20)(60) 𝑄𝐻 𝑄𝐻 Δ𝑇 𝑘𝐽 = − =− = −133.3 𝑇𝐻 𝑇𝐻 1273 𝐾 (41.4)(20)(60) 𝑄𝐿 𝑄𝐿 Δ𝑇 Δ𝑠𝐿 = = − =− = 133.2 𝑘𝐽/𝐾 𝑇𝐿 𝑇𝐿 373 The net entropy change of the two reservoirs is zero. Δ𝑠𝐻 =

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QUESTION 45. The initial temperature is: 𝑇1 =

𝑃1 𝑉1 (200)(0.8) = = 278.7𝐾 (2)(0.287) 𝑚𝑅

The entropy change can be found: 𝑇2 273 𝑘𝐽 Δ𝑠 = 𝑚 [𝐶𝑝 ln − 𝑅 ln(1)] = 2(1.00) ln = 2.040 𝑇1 278.7 𝑘𝑔 QUESTION 46. The air mass is: 𝑚=

𝑃𝑉 (600)(200 𝑥10−6 ) = = 0.001427 𝑘𝑔 (0.287)(293) 𝑅𝑇

The final temperature is found using ideal gas: 𝑇1 𝑉2 1000 = (293) ( ) = 1465 𝐾 𝑉1 200 The heat transfer (constant pressure process): 𝑇2 =

𝑄 = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) = (0.001427(1465 − 293) = 1.672

𝑘𝐽 𝑘𝑔

QUESTION 47. For a cycle, the work output equals the net heat input so that: 𝑊 = Δ𝑇Δ𝑠 500 = (275.6 − 45.8)(𝑠2 − 3.0273), 𝑠2 = 5.203 𝑘𝐽/𝐾 The s2 is the entropy at the end of the isothermal expansion. Using the values of sf and sfg at 6MPa,we have: 5.203 = 3.0273 + 2.8627 𝑥2 Therefore the quality is 0.760 QUESTION 48. The exit state is at the same entropy as the inlet. This allows us to determine the exit quality as follows: 𝑠2 = 𝑠1 = 7.1685 = 0.6491 + 7.5019𝑥2 ; 𝑥2 = 0.8690 The exit enthalpy is: ℎ2 = ℎ𝑓 + 𝑥2 ℎ𝑓𝑔 = 191.8 + (0.8690)(2392.8) = 2271 𝑘𝐽/𝑘𝑔 The power output is: 𝑊𝑇 = −𝑚(ℎ2 − ℎ1 ) = −(2)(2271 − 3658.4) = 2774 𝑘𝑊 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

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QUESTION 49. An isentropic compression requires the minimum power input for an adiabatic compressor. The outlet temperature for such process is: 𝑃2 𝑘−1 10000 0.2857 𝑘 𝑇2 = 𝑇1 [ ] = 293 ( ) = 1092𝐾 𝑃1 100 To find the mass flux, we must know the density. 𝑃 100 𝑘𝑔 𝜌= = = 1.189 3 𝑅𝑇 (0.287)(293) 𝑚 The mass flux is then 20 𝑘𝑔 𝑚 = 𝜌𝑉 = (1.189) ( ) = 0.3963 60 𝑠 The minimum power requirement is: 𝑊 = 𝑚(ℎ2 − ℎ1 ) = (0.3963)(1.00)(1092 − 293) = 317 𝑘𝑊 QUESTION 50. The compression ratio of 8 allows us to calculate: 𝑉1 𝑇2 = 𝑇1 [ ]𝑘−1 = 293(8)0.4 = 673.1 𝐾 𝑉2 The heat supplied is then: 𝑞𝑖𝑛 = 𝐶𝑣 (𝑇3 − 𝑇2 ) = 0.717 (1773 − 673.1) = 788.6

