SM_XI_Physics_Unit-1_Section-A.pdf

Mechanics

UNIT

1

Section - A : Straight Objective Type 1.

2.

Answer (2)

(1)

⎡ Potential energy ⎤ ⎢ ⎥ = [L] weight ⎣ ⎦

(2)

[Pressure] = ML–1T–2

(3)

⎡ Force ⎤ ⎢ ⎥ = [L] ⎣⎢ Pressure ⎥⎦

(4)

⎡ ( velocity)2 ⎤ ⎥ = [L] ⎢ ⎣⎢ Acceleration ⎦⎥

Answer (4) Least count = 1MSD – 1VSD = 1MSD −

3.

1 1 49 MSD = MSD = = 0.02 mm 50 50 50

Answer (3) Relative velocity has the dimensions of velocity.

4.

Answer (3) ⎡ hc ⎤ 0 0 0 ⎢ x ⎥ = [M L T ] ⎣ ⎦

⇒

[x] = [h][c] = [ML2T–1][LT–1] = [ML3T–2]

5.

Answer (1) Using n1u1 = n2u2, we get 1 dyne = n2 dyne/cm2 1(mg)(mm)–1 (ms)–2 = n2 gcm–1s–2

⇒

n2 = (10 −3 )(10 −1 )−1(10 −3 )−2 = 104

6.

Answer (1) Δx 0 .1 × 100 = × 100 ≈ 1% x 10.1

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Answer (4) Solid angle has unit steradian, but no dimensions.

8.

Answer (2) Least count is 100 ms = 0.1s. So, reading should be upto one place after decimal.

9.

Answer (3) m=

F 2.0 = = 1.0 a 2.0

Δm ΔF Δa 0.1 0.1 0.2 = + = + = m F a 2. 0 2. 0 2 . 0

0.2 = 0.1 2.0

⇒

Δm = 1.0 ×

⇒

m = (1.0 ± 0.1) kg

10. Answer (2) x + y is meaningless. 11. Answer (3) x=y×z 10 y ⎞⎛ 10 z ⎞ ⎛ x′ = ⎜ y + ⎟⎜ z − ⎟ 100 100 ⎠ ⎝ ⎠⎝ x ′ = (1.1× 0.9)yz

⇒

x ′ = 0.99 x

∴

Δx 0.99 x − x × 100 = × 100 = −0.01× 100 = −1% x x

12. Answer (2) Let F = [ρ]α [a]β [f]γ ⇒

MLT–2 = [ML–3]α [LT–2]β [T–1]γ

⇒

α = 1, β = 4, γ = –6

⇒

F = [ρa4f–6]

13. Answer (1) For real image

1 1 1 = + ( u = – x, v = + y) f x y ∵ 1 1 1 = + y 0.30 0.60 ⇒

y = 0.20 cm

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Also −

1 f

2

df = −

1 x

2

dx −

1 y2

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3

dy

⇒

⎡ dx dy ⎤ df = f 2 ⎢ 2 + 2 ⎥ y ⎥⎦ ⎢⎣ x

⇒

⎡ 0.01 0.01 ⎤ + df = (0.20 )2 ⎢ ⎥ 2 (0.60)2 ⎥⎦ ⎢⎣ (0.30)

df = 0.0055 ≈ 0.01 m (∵ f has two places after decimal)

14. Answer (3)

S ∝ t2

S ′ ∝ (1.3t )2 ⇒

S′ − S × 100 = 69 % S

15. Answer (4) A = 5 units = constant in all coordinate systems.

16. Answer (2) The new vector formed lies in x-z plane at angle θ above x-axis as shown. The magnitude remains 3 units. 17. Answer (3) Accuracy is determined by relative error. In B, relative error is

1 , which is greatest. 20

x

θ 3

z

18. Answer (2) 1ly = 9.46 × 1015 m

⇒

1m = 1.05 × 10–16 ly = 10–16 ly (one significant figure)

19. Answer (3) (a + b ) || (a − b )

⇒

(a + b ) × (a − b ) = 0 (This implies that the angle between (a + b) and (a − b) is 0 or 180°)

and (a + b )·(a − b ) > 0 (This means angle is not 180°) 20. Answer (2) Only the non-zero digits are significant in this case. Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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21. Answer (3)

L g

T∝ g∝

L

T2 Δg ⎛ ΔL 2ΔT ⎞ × 100 = ⎜ + ⎟ × 100 ⇒ g T ⎠ ⎝ L 1 ⎞ ⎛ 0.1 =⎜ + 2× ⎟ × 100 90 ⎠ ⎝ 20.0

≈ 3%

21(a).

Answer (2)

IIT-JEE 2008

Δg ΔT Δl + = 2 g l T For student I, T = 128.0, ΔT = 0.1 s. This has the least error. 22. Answer (1) ν2 = 2ν

25

(4 + x )2

25 dv (– 2) = dx (4 + x )3

⇒a= –

25

(4 + x )3

23. Answer (1)

y v φ

g x Radial acceleration aR at any point P is given by aR = gcosφ Where φ is the angle of tangent at point P. As tan φ =

Vy Vx

=

V0 sin θ − gt 1 ⇒ cos φ = V0 cos θ 1 + tan 2 φ

Clearly gcosφ does not vary Linearly with time Also At t = 0 At t =

T 2

At t = T

φ = θ, |aR| = |gcosθ| φ = 0, |aR| = |g| φ = 180 – θ, |aR| = |gcosθ|

Now we conclude from above statement, that graph ‘1’ is the best possible graph. Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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24. Answer (1) Tangential acceleration |at| = |gsinφ| gsinφ does not vary linearly with time. Also At t = 0 At t =

T 2

At t = T

φ = θ, |at | = |gsinθ| φ = 0, |at | = 0 φ = 180 – θ, |at | = |gsinθ|.

∴ Graph ‘1’ is the best possible graph. 25. Answer (3) a= v

dv dx

For 0 < x < 100 m v = 0.08 x a = 0.08 [0.08 x] and – 100 m < x < 0 υ = – 0.08 x + 16 a = – 0.08 (– 0.08x + 16) 26. Answer (4) Given that ar ∝ t, then

V2 ∝ t r

V∝ t Tangential acceleration at =

dv 1 ∝ dt 2 t

Now ar · at2 will be independent of time. 27. Answer (1)

y O

B 45º

x A Average velocity from 0 to Vavg =

T is given by 4

AB T 4

When particle reach at point B, displacement vector is AB as shown in figure and acceleration vector is BO . Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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The angle between AB and BO is 135º.

135º B

O A

Δv

Change in velocity from A to B is Δv = vˆj − viˆ Angle between a and Δv

is 45º.

45º

a

28. Answer (1) Given,

Kinetic energy at highest point 3 = Kinetic energy at time of projection 4

1 m (V0 cos θ)2 2 1 2 Kinetic energy at time of projection = m V0 2 Kinetic energy at highest point =

V02 (cos θ)2 V02

cos θ =

=

3 4

3 ⇒ θ = 30º 2

Velocity at projection point = V0 cos 30º iˆ + V0 sin 30 ˆj = V0 Velocity at highest point = V0 cos 30º iˆ = V0 Change in velocity = −

3 ˆ V0 ˆ i+ j 2 2

3ˆ i 2

V0 ˆ j 2

29. Answer (3) Normal reaction is perpendicular to tangent at A. Given y = tan θ = tan θ =

x2 , slope at any point on the wire track is a

dy dx

ω

N

90º – θ

2x a

A θ

At point (a, a), slope tanθ = 2

x

Now From Newton’s Law towards centre, Nsinθ = mω2a

.... (i)

and in vertical direction. Ncosθ = mg

.... (ii)

N

divide equation (i) by equation (ii) tan θ =

⇒ ω=

ω 2a g 2g a

θ acc. = ω a 2

mg

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29a.

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Answer (2)

7

(IIT-JEE 2009)

For equilibrium

y

a dy = tanθ = g dx ⇒

2kx =

⇒

x=

a g

θ

x

a 2gk

30. Answer (4) Maximum friction force = μsmg

f max

2 kg

= (0.2) (2) (10)

F(N) = 2t

=4N At t = 2s, F(N) = 4 N Upto t = 2 s, block is in rest. After that relative motion starts between block and surface and kinetic friction acts between surfaces. Now from figure, force due to friction is Ff = 4 – 1 = 3 Acceleration after two second is a =

a=

F m

2t − 3 2

dv 2t − 3 = dt 2 v

4

⎛ 2t − 3 ⎞ dv = ⎜ ⎟dt 2 ⎠ ⎝ 0 2

∫

∫

v = 3 m/s 31. Answer (1)

y 30º

60º

v 30º

x x

As shown in figure, Slope at point B is = –tan60º vy vx

= − tan 60º

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v 0 (sin θ) − gt =− 3 v 0 cos θ

θ = 30º

v0 = 20 m/s, 10 – 10t = –30 t=4s x coordinate

= (v0cosθ)t = 20 ×

3 ×4 2

= 40 3 m 32. Answer (4) (P × Q ) is always perpendicular to P and Q .

33. Answer (3) Initial velocity of particle is v1 Retardation = a Total time to come to rest is t Total distance s t = v 1t −

1 2 at 2

Distance travelled in time (t – 1) is s t −1 = v 1(t − 1) −

1 a( t − 1) 2 2

Distance travelled in last second s = st – st–1 = (v 1t −

= v1 +

As ⇒

t=

a − at 2

v , a

s = v1 + s=

1 2 ⎛ 1 ⎞ at ) − ⎜ v 1(t − 1) − a(t − 1) 2 ⎟ 2 2 ⎝ ⎠

a v − a⎛⎜ ⎞⎟ 2 ⎝a⎠

a 2

So distance travelled does not depend on initial velocity, hence s1 = s2 = s3. 34. Answer (2) Displacement = Algebraic sum of area under curve. Distance = Sum of magnitude of area 35. Answer (1) The man and the dog are moving with constant velocities, hence the relative acceleration is zero. So with respect to each other, they will be moving in a straight line. Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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36. Answer (4) For same range in projectile motion, θ + φ = 90º given φ = 2θ ⇒

θ = 30º

and

φ = 60º

T1 sin θ 1 = = T2 sin φ 3 ⎛ 2u sin θ ⎞ ⎜⎜∵ T = ⎟⎟ g ⎝ ⎠

37. Answer (3)

1 k d 2 > μ (m1 + m2 ) gd 2 m1

d>

k l0

m2

2μ (m1 + m2 ) g k

38. Answer (2)

C

y

D

B x

A

v

Velocity of A = v iˆ Velocity of B = v ˆj Velocity of C = −v iˆ Velocity of D = −v ˆj v BA = v B − v A = v jˆ − v iˆ v CA = v C − v A = −v iˆ − v iˆ v DA = v D − v A = −v jˆ − v iˆ Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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39. Answer (3) Slope of x -t graph gives velocity v A = tan 30 º iˆ − tan 30 º ˆj v B = − tan 30 º iˆ + tan 30 º ˆj v AB = 2 tan 30 º iˆ − 2 tan 30 º ˆj

=

2 ˆ 2 ˆ i − j 3 3

40. Answer (1)

v max t1 v max t 2 + 2 2 Average speed = t1 + t 2

=

vmax

v

v max 2

t1

t2

t

v max =2 avg. speed

41. Answer (2) (n ≠ –1)

a = btn dv = bt n dt

v=

bt n +1 (assuming that initial velocity is zero) n +1

dx bt n +1 = dt ( n + 1)

r =

bt n +2 (assuming that initially, position vector is zero) (n + 1)( n + 2)

v ( n + 2) = r t

At t = 1 s

v = (n + 2) r

or v = r(n + 2) 42. Answer (4) r = 15(cos pt iˆ + sin pt jˆ )

Radius of circle = 15 Angular velocity ω = p rad/s ac = ω2r = 15 p2 m/s2 Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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43. Answer (4) y =

c 6 t 6

v=

dy = ct 5 dt

At t = 5 s,

v = c(5)5 d 2y

a=

dt 2

= 5 ct 4

a = 5c(5)4 ⇒ v=a 44. Answer (1) When the seat is broken, motion will be governed by gravity and the general motion will be on a parabolic path. 45. Answer (2) F 1 + F 2 + F 3 ...F n = 0 F 2 + F 3 ...F n = – F1

a=

– F1 m

46. Answer (3)

b

Snell law is given by

r

sini = x sinr Now

i a × c =| a || c | sin i ( −nˆ )

c Incident medium a

nˆ = unit vector along outward normal to boundary

c × b =| a || b | sin r ( nˆ ) a × c = sin i ( − nˆ )

c × b = sin r (nˆ )

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47. Answer (1) The general equation of trajectory is y = x tan θ −

gx 2

Slope

2u 2 cos 2 θ

0

dy gx = tan θ − 2 dx 2u cos 2 θ

t

As x varies linearly with t ∴ slope also varies linearly with t. 48. Answer (4) Angular momentum about O is L = m(vsinθ)R = mv sin θ

v 2 sin 2θ g

v cosθ

O

R v sinθ

=

2mv sin θ cos θ g 3

2

49. Answer (4)

u θ

At the point of projection

dv = g sin θ dt

Radius of curvature at the point of projection R =

v2 u2 = aN g cos θ

g sin θ.g cos θ g 2 sin θ.cos θ = u2 u2 50. Answer (1)

r B = (2tiˆ + 40 ˆj ) 1 ⎛ ⎞ r s = v1tiˆ + v 2 jˆ + ⎜ 20t − gt 2 ⎟ kˆ 2 ⎝ ⎠ Condition for stone to hit the block is 2t = v1t, 40 m = v2t Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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20t −

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1 2 gt = 0 2

v2 = 10 m/s 51. Answer (2) vy = vsin30 – gcos30ºt

t=

v sin30º v 1 2 = = s g cos30º g 3 3

52. Answer (3) Time for A to move from (0, 50) to (50, 0)

=

2π × 50 1 × =5s 4 5π

Displacement of B = 50 iˆ − 50 jˆ ∴ Velocity of B = 10( iˆ − ˆj ) m/s

Rdθ

53. Answer (3) Take a small element of mass dm

dθ

Use Newton’s Law in Radial direction, 2T sin

ω R

dθ = (dm )ω2R 2

dm = λRdθ

As angle is small, So sin

dθ dθ = 2 2

⇒ 2T

dθ = dmω2R 2

2T

ω

T

dθ dθ 2 2

T

dθ = λR 2dθω2 2

T = λR2ω2 54. Answer (1) Tcos30º + Tcos60º = mω2r

.... (i)

Tsin30º + Tsin60º = mg

.... (ii)

⇒ ω=

T T

Divide equation (i) by (ii) mω2 r =1 mg

30º 60º

g r

30º 30º

r

C

v = ωr = gr Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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55. Answer (2) When it breaks off, normal reaction N = 0

1

2 v

θ Fixed

Mg

Mass m is in circular motion, so use Newton’s Law in Radial direction.

mg cos θ =

mv 2 R

.... (i)

Energy conservation between point (1) and (2) yields mg (R − R cos θ ) =

1 mv 2 2

.... (ii)

From equation (i) and (ii) cosθ = 2(1 – cosθ) cosθ = 2 – 2cosθ cos θ =

at = g sin θ =

2 ⇒ sin θ = 3

5 3

5g 3

56. Answer (1) 57. Answer (4)

θ B A 3gl

By energy conservation between A and B, VB = gl + gl cos θ at As tan45º = a c

θ ac

=

g sin θ g + g cos θ

a

45º at = gsinθ

we get, θ = 90º Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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58. Answer (4) v2 v1

Relative velocity = v1 + v2 = 4 v1

.... (i)

v2

Relative velocity v 1 − v 2 =

4 10

.... (ii)