𝑘𝐽 𝑘𝑔

The mass of air in six cylinders is: 𝑃1 𝑉1 (100)(600𝑥10−6 ) 𝑚= = = 0.004821 𝑘𝑔 (0.287)(293) 𝑅𝑇1 The heat supplied per cycle is: 𝑄 = 𝑚𝑞𝑖𝑛 = (0.004281)(788.6) = 3.376 𝑘𝐽 QUESTION 51. Assuming isentropic compression and expansion, we find: 𝑃3 𝑘−1 𝑇3 = 𝑇2 [ ] 𝑘 = 263(10)0.2857 = 508𝐾 𝑃2 𝑃1 𝑘−1 1 0.2857 𝑇1 = 𝑇4 [ ] 𝑘 = (303) ( ) = 157𝐾 𝑃4 10 The COP now calculated as: 𝑞𝑖𝑛 = 𝐶𝑝 (𝑇2 − 𝑇1 ) = (1.00)(263 − 157) = 106 𝑘𝐽/𝑘𝑔 𝑤𝑐𝑜𝑚𝑝 = 𝐶𝑝 (𝑇3 − 𝑇2 ) = (1.00)(508 − 263) = 245 𝑘𝐽/𝑘𝑔

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600

𝑤𝑡𝑢𝑟 = 𝐶𝑝 (𝑇4 − 𝑇1 ) = (1.00)(303 − 157) = 146 𝑘𝐽/𝑘𝑔 𝐶𝑂𝑃 =

𝑞𝑖𝑛 106 = = 1.07 𝑤𝑐𝑜𝑚𝑝 − 𝑤𝑡𝑢𝑟 245 − 106

QUESTION 52. To find the partial pressures we need the mole fraction. The moles of nitrogen and CO2 are: 𝑚1 2 = = 0.0714 𝑚𝑜𝑙𝑠 𝑀1 28 𝑚2 4 𝑁2 = = = 0.0909 𝑚𝑜𝑙 𝑀2 44

𝑁1 =

Total moles N = 0.1623 moles The mole fractions are: 𝑦1 =

𝑁1 0.0714 = = 0.440 𝑁 0.1623

The partial pressure of nitrogen is: 𝑃1 = 𝑦1 𝑃 = (0.44)(2) = 0.88 𝑀𝑃𝑎 QUESTION 53. The heat transfer needed is given by the first law (the work is zero for a rigid tank) 𝑄 = ∆𝑈 = 𝑚∆𝑢 = 𝑚𝐶𝑣 ∆𝑇 𝐶𝑣 = 𝑚1 𝐶𝑣,1 + 𝑚2 𝐶𝑣,2 + 𝑚3 𝐶𝑣,3 = (0.2)(0.177) + (0.4)(0.158) + (0.4)(0.157) = 0.161 𝑄 = (20)(0.161)(300 − 80) = 708 𝐵𝑡𝑢 QUESTION 54. The reaction equation for theoretical air is: 𝐶4 𝐻10 + 6.6(𝑂2 + 3.76𝑁2 ) → 4𝐶𝑂2 + 5𝐻2 𝑂 + 24.44𝑁2 The air to fuel ratio for theoretical air is: (6.5)(4.76)(29) 𝑚𝑎𝑖𝑟 𝑘𝑔𝑎𝑖𝑟 𝐴𝐹𝑡ℎ = = = 15.47 (1)(58) 𝑚𝑓𝑢𝑒𝑙 𝑘𝑔 𝑓𝑢𝑒𝑙 This represents 100% theoretical air. The actual air-fuel ratio is 20. The excess air is then: 𝐴𝐹𝑎𝑐𝑡 − 𝐴𝐹𝑡ℎ 20 − 15.47 % 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟 = 𝑥 100 = 𝑥 100 = 29.28% 𝐴𝐹𝑡ℎ 15.47

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601

QUESTION 55. The problem is solved assuming that there is 100 moles of dry products. The chemical equation is: 𝑎𝐶4 𝐻10 + 𝑏(𝑂2 + 3.76𝑁2 ) → 𝐶𝑂2 + 1𝐶𝑂 + 3.5𝑂2 + 84.5𝑁2 + 𝑐𝐻2 𝑂 We perform the following balances: 𝐶: 4𝑎 = 11 + 1, 𝑎 = 3 𝐻: 10𝑎 = 2𝑐, 𝐶 = 15 𝑂: 2𝑏 = 22 + 1 + 7 + 𝑐; 𝑏 = 22.5 Dividing through the chemical equation by the value of a so that we have 1 mol of fuel: 1𝐶4 𝐻10 + 7.5(𝑂2 + 3.76𝑁2 ) → 3.67𝐶𝑂2 + 0.33𝐶𝑂 + 1.17𝑂2 + 28.17𝑁2 + 5𝐻2 𝑂 The air-fuel ratio is: % 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑖𝑟 = (