From equation (i) and (ii) v1 = 2.2 m/s v2 = 1.8 m/s 59. Answer (1) 1

⎛ m 2v 4 ⎞⎟ 2 vdv – μ ⎜ m 2g 2 + = m 2 ⎟ ⎜ ds r ⎝ ⎠

s=

∫

0

ds =

∫ v

[– v d v ] μ g2 +

v4 r2

Solving integral we get s 60. Answer (3)

y =

x2 t2 , x= 2 2

⇒

y =

t4 8

Now,

dy t 3 = dt 2 dx =t dt

⇒

At t = 2 s, v = 2iˆ + 4 ˆj

61. Answer (4) At same horizontal level speed is 10 m/s. Angle between g and u is 30º. Now, ρ =

u2 102 × 2 = = 20 m g sin 30 10 × 1

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62. Answer (3) ux = 8 π sin 2πt vy = 5 π cos 2πt This means that x and y coordinate will also be sine and cosine functions but with different amplitudes. So, path is an ellipse. 63. Answer (3) dv = a − bv dt

⇒

dv = dt a − bv

or

[ln (a − bv )]v0 = −b

⇒

v=

∫

t

dt

0

a [1 − e −bt ] b

64. Answer (2) The maximum speed will be attained at t = 8s. Δv = Area of graph for 0 – 8s v–0=

⇒

1 × 4 × 5 + 4 × 5 = 10 + 20 = 30 2

vmax. = 30 m/s

65. Answer (4) a =u×p+

a

1 a0 p 2 2

a + b = u × (p + q) +

…(1) a0 ( p + q )2 2

t=0

b t=p

t = p+q

…(2)

from (1) & (2), we obtain

a0 =

2(bp − aq ) pq( p + q )

66. Answer (1) v=

2gc

N – mg =

mv 2 r

x2 =

b2 y c

3/2

⎡ ⎛ dy ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ dx ⎠ ⎥⎦ r= d 2 y / dx 2

=

1 b2 = 2c 2c b2

dy 2 xc = 2 dx b

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d 2 y 2c = dx 2 b 2

2 g cm b2 2c

N = mg +

17

⎛ b 2 + 4c 2 ⎞ ⎜ ⎟ = mg ⎜ 2 ⎟ b ⎝ ⎠ 67. Answer (4) Let the target be at a distance R. Let θ be required angle. Then, R =

u 2 sin 2θ g

Now, R − 6 =

u 2 sin 60° g

…(1)

sin 90° g

…(2)

R + 9 = u2

9 3 + 12

From (1) & (2)

R=

and

u2 =R +9 g

Now, R =

⇒

2− 3

u 2 sin 2θ (where θ is the required angle of projection) g

sin 2θ =

R u2 g

Solving sin 2θ =

θ=

=

R R +9

3 3+4 10

⎛3 3 +4⎞ 1 ⎟ sin −1⎜ ⎜ 10 ⎟ 2 ⎠ ⎝

68. Answer (3) Maximum acceleration a1 = (n – 1)g Maximum retardation a2 = g S=

1 ⎛ αβ ⎞ 2 ⎜ ⎟t 2 ⎜⎝ α + β ⎟⎠

a=

1 ⎡ ( n − 1)g × g ⎤ 2 ×⎢ ⎥t 2 ⎣ ( n − 1)g + g ⎦

⎡ 2na ⎤ t=⎢ ⎥ ⎣ (n − 1)g ⎦

12

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69. Answer (3) The velocity-time graph is shown.

S=

v

1 1 × t 0 × ft 0 + (t − 2t 0 )ft 0 + t 0 × ft 0 2 2

S = ftt 0 − ft 02 . This gives t 0 =

t–

t2 –

ft0

4s f

2

t0

The time for constant velocity is t – 2t 0 =

t2 –

t – 2t0

t0

t

4s f

70. Answer (2) dv = −kv dt

⇒

dv = −k dx

⎛ dx ⎞ =v⎟ ⎜∵ ⎝ dt ⎠

dx = – kdv

⇒

Δx = –kΔv

71. Answer (3) The velocity-time graph for the train is shown. v0 = xt1 = yt2 As

t1 + t2 = 4

⇒

v0 v0 + =4 x y

Also, 2 =

⇒ ⇒

v v0

1 × t × v0 2

t1

t2

t

v0 = 1

1 1 + =4 x y

72. Answer (2) Range R =

u 2 sin 2θ 15 × 15 3 45 3 = × = g 10 2 4

Time of flight T =

2 × 15 × 1/ 2 = 1. 5 s 10

Distance to be moved by the fielder is 70 −

Speed =

45 3 . 4

45 3 4 ≈ 33 m/s 1.5

70 −

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73. Answer (4)

N

For the block to be at rest in car, ma ≥ mg sin θ − μmg cos θ

ar fc o θ r n o si flo mg a ar) mg cos θ (c

and ma ≤ mg sin θ + μ mg cos θ

⇒

a≥

g 3

and

a≤

ma

2g 3

74. Answer (3) N = mgcosβ + Fsinθ

... (1)

mN + Fcosθ = mgsinβ

... (2)

From (1) and (2) we get

F=

mg (sin β − μ cos β) cos θ + μ sin θ

N

μN θ

d (cos θ + μ sin θ ) = 0 dθ

⇒ μ = tanθ ⇒ Fmin =

F mg

mg (sin β − μ cos β) 1 + μ2 1+ μ

75. Answer (1) a1 =

P – ( μmg + 2μmg ) m

T ⎛ P – 3μmg ⎞ = ⎜ ⎟ m ⎝ ⎠

T – f1 = ma2 P – T – f2 – f1 = ma2

T f1

f1

f1

f1 P

f2

P

T f2

P – f2 – 2f1 = 2ma2 ⎛ P – 4μmg ⎞ a2 = ⎜ ⎟ 2m ⎝ ⎠

For 2P – 6μmg = P – 4μmg P = 2μmg 76. Answer (2) a1 =

F − f1 F − μm1g = = 10 m/s2 m1 m1

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−F + μm2 g = −1 m/s2 m2

a2 =

∴

Success Magnet-Solutions (Part-I)

s=

1 1 [10 − ( −1)]t 2 = × 11× 4 = 22 m 2 2

77. Answer (2) The tension in the rope is t(N). When t = 10s, 1kg will start moving up and at t = 20 s, 2 kg will start moving. As force increasing with time, acceleration will increase with time. 78. Answer (3) When they move together, a =

F F = 2+3+ 4 9

Now, this ‘a’ should be less than or equal to maximum possible acceleration of the blocks. For 4 kg,

amax . =

0.1× (2 + 3) × 10 = 1.25 m/s 2 4

For (3 + 4) kg block

amax . =

0.5 × 2 × 10 = 1.4 m/s 2 3+4

∴

F ≤ 1.25 9

⇒

F ≤ 11 .25 N

79. Answer (1) a = gsinθ – μgcosθ

2 sec θ = a

t=

2 sec θ g sin θ − μg cos θ

dt =0 dθ

⇒

tan2θ = −

1 μ

80. Answer (3) When m1 remains at rest, the T = m1g Also,

4m2 m3 g = m1g T = 2T0 = ( m2 + m3 ) 4 1 1 ⇒ m =m + m 1 2 3

T

T

m1 T0 m2

T0 m3

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Machanics

21

81. Answer (2) The accelerations of the blocks w.r.t. pulley are equal and opposite.

⇒

a1 − aP = −( a2 − aP )

⇒

a1 + a2 = 2aP N

82. Answer (1) cos θ =

12 20

Also N cos θ = W N=

12

W W × 20 = = 1.67W cos θ 12

36 cm

T2

The C.G. of the triangle will lie on the median. The resultant of T2 and T3 must pass through A. T1 = T2 = T3 =

20

16

83. Answer (3)

⇒

θ

T1

l1 2 A

mg 3

T3

84. Answer (4)

l1 2

mg V H

T

T1 500 N

100 N Tsin(37º) × 1.0 – 50 × 0.7 – 100 × 1.4 = 0

... (i)

H – Tcos(37º) = 0

... (ii)

V + T sin37º – 50 – 100 = 0

... (iii)

On solving equations we get H = 234 N 85. Answer (4) The sand leaks out vertically. It will not impart any horizontal force. 86. Answer (4) Force required = v dm = λ

⇒

dm dt

dx , where λ is mass/length 2

F =

λv 2 = 128 N 2

(when free end of folded part moves by dx, the increment in mass is

λdx ) 2

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Machanics

Success Magnet-Solutions (Part-I)

87. Answer (3) F = T2 T1 = mg = 2g As

T2 = T1eμπ

⇒

F = 2geμπ

T2

T1 T1 2kg

T2

88. Answer (2)

F The maximum tension (at bottom) is

T = mg +

mv 2 l

By energy conservation

1 mv 2 = mgl (1 − cos 60 ) 2 ⇒ T = 2mg = 2 × 2 × 10 = 40 N T ≤ μMg

For no slipping,

40 ≤ μ × 8 × 10 μ ≥ 0.50 89. Answer (1)

rg (sin θ + μ cos θ) (cos θ − μ sin θ)

Vmax =

= 48.38 m/s 90. Answer (2) At mid point, tension is horizontal. Consider the FBD of half part,

Tsinθ θ

T Tcosθ

T0 mg 2 T sin θ =

mg 2

Tcosθ = T0 ⇒ T0 =

mg 2 tan θ

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23

92. Answer (2) Considering vertical equilibrium,

TAO θ θ TBO TAOcosθ = TBOcosθ + Mg ⇒ TAO > TBO 93. Answer (3) The force on the wedge is given to be F = Mgcotθ

m

F M

θ

In this case, m falls freely. ∴ a = 10 m/s2 94. Answer (3) Fav =

Δp m × 2 × 2gh = = mg Δt 2h 2× g

95. Answer (2) Ncosθ = Mg Nsinθ = mω2 (Rsinθ)

⇒ cos θ =

If

g ω 2R

θ

R

N

r

g

Mg

ω2 R

≥ 1, it will remain at lowest point.

96. Answer (1) Consider the FBD of circular part For the element dθ, dT = dmg cosθ TH

∫ dT = λRg cos θdθ

TH

T + dT dθ θ

T dmg cosθ

TL

TH = TL + λRg ⇒

TH =

TL =

λg 2

λg 2 + λg × 2 π

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Machanics

97.

Answer (3)

Success Magnet-Solutions (Part-I)

F =

Fc2 + Ft2

Fc =

mv 2 4 × (2)2 = = 16 N r 1

Ft =

mdv = 4 × 3 = 12 N dt

⇒ F = 20 N 98.

Answer (1) For 4 kg block, kx = ma 20 = 2 m/s2 4+6

Also a =

⇒ 200 × x = 4 × 2 x =

8 200

⇒ x = 4 cm 99.

Answer (1) Just before breaking, 4 kg block is in equilibrium ⇒ kx = 4 × 10 + 6 × 10 kx = 100 N When the lower spring breaks, kx – 40 = 4a 60 = 4a ⇒ a = 15 m/s2

100.

Answer (4) Given tan

3 = 37° 4

⇒ angle of repose is 37° At rest, net force is zero. When θ < angle of repose, f = mg sinθ When θ > angle of repose, body slides and f = μN = μ mg cosθ When body is at rest,

N 2 + f 2 = mg

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101.

25

Answer (1) F =

102.

Machanics

T 2 + T 2 + 2T 2 × cos 60

As T =

3 mg 4

⇒ F =

3 3 mg 4

Answer (2) Let force is applied at an angle θ above horizontal At the verge of moving, F cosθ = μN, N = mg – F sinθ

N F sinθ

μ mg F = cos θ + μ sin θ

cosθ + μ sinθ has a maximum value ⇒ Fminimum =

103.

θ

μN

F F cosθ

mg

1+ μ2

μ mg 1 + μ2

Answer (1) As μ > tan30°, therefore, the block does not have a tendency to slide down. To make it move up, F = mg sinθ + μ mg cosθ

⇒ F = 10 × 10 ×

1 3 3 + × 10 × 10 × 2 4 2

= 115 N ⇒ 0 < F < 115 N, block remains at rest 104.

Answer (3) By drawing F.B.D of both block it can be seen. Mg sin30° = mg M 2

m= 105.

Answer (3) F = – kv mv

v0 / 2

x

v0

0

∫ dv = ∫ – kdx

m

−

dv = – kv dx

mv 0 = – kx 2

x=

mv 0 2k

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106.

Answer (3)

Success Magnet-Solutions (Part-I)

By conservation of linear momentum. ( 2i + 4 jˆ ) + ( 6 ˆj ) = 10 ˆj + p A ⇒ p A = 2iˆ

107.

Answer (2) The change in velocity takes place in a direction perpendicular to wall.

nˆ =

v 2 – v1

=

(–2iˆ + 6 ˆj ) – (3iˆ + 4 ˆj )

v 2 – v1 = 108.

v 2 – v1

– 5iˆ + 2 jˆ 29

Answer (1) 40 × 4iˆ + 40 × 3 ˆj = 80v

3ˆ j 2

⇒ v = 2iˆ +

1 1 1 3 × 40 × 4 2 + × 40 × 3 2 – × 80 2iˆ + ˆj 2 2 2 2

K Loss =

2

= 320 + 180 – 250 = 250 J 109.

Answer (3) When string becomes taut, displacements are same in magnitude ∴ 4t = 2t + 5t2 ⇒ t = 0.4 The velocity of hanging block is v = 2 + 10 × 0.4 = 6 m/s is when string is taut speeds become same ⇒ By impulse momentum theorem 4v + 4v = 4 × 6 + 4 × 4 ⇒ v = 5 m/s ⇒ I = Δp = 4 × 5 – 4 × 4 = 4 N-s

110.

Answer (1) The resultant of centripetal and tangential force is net force, equal to mg.

111.

Answer (1) Consider a small section of length dl subtending angle dθ at the centre ⎛ dθ ⎞ Along radial direction, 2T sin ⎜ ⎟ = dm ω2 R ⎝ 2 ⎠

also, dm = ⇒ T =

m dθ 2π

T

dθ 2

dθ 2 O ω

T

mω 2 R 2π

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112.

Machanics

27

Answer (1)

T

T + F – Mg = Ma

T

F

mg – T = ma

F + mg – Mg (M + m ) = a

mg

mg

⎛ F + (m – M ) g ⎞ ⎟⎟ t v = ⎜⎜ M+m ⎝ ⎠

113.

Answer (2) As friction between 2 kg block and incline is less it moves down with acceleration, while 4 kg is not able to move f2 kg = 0.6 × 2 × 10 × cos37 =

48 N 5

f4 kg = mg sinθ (∵ it does not move down) = 4 × 10 × 114.

3 = 24 N 5

Answer (3) As the block moves with constant speed, friction is 12 N. Normal reaction is 5 N ⇒ Fc =

115.

FN2 + f r2

= 13 N

Answer (3) t

k mean

1 = kdt t

∫ 0

t

k mean

1⎛ 1 ⎞ = ⎜ m ⎟ v 2 dt t⎝2 ⎠

∫ 0

Take v = u + at and integrate. k mean =

1 m[u 2 + v 2 + uv ] . Here u represents u and v represents u 1 2 6

116. Answer (2) When the particle leaves the circular motions, T = 0

v mgcosθ

∴ v = ga cos α , where α = 180 – θ... (i)

Also v = v L2 – 2ag (1 + cos α ) from (i) & (ii) cos α = ⇒ cos θ = −

... (ii)

α

α

θ a vL

1 3 1 3

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Success Magnet-Solutions (Part-I)

117. Answer (2) v L = 2 gl

v = v L2 − 2gl (1 + cos θ)

.... (i)

As T = 0

v

mv 2 ⇒ mg cos θ = l

.... (ii)

θ

h=lcosθ

2 ⇒ cos θ = 3

l

l

2l ⇒h= 3

vL

118. Answer (3)

a O

VL2 − 2ga

b O′

VL For circular motion about O′,

VL2 − 2ga ≥ 3g (b − a )

VL ≥ ( 5a − 3b )g

119. Answer (3) TL – TH = 6 mg 120. Answer (1) N = m(g – a sinθ) = 10 (10 – 2 × sin 30°) = 10 (10 – 1) = 90 N 121. Answer (1)

vsinθ θ

v v1

Before collision

v2 After collision

Motion wil be orthogonal if v1 = 0

⇒

v1 =

(M − em ) v cos θ M +m

e=

M m

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29

122. Answer (2) T =

πr u

After impact, relative speed = eu t=

2πr 2T = eu e

122a. Answer (3)