7.5 ) 𝑥 100 = 108% 6.5

QUESTION 56. The stoichiometric equation is: 𝐶2 𝐻6 + 3.5(𝑂2 + 3.76𝑁2 ) → 2𝐶𝑂2 + 3𝐻2 𝑂 + 6.58𝑁2 The required combustion is: 𝐶2 𝐻6 + 5(𝑂2 + 3.76𝑁2 ) → 2𝐶𝑂2 + 3𝐻2 𝑂 + 18. 𝑁2 + 1.5𝑂2 The excess air since the actual reaction used 5 mol of oxygen rather than 3.5 mol. The percent excess air is: % 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑖𝑟 =

5 − 3.5 𝑥 100% = 42.9% 3.5

QUESTION 57 The heat transfer per unit area is given by: 𝑞 𝑇2 − 𝑇1 0.19(21 − 38) 𝑊 = −𝑘 ( )= − = 80.75 2 𝐴 𝑥2 − 𝑥1 0.04 𝑚 QUESTION 58 The heat transfer rate per unit surface area: 𝑞 𝑊 = ℎ(𝑇𝑠 − 𝑇∞ ) = 225(120 − 10) = 24,750 2 𝐴 𝑚

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602

QUESTION 59 The heat transfer rate is: 𝑞 = ℎ(𝑇𝑠 − 𝑇∞ ) 𝐴 The surface area: 0.5 1.0𝑚 )( ) = 10−4 𝑚2 100𝑚 100 𝑞 3𝑥103 𝑇𝑠 = 𝑇∞ + = 50𝐶 + = 50 + 3.33 = 53.33𝐶 ℎ𝐴𝑠 9 𝑥10−4 2(

QUESTION 60 The heat transfer rate per unit area is given by: 𝑞 𝑇1 − 𝑇3 1250 − 310 𝑊 = = = 1495 2 0.2 0.03 ∆𝑥𝑎 ∆𝑥𝑏 𝐴 𝑚 1.0 + 0.07 𝐾𝑎 + 𝐾𝑏 The interfacial temperature can be found using: 𝑞 𝑇1 − 𝑇2 𝑞 ∆𝑥𝑎 = ; 𝑇2 = 𝑇1 − ∆𝑥𝑎 𝐴 𝐴 𝐾𝑎 𝐾𝑎 0.2 = 1250 − 1495 ( ) = 951𝐾 𝑜𝑟 678 𝐶 1.0 QUESTION 61 The heat transfer per unit area of a composite wall is: 𝑞 400 − 50 𝑊 = = 920.95 2 −2 −2 −2 0.5𝑥10 0.25𝑥10 2.0𝑥10 𝐴 𝑚 [ 249 + 0.166 + ] 0.0548 QUESTION 62 The heat transfer per unit length is given by: 𝑞 2𝜋(370 − 305)𝐾 2𝜋(65) 𝑊 = = = 18.04 1 6.7 1 10.1 𝐿 𝑚 [0.07 ln ( ) + 0.048 ln ( 6.7 )) 14.08 + 8.55 2.5 The interfacial temperature desired is T2: 2𝜋(0.07)(370 − 𝑇2 ) 18.04 = ; 𝑇2 = 329.6𝐾 6.7 ln( ) 2.5 QUESTION 63 The heat transfer per unit length is given by: 𝑞 𝑇2 − 𝑇1 = −2𝜋𝐾 𝑟 𝐿 ln(𝑟2 ) 1