(IIT-JEE 2009) rd

A 3 collision

60°

st

1

2 nd col lisio n

60° 60°

ion llis co

123. Answer (3) v sin α − v 1 u cos θ

.... (i)

I = mucosα – mv1

.... (ii)

I sinα = Mv

.... (iii)

e=

⇒

Mv = mu cos α − mv 1 .... (iv) sin α

⇒

Mv = mu cos α − m[v sin α − eu cos α] sin α

(M + m sin α )v = mu cos α(1 + e) sin α

α u

I (impulse)

v α

α α v1

v2

2

v=

α I (impulse)

mu cos α sin α(1 + e ) M + m sin2 α

124. Answer (3) P = F .V

= 10 × 5 – 10 × 3 + 20 × 6 = 50 – 30 + 120 = 140 J/s

124a. Answer (5)

JEE (Advanced)-2013

As power is constant, P × t = ΔkE

⇒

1 × (0.2)v 2 − 0 2 2.5 = 0.1 v2

⇒

v = 5 m/s

⇒

0.5 × 5 =

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Success Magnet-Solutions (Part-I)

125. Answer (3) Along horizontal, there is no external force. So, centre of mass continues to remain at rest.

m × v sin θ . M+m

Along vertical v cm = 126. Answer (4) Fav =

vf − vi ×m Δt

= 15 ×

(6i + 4 j + 5k ) − (i − 2 j ) = 150 [5iˆ + 6 jˆ + 5kˆ ] 0. 1

127. Answer (2) When the man moves through L on the plank, Plank moves back by x =

∴ Net displacement is L −

M ×L 3L = M 4 M+ 3

3L L = . 4 4

128. Answer (1) x

Using Δk = w = –

k

∫x

2

dx

a

x

1 ⎛ 1⎞ mv 2 = +k ⎜ ⎟ 2 ⎝ x ⎠a

1

2k ⎛ a − x ⎞ 2 ⎜ ⎟ m ⎝ ax ⎠

⇒v = 129. Answer (3) 5

W = ∫ (4 x 2 + 115) dx − 0.5 × 20 × 9.8(5) 0

=

755 J 3

130. Answer (4) Speed of M (>>m) remains u. Let v be the speed of m after collision, then (v – u) = e(u) as

e=1

⇒ v = 2u. 131. Answer (1) When the bar 1 breaks, spring is just relaxed. ∴

1 2 1 kx = m2v 2 ⇒ v = 2 2

Now v cm =

k x m2

m2v m1 + m2

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132.

Machanics

Answer (2)

y

The second range is eR = H Also tan α =

31

4H R

u α

⇒ tanα = 4e

H R

x

eR

or α = tan–1 (4e) 133. Answer (2) 0 + mgsinθ – μmgcosθ – μmgs = 0 134.

Answer (2) As k ↑ F . ds > O

135. Answer (1) f = μmg

a = – μg

Time taken by the body to come to rest is t =

4 4 = a μg

Total change in kinetic energy due to friction is ΔE =

1 mu 2 2

1 mu 2 ΔE 2 1 = = muμg u Mean power = t 2 μg 136.

Answer (4) Till the ball reaches the lowest point, the wedge does not move. So v = 2gr is the speed of the ball at lowest point. Now v cm =

137.

m 2gR remains constant. M +m

Answer (2)

⎡ 1 x2 ⎤ 1 (2k1 + 2k 2 ) ⎢ x 2 – mv 2 ⎥ = 2 4 2 ⎥⎦ ⎣⎢ 138.

Answer (2) When block moves down by x, let the upper string stretches by y. The restoring force will be ky. This implies that the lower spring strectes by

y 2

ky

For the pulley, constraint relation gives, x−

x=

y = 2y 2

5y 2

ky 2

ky 2

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Machanics

Or y =

Success Magnet-Solutions (Part-I)

2x 5

By energy conservation,

1 2 1 ⎛y⎞ ky – x ⎜ ⎟ 2 2 ⎝2⎠

KE = mgx – =

139.

2

10mgx – kx 2 10

Answer (2) dU = – F . dx

dU = –(ydx + xdy) dU = –d(xy) ⇒ U = –xy + c 140.

Answer (3) The work done by the force on different paths is different For path AC, y = x W

=

∫ F . dr 1

=

∫

1

2xydx +

0

2

dy

0

1

=

∫y 1

∫ 2x dx + ∫ x dx 2

(∵ y = x )

2

0

0

1

=

∫ 3x dx + [ x 2

= 1J

3 1 ]0

0

140(a). Answer (4)

JEE (Advanced)-2013

⎡ ⎤ xiˆ yjˆ F =k⎢ 2 + 2 2 3/2 2 3/2 ⎥ (x + y ) ⎦ ⎣(x + y )

dW = F .dx ⎡ xdx + ydy ⎤ dW = k ⎢ 2 2 3/ 2 ⎥ ⎣(x + y ) ⎦

let x2 + y2 = r2 xdx + ydy = r2 krdr

⇒ dW =

r3

=

k r2

dr

r

⎡ −k ⎤ 2 W =⎢ ⎥ ⎣ r ⎦ r1 Now, r1 = a, r2 = a ⇒ W = 0 Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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141.

Machanics

33

Answer (2)

mv 2 = v ⇒ v =0 R By conservation of energy, At P, N =

142.

1 2 kx = mgR 2 Answer (2)

v

After collision, by momentum conservation along common tangent mu sinθ = mv

⇒

v = u sinθ

u

⇒ v = (u sin θ) cos (90 – θ) iˆ + (u sin θ) sin (90 – θ) ˆj

Common tangent

Common normal

v = u sin 2 θiˆ + u sin θ cos θjˆ

143.

90–θ θ

Answer (3) The centre of mass moves under the action of gravity alone.

144. Answer (4) By symmetry dm = λRdθ

xcm = 0

y cm

1 = M

R 2λ R cos θ λ Rd θ = ∫ M

+

π 4

∫ cos θ d θ = −

π 4

2 2R π

145. Answer (1)

ax =

3×

3g g cos60º −2 × cos30º 3g 2 2 = 5 20

146. Answer (3) (mgsinθ – μmgcosθ) t = mv0 (mgsinθ – μmgcosθ) hcosecθ =

1 mv 0 2 2

t 2 = h cosec θ v 0 t=

2h cosec θ v0

147. Answer (2) t

(2 + 3)v CM = ∫ 5 tdt = 0

5t 2 2

At t = 10 s, we have

v CM

5[10]2 = 2 = 50 m/s 5

v CM =

3 × 30 + 2 × v 2 5

v2 = 80 m/s Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Machanics

148.

Answer (2)

Success Magnet-Solutions (Part-I)

–40 × 10 + 20 × 10 40 + 20 + 100

x Boat =

–200 –5 m = 160 4

=

5 m towards left 4 Answer (3) ⇒

149.

x3 5x 2 – + 6x + 3 3 2

U =

dU = x 2 – 5x + 6 = 0 dx

x = 3 and x = 2 d 2U

= 2x – 5

dx 2

As x = 3, U min =

d 2U dx 2

> 0 ⇒ U is min

9 27 – 5× + 6×3 + 3 2 3

45 15 = = 7.5 J 2 2 = Total energy–Umin

= 30 – KEmax

= 17–7.5 = 9.5 J 150.

Answer (2) L = mvr m⎛ L ⎞ mv 2 ⎜ ⎟ F= = r ⎝ mr ⎠ r

=

2

L2 mr 3

151. Answer (3)

v = (u cos θ) iˆ + (u sin θ − gt ) jˆ F = −mg jˆ P = F.v = mg (gt − u sin θ) 152.

Answer (4) Magnitude of angular momentum changes as vt is decreasing Net acceleration is resultant of at & ac , not towards centre Angular momentum is along the axis of rotation, So its direction remains same.

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153.

Machanics

35

Answer (1) It will perform curvilinear translatory motion

1 mv 2 2 154. Answer (2)

so k =

L = mu cosθ H

u cosθ

⎡ u 2 sin2 θ ⎤ mu cosθ ⎢ 2g ⎥ ⎦⎥ ⎣⎢

=

mu 3 sin2 θ cos θ 2g

tanθ =

4H R mHu

L=

1 + tan 2 θ

mHu

=

1+

16H 2 R2

⎛ R 2 ⎞⎟ m ⎜⎜ H 2 + ω = 4 ⎟⎠ ⎝ ω=

(4H

mHRu R 2 + 16H 2

4HRu 2

+ R2

)

R 2 + 16H 2

155. Answer (4) 2 mg = mgeμ(5π) μ=

1 ln (2) 5π

156. Answer (4) m1g – T1 = m1a T2 – m2g = m2a If spring start slipping we have T1 = T2 eμπ m1 0.2 π e m2

Using a = 0 for string just slip we have 157. Answer (3) +

E=

2

1⎛ m

∫ 2 ⎜⎝

−

1 ⎞ dx ⎟ ( x sin θ ω)2 = mω2 24 ⎠

2

sin2 θ

2

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Success Magnet-Solutions (Part-I)

158. Answer (1) Vcm = rω′ = (R + r ) k =

=

dθ dt

2 mr 2ω′2 5

2 ⎡R + r ⎤ mr 2 ⎢ ω⎥ 5 ⎣ r ⎦

2

159. Answer (1) Centre moves in straight line with constant speed V 160. Answer (1)

M 2 M 2

( 2gR ) R

=

1 M ω R 2 + MR 2ω 2 2

( 2gR ) R = M R ω

ω=

2

g 2R

161. Answer (4) I

⎡ 2m (2r )2 ⎤ ⎥ = 2 mr2 + 3 ⎢ 3 ⎦⎥ ⎣⎢ = 2 mr2 + 8 mr2 = 10 mr2

162. Answer (1) v A = 2v 0 iˆ v B = v 0 iˆ – v 0 ˆj v AB = v 0 iˆ + v 0 ˆj

∴ v AB =

2v 0

2v 0

So ω AB =

2R

=

v0 R

163. Answer (2) mv0 – ft = mv

0 + fRt = I

v R

mv 0 − ft mv = f Rt ⎛2 2⎞ v ⎜ 5 mR ⎟ R ⎝ ⎠ Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Success Magnet-Solutions (Part-I)

Machanics

37

2v 0 ⇒ t = 7 μg

1 s = v 0 t − μgt 2 2 s=

12 v 02 49 μ g

164. Answer (2) Apply energy conservation, we get

mg R cos 30° =

1 IC ω2 (Here C is instantaneous centre of rotation) 2

Initial R cos30°

⎛ 4× 4 ⎞ + 4 × 3 ⎟ ω2 40 3 = ⎜ 12 ⎝ ⎠

C R cos30°

Final

⇒ ω = 3.2 rad/s 165. Answer (2)

l =

1 2 at 2

⎛ ⎜ 1 ⎜ g sin θ l = I 2 ⎜ ⎜1+ mR 2 ⎝

⎞ ⎟ ⎟t2 ⎟ . ⎟ ⎠

For maximum time, I = mR2 (maximum) 166. Answer (3) ma R sinθ = mg (1 – cos θ) R a ⇒ = g

2 sin 2

θ 2

θ θ 2 sin cos 2 2

= tan

θ 2

–1 ⎛ a ⎞ ⇒ θmax = 2 tan ⎜⎜ ⎟⎟ ⎝g ⎠

167. Answer (2) In case (2), body will definitely slip on BC, loosing energy against friction while in case 1, it will roll down. 168. Answer (2) Since τ = Fx = Iα ⇒ α = ∴a =

i.e.

Fx I αL 2

a∝x

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Success Magnet-Solutions (Part-I)

169. Answer (4) N 1 = 100 f = N2 By taking moments about A, 100 ×

4 5

= N2

3 ×2 5

200 3

⇒ N2 = ⇒ f =

200 = μN1 3

⇒ μ =

2 = 0.67 3

N2 N1 A

100 f

170. Answer (2) 2

IP = ICM

2

2R ⎞ ⎛ ⎛ 2R ⎞ ⎟ , ICM = I 0 – m ⎜ ⎟ . I0 = mR2 + m ⎜R – π ⎝ ⎝ π ⎠ ⎠

(I0 is the moment of inertia about centre of semicircle)

171. Answer (1) J = Icm ω 1 × 15 = 2 (1) (1)2 ω ⇒ ω = 7.5 rad/s 172. Answer (3) τ = I0 α τ = Mg x I0 = IC + M (OC)2 =

M 12

⎛ 2 a2 ⎞ ⎜a + ⎟+M ⎜ 4 ⎟⎠ ⎝

⎛ a2 ⎞ ⎜ + x2 ⎟ ⎜ 16 ⎟ ⎝ ⎠

1 1⎞ ⎛ 1 + + = Ma 2 ⎜ ⎟ + Mx 2 12 48 16 ⎝ ⎠ 2 ⎛ 4 + 1+ 3 ⎞ ⎟ + Mx 2 = Ma ⎜ 48 ⎝ ⎠

=

Ma 2 + Mx 2 6

⎞ ⎛ a2 So Mgx = M ⎜ + x2 ⎟ α ⎟ ⎜ 6 ⎠ ⎝

⇒α=

gx g = 2 ⎛ a2 ⎞ ⎛a ⎜ ⎜ + x2 ⎟ + ⎟ ⎜ 6 ⎜ 6x ⎝ ⎝ ⎠

⎞ x⎟ ⎟ ⎠

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For α to be minimum or –

a2 6x 2

⇒x=

Machanics

39

a2 + x = maximum 6x

+1=0

a 6

173. Answer (4) T =

4π R 2π (2R) ⇒V= V T

ω=

4π T 2

1 ⎛ 4πR ⎞ 1 ⎟ + Iω 2 K = m⎜ 2 ⎝ T ⎠ 2 2

1 ⎛ 4πR ⎞ 1 MR 2 ⎛ 4π ⎞ + ×⎜ ⎟ = m⎜ ⎟ 2 ⎝ T ⎠ 2 2 ⎝T ⎠

2

12mπ2 R 2 T2 174. Answer (4)

=

After collision v = Rω ⇒ ω > 4 rad/s

Before Collision

l = m (vf – vi )

I = 2 [4 – (– 4)]

4 m/s

⇒ I = 16 N–s

During Collision

9 rad/s

Also –μ I × R = Lf – Li

After Collision 4 rad/s

I (impulse)

4 m/s

μI

2 2 – μ × 16 × 1 = mR ( 4 – 9 ) 5

1 4 175. Answer (2) ⇒ μ =

v

Now after collision, v becomes zero. At pure rolling,

ω

ω′ = ω – α t ω′ = ω –

ωR = v 5f t 2mR

⎛ ⎜∵ α = ⎝

ft m As v′ = ω′ R for pure rolling, v′ =

ω′ = ω –

fR ⎞ ⎟ I ⎠

ω

v f

f

5 ω′ 2

2ω 7 2 ωR ∴ v′ = 7 ω′ =

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Machanics

Success Magnet-Solutions (Part-I)

176. Answer (3) Net force on the cylinder passes through the topmost point. So torque of the net contact force is zero. 177. Answer (4) M = I1 + I2 =

1 1 m1l12 + m2 l 22 3 3

177(a). Answer (5)

IIT-JEE 2011

F1 = mg(sin θ – μcosθ) F2 = mg(sinθ + μcosθ) = 3F1

1 tan θ 2 ⇒ 10μ = N = 5 ⇒ μ=

178. Answer (3) The velocity of B is v sinθ along BO ω =

v sin θ AB

⎛θ⎞ tan⎜ ⎟ = ⎝2⎠

R AB

θ /2

⎛θ⎞ AB = R cot ⎜ ⎟ ⎝2⎠

ω =

θ v

B O R

A

θ θ cos 2 2 θ R cot 2

2v sin

=

2v sin 2

θ 2

R

179. Answer (3)

IT 0 =

⎛M ⎞ ICM = ⎜ ⎟ ⎝4⎠ I0 =

m′ 1 = M 4

MR 2 , 2

MR 2 32

R2 (for the cut out disc) 4×2 +

MR 2 6

I R 0 = IT 0 – T0 =

=

3MR 2 32

MR 2 3MR 2 13MR 2 = – 2 32 32

179a. Answer (3)

IIT-JEE 2012

Let the mass density be σ. σπ(2R )2 (2R )2 3 − σπ(R )2 (R )2 2 2 13 σπR 4 = 2

IO =

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41

⎡ σπR 2 (R )2 ⎤ 3 2 2 + σπR 2 × ( 5R )2 ⎥ IP = 2 σπ(2R ) (2R ) − ⎢ 2 ⎣ ⎦ =

37 σπR 4 2

Ratio

IP 37 ≈3 = IO 13

180. Answer (3) Since plank slides forward ∴ friction on it acts backward ∴ on A and B, it is in forward direction.

2R π

181. Answer (3) 2R from O 3π The ring will rotate about point of contact A

The position of centre of mass will be

α =

α =

4 kg

2R π

C O CM

τA IA

2 kg

N 2R 3π C

(60 sin 30°)OA + (60 cos 30°)CO (2m A + 2mB )r

α = 6.84 rad/s

O θR

2

A

60 N

2

36°

182. Answer (4) α = – 7 rad/s2

∴ Slowing down

and ω > 0

∴ Anticlockwise rotation

183. Answer (3)

τ = F (r + R) dL dt 2 L = t (r + R)

2t (r + R ) =

184. Answer (1) F= f

mg = N

F

For toppling about A,

F 3a ≥ mg μ ≥

1 2 3

N mg

a 2

f

A

(∴ f = F = μN = μ mg)

185. Answer (4) Angular velocity and angular acceleration are either parallel or antiparallel in circular motion 186. Answer (4) Energy conservation yields mg 5R =

At P, N =

1 1 mv 2 + Iω2 and v = rω 2 2

mv 2 Force = R

(mg )2 + N 2

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Machanics

Success Magnet-Solutions (Part-I)

187. Answer (1) As point of contact with the ground is moving towards left, a frictional force acts towards right. 188. Answer (2)

∵ fR = Iα mg sin θR =

mR 2 α 2

2g sin θ = α R It will fall when ω = 0

∴

ω0

ω0 R = t 2g sin θ

θ

189. Answer (2) mgh =

f

mg sinθ

1 mv 2 (On a smooth incline, rotational energy will not be converted into potential energy) 2

∴h=5m 190. Answer (3) ω =

15 m/s 20–x

15 10 = 20 – x x

x

Solving, we get x = 8 cm

Q

191. Answer (4) 2

1 1 ⎛v ⎞ 2 By energy conservation mgh = mv + I ⎜ ⎟ 2 2 ⎝R⎠

v =

2gh I 1+ mR 2

As I depends on R 2,

I R2

is a constant for a given shape.

So v is independent of R 192. Answer (2) In this situation, friction adds to force in forward direction.

F

f 193. Answer (4)

F

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43

194. Answer (4) τ = mg x τ = mg u cosα t ∴ τ ∝ t. Also τ =

mv u α x

dL dt

⇒ L ∝ t2

mg

195. Answer (1) ωA =

θ 2t

ωC =

θ t

θ/2

θ

ωC 2

∴ ωA =

196. Answer (2) For AB,

1 2 1 ⎞ 1 ⎛ 2 mgh = KT + K R ⎜ KT = mv ; K R = Iω ; KT = k R ⎟ 2 2 2 ⎝ ⎠ 2

mgh = K T ′ –KT 2

⇒

KT + K T ′ KR

= 5

197. Answer (4) L = r ×P L = ( iˆ + jˆ ) × 4 ( iˆ – jˆ + kˆ ) L = 4 [i – j – 2k ]

198. Answer (2) f 2R = α1 for big cylinder 4I

α1 =

fR . 2I

Similarly,

fR = α 2 (small cylinder) I

ω2 = ω0 –

fR t 2I

fR t I For no relative motion finally, ω1 =

Rω1 = 2Rω2

⇒ ω1 = ω0

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Machanics

198(a).

Success Magnet-Solutions (Part-I)

Answer (8)

JEE (Advanced)-2013

By conservation of angular momentum,

1 ⎡1 ⎤ × 50 × (0.4)2 × 10 = ⎢ × 50 × (0.4)2 + 2 × 2 × 6.25 × (0.2)2 ⎥ ω 2 ⎣2 ⎦ ⇒ ω=

40 = 8 rad/s 4 +1

199. Answer (3)

ω0 2

⇒ ω2 =

As Rω1 = 2Rω2 200. Answer (1) mv1(2R) = mv2(4R) v1 = 2v2

... (i)

From C.O.E 1 GMm 1 GMm mv12 − = mv 22 − 2 2R 2 4R

... (ii)

From (i) and (ii)

v2 =

GM 6R

v1 =

mv12 GMm = r (4R )2

⇒

r =

2 GM 3R

8R 3

201. Answer (4) The packet will itself become a satellite. 202. Answer (3) Gravitational potential at the mid-point P = v1 + v2 = − Gravitational potential energy u = − 1 ⎧ 2⎫ ⎨0 − m0v ⎬ + 2 ⎩ ⎭

⇒ v =2

4Gm d

4Gmm0 d

4 Gmm0 ⎫ ⎧ ⎨0 − ⎬=0 d ⎩ ⎭

2 Gm

203. Answer (1) E=−

GMm 2a

1 GMm GMm mv 2 − =− 2 r 2a

⎡ 2 1⎤ v 2 = GM ⎢ − ⎥ ⎣r a⎦ Re

204. Answer (1) Re sin λ = Re + h

⇒

1 7

90

λ

h = 6 Re

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45

205. Answer (2) U=

−GMm r

As r increases, potential energy becomes less negative. 206. Answer (2) By momentum conservation, 0 = Mv1 – mv2 Mv1 = mv2 = p (say) By mechanical energy conservation 0=

1 −GMm 1 + Mv 12 + mv 22 2 2 d

GMm p 2 p2 = + d 2m 2M

⇒ p = mM

2G D(M + m )

v r = v1 + v 2 =

p p + m M

207. Answer (2) The torque due to gravity increases the angular momentum about point of projection. 208. Answer (4) CM divides the line in the ratio m : M. Whereas the null point divides the line in the ratio

m

:

M : m.

M

CM M

m m

m

209. Answer (1) v = v1 − v 2

In centre of mass frame, v cm =

M v 1 + mv 2 M +m

v 1 cm = v 1 − v cm

=

m(v1 − v 2 ) mv = M +m M +m

v 2 cm =

M (v 2 − v1 ) − M v = M +m M+m 2

1 ⎛ mv ⎞⎟ 1 ⎛ M v ⎞⎟ k = M⎜ + m⎜ 2 ⎜M + m ⎟ 2 ⎜M +m⎟ ⎝ ⎠ ⎝ ⎠ k=

2

1 mM 2 v 2M +m

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Machanics

Success Magnet-Solutions (Part-I)

210. Answer (2) ve =

v=

2GM R

2G(2M ) = 2v e ⎛R⎞ ⎜ ⎟ ⎝2⎠

211. Answer (2) Use T 2 ∝ R 3. 212. Answer (3) The speed of satellite becomes more than the orbital speed. It will have elliptical orbit with minimum distance from earth R. ∴ Minor axis = 2R. 213. Answer (3) Total energy = −

GMm 2r

Kinetic energy =

GMm 2r

To escape, energy needed = –TE =

GMm = Kinetic energy. 2r

214. Answer (3) For a geostationary satellite, time period of earth’s rotation is equal to time period of revolution of satellite around earth. 215. Answer (4) L = mr2ω = 6 × 10 24 × (1.5 × 1011 )2 ×

2π 365 × 24 × 3600

≈ 2.7 × 1040 kg m2/s. 216. Answer (1) V =

2GM = R

8 GπR 2ρ 3

⇒V∝R ⇒

VA R A = =2 VB RB

217. Answer (3) Satellite and person are in a state of free fall, moving in circular orbit under the gravitational force. Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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47

218. Answer (4) V =

2GM . Here M is the mass of the planet r

219. Answer (3) At surface, potential is GM = V =− R

−G×

4 πR 3 ρ 4 3 = − GπR 2ρ 3 R

220. Answer (2) Buoyant force = v ρ g = loss in weight =v=

(264 − 221)g = 43 cc 1× g

Volume of material of body =

264 = 30 cc 8.8

So, volume of cavity = 43 – 30 = 13 cc 221. Answer (1) Force on the base = ρgh × πr 2 Weight w = ρ ×

1 2 πr h × g 3

⇒ F>W 222. Answer (3) Weight of excess of water displaced = Vd 2 g =

M d 2g d1

∴ Excess of pressure at the bottom of vessel =

Md 2 g d1A

223. Answer (3) Pressure at E = pressure at B (h × 0.8 × g) + (10 – h) × 13.6 × g = 10 × 1.3 × g ⇒ h = 9.6 cm 224. Answer (2)

1 From Bernoullie’s theorem between 1 & 2

h

1 1 2 2 P0 + h1 ρg + ρv 1 = P0 + ρv 2 2 2

Before the pressure is added, v1 = 0 and v2 = v ⇒ v =

2gh =

2 × 10 × 2 =

2

40 m/s

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1 ρv ′ 2 2

2 ×10 4 + hρg = v′ =

4 × 10 4 10 3

+ 2 × 20 =

80 = v 2

Rate of flow Q ∝ v ⇒ Q′ = Q 2 = 10 2 cc/s

225. Answer (3) In accelerating lift T = m(g – a) = 3 kg wt

…(i)

From equation (i) 3 3 = (g – a) = m 4

...(ii)

When immersed in water T ′ = (m – Vd)(g – a)

…(iii)

From equation (ii) and (iii) ⎛3⎞ T ′ = (4 – 5 × 10 – 4 ×10 3 ) ⎜ ⎟ = 2.625 kg wt ⎝4⎠

226. Answer (3) vt =

2r 2 [(ρ − σ)g ] down 9η

vt =

40 m/s 9

Distance travelled =

40 8 × 0.2 = m 9 9

227. Answer (1) Let us take a small element of length dr from axis of rotation.

(P + dp ) − P = ρ ω2 r dr r1

⇒ P1 − P2 = ρ ω

2

∫ r dr

− r2

⇒ P1 − P2 =

ρω (r12 − r22 ) 2

⇒ h1 − h2 =

ω2 2 ( r1 − r22 ) 2g

2

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49

228. Answer (1) F = μ Δv

v ⎛v F = μ ⎜ 0 iˆ + 0 2 ⎝ 2

⎞ 3 ˆj – v 0 iˆ ⎟ ⎠

⎛ – v0 v 3 F = μ⎜ iˆ + 0 ⎜ 2 2 ⎝

v0 θ

⎞ ˆj ⎟ ⎟ ⎠

i k

= μv 0

F

j

v0

229. Answer (2) When block is floating, 0.02 × b ×

ρg = 1 × g

When stone is placed on block 0.024 × b × ρg = (1 + m)g 0.020 1 = 0.024 1+ m

20 + 20 m = 24 20 m = 4 1 4 kg = 5 20

m=

= 200 gm 230. Answer (3) F

P

=

dp d = (mv ) dt dt

=

dm v = ρπr 2v 2 dt

= F.v = ρπr 2v 3

231. Answer (2) Th = T0 + mg i.e., T0 = Th – mg = v(σ – ρ)g. When the lift is accelerated up, geff = g + a so, that T = v(σ – ρ)(g + a). Therefore

T g +a = T0 g

232. Answer (1) Equating pressure on both limbs R(1 – sinθ)ρ1 g = (Rsinθ + Rcosθ)ρ2g + R(1 – cosθ)ρ1g tan θ =

ρ1 – ρ 2 ρ1 + ρ 2

θ

θ θ R

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Success Magnet-Solutions (Part-I)

233. Answer (2) (Neglecting density of air)

3

U = 4 π D ρg 3 8

4 D3 ⎛D⎞ π ρg = 6πη ⎜ ⎟v 3 8 ⎝2⎠

6πη D v 2

D2 ρg = 6 ηv 3

η=

D 2 ρg 18 v

234. Answer (2) Force exerted on the water by the vane F =

ρQ ρQ (v cos 60° – v ) + (v cos 60° – v ) 2 2

= –

ρQ v ρQ v ρQ v – =– 4 4 2

= –

10 +3 × 0.03 × 20 2

v2 60° 60°

v

v2′

⇒ F = 300 N towards left 235. Answer (3) Acceleration of the container a = g sinθ – μg cosθ =

g sin 45° 2

Resultant of – ma and mg should be perpendicular to surface of water Hence along the surface of water

α–

mg sinα = ma cos(α – 45°)

α

mg sin 45 ° cos(α – 45°) = 2 sinα =

α

sin 45° cos(α – 45°) 2

1 ⎛ cos α sin α ⎞ ⎜⎜ ⎟⎟ + = 2 2⎝ 2 2 ⎠

=

° ma 45

45° mg

1 (cos α + sin α ) 4

4 sinα = cosα + sinα tan α =

1 ⎛ 1⎞ ⇒ α = tan –1⎜ ⎟ 3 ⎝3⎠

236. Answer (2) Mass of balloon =

V0 × 1 ; gm 2

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Machanics

PV = P0V0

51

...(i)

Mass of balloon remains unchanged

V0 ×1 = V × 1 2 ⇒ V =

V0 2

...(ii)

P = 2 P0 = 20 m of water height ∴ Required depth = 10 m 237. Answer (2)

⎡1 1 ⎤ Excess of pressure P =T ⎢ + ⎥ ⎣ R1 R2 ⎦ R1 =

d and R2 = ∞ 2

⎡ 2 1 ⎤ 2T P= T⎢ + ⎥= d ⎣d ∞ ⎦

∴ The outside pressure P0 being in excess of the inside pressure by

2T , tends to press the two plates d

⎛ 2T ⎞ 2TA2 ⎟A = by a force = ⎜ ⎝ d ⎠ V

238. Answer (1) Weight of liquid column raised = Force due to surface tension π(R – r )yρg = T(2πR + 2πr) 2

y=

r

R

2

2T (R + r ) (R – r )ρg 2

2

=

2T (R – r )ρg

y

239. Answer (4) Energy released = σΔA = σ[N(4πa 2 ) – 4πb2 ] Energy released = Kinetic energy 1 4 3 2 ρ πb v = σ 4π [Na 2 – b 2 ] 2 3 v =

6σ (Na 2 – b 2 ) ρ b3

as total volume is same = N

4 3 4 3 πa = πb 3 3

Na 3 = b 3 v =

6σ ⎛ 1 1 ⎞ ⎜ – ⎟ ρ ⎝a b⎠

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Success Magnet-Solutions (Part-I)

240. Answer (1) Work done by upthrust force = gain in potential energy of ball

h′

(vσg)h = vρg(h + h ′) h′ =

( σ – ρ)h ⎛ σ ⎞ = ⎜⎜ – 1⎟⎟h . ρ ⎝ρ ⎠

h

241. Answer (2) P0 AH = P1A(H – h), where P1 is the pressure of air column trapped in the tube. P1 =

P0 H (H – h )

Also P1 = P0 + (H – h) P1 = 2H – h ⇒

H2 + h = 2H H –h

⇒ (2H – h)(H – h) = H 2 242. Answer (4) The contact point is in equilibrium when subjected to three identical forces. Hence angle between any two surfaces is 120o. 243. Answer (4) Let true mass of body is m, then ⎛m⎞ ⎛M mg – ⎜⎜ ⎟⎟dg = Mg – ⎜⎜ d ⎝ 1⎠ ⎝ d2

⎞ ⎟⎟dg ⎠

⎛ d ⎞ ⎟⎟ M ⎜⎜1– d