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295 − 305 = −2𝜋(367) ( ) = 39.20 𝑘𝑊/𝑚 1.8 ln (1.0) QUESTION 64 The heat transfer per unit area is: 𝑞 𝑇𝑖 − 𝑇𝑜 70 − 20 𝐵𝑡𝑢 = = = 31.25 1 𝐿𝑎 1 1 0.5 1 𝐴 ℎ𝑓𝑡 2 + + ℎ𝑖 𝐾𝑎 ℎ𝑜 [2 + 0.50 + 10] QUESTION 65 The heat transfer from the water is given by: 𝑞 = 𝑚𝑐𝑝 ∆𝑇 𝑞 = (0.166)(4.180)(50 − 40) = 6.967𝑥103 𝑊 The log-mean temperature difference is given by: ∆𝑇2 − ∆𝑇1 ∆𝑇𝑙𝑚 = ln(∆𝑇2 /∆𝑇1 ) ∆𝑇1 = 50 − 6 = 44𝐶; ∆𝑇2 = 40 − 12 = 28𝐶 28 − 44 = 35.4𝐶 28 ln (44) 6.967𝑥103 𝑊 = = 0.231 𝑚2 850 𝑥 35.4

∆𝑇𝑙𝑚 = 𝐴=

𝑞 𝑈∆𝑇𝑙𝑚

QUESTION 66 The total heat transfer to the water is: 𝑞 = 𝑚𝑐𝑝 ∆𝑇 𝑙𝑏𝑚 𝐵𝑡𝑢 = (60 = 108,000 60𝑚𝑖𝑛 1.0𝐵𝑡𝑢 ℎ min) ( )( ) (95 − 65) ℎ 𝑙𝑏𝑚𝐹 The log-mean temperature difference is given by: ∆𝑇2 − ∆𝑇1 ∆𝑇𝑙𝑚 = ln(∆𝑇2 /∆𝑇1 ) ∆𝑇1 = 200 − 95 = 105𝐹; ∆𝑇2 = 140 − 65 = 75𝐹 75 − 105 ∆𝑇𝑙𝑚 = = 89.16𝐹 75 ln ( ) 105 FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com

604

𝐵𝑡𝑢 108,000 𝑞 ℎ 𝐴= = = 24.23𝑓𝑡 2 𝐵𝑡𝑢 𝑈∆𝑇𝑙𝑚 50 𝑥 89.16𝐹 ℎ. 𝑓𝑡 2 QUESTION 67 The heat exchanger ratio may be written as: 𝑞𝑐𝑙𝑒𝑎𝑛 𝑈𝑐𝑙𝑒𝑎𝑛 𝐴∆𝑇 𝑈𝑐𝑙𝑒𝑎𝑛 = = 1.10 𝑜𝑟 𝑈𝑑𝑖𝑟𝑡𝑦 = 𝑞𝑑𝑖𝑟𝑡𝑦 𝑈𝑑𝑖𝑟𝑡𝑦 𝐴∆𝑇 1.10 The fouling factor is given by: 𝑅𝑓 =

1 𝑈𝑑𝑖𝑟𝑡𝑦

−

1 𝑈𝑐𝑙𝑒𝑎𝑛

=

1.10 1 0.10 − = 𝑈𝑐𝑙𝑒𝑎𝑛 𝑈𝑐𝑙𝑒𝑎𝑛 𝑈𝑐𝑙𝑒𝑎𝑛

QUESTION 68 The total heat transfer to the water is: 𝑞 = 𝑚𝑐𝑝,𝑤𝑎𝑡𝑒𝑟 = (0.75

𝑘𝑔 𝑘𝐽 𝑘𝐽 ) (4.18 ) = 3.14 𝑠 𝑘𝑔𝐾 𝑠. 𝐾

The heat transfer from oil is: 𝑚𝑐𝑝,𝑜𝑖𝑙 = (1.5

𝑘𝑔 𝑘𝐽 𝑘𝐽 ) (1.88 ) = 2.82 𝑠 𝑘𝑔. 𝐾 𝑠. 𝐾

𝐶𝑚𝑖𝑛 2.82 = = 0.90 𝐶𝑚𝑎𝑥 3.14 𝑁𝑇𝑈 =

𝑈𝐴 340 𝑥 13 = = 1.57 𝐶𝑚𝑖𝑛 2.82

QUESTION 69 𝜏 =1−𝜌−𝛼 =1−

450 900 − = 0.386 2200 2200

QUESTION 70 𝐸𝑎 = 𝜎𝑇 4 = 5.6697 𝑥 10−8

𝑊 𝑊 𝑘𝑊 (1273𝐾)4 = 148 967 2 = 149 2 2 4 𝑚 𝐾 𝑚 𝑚

FE-CBT-Mechanical/Dr. Sri Susarla www.licensingexamprep.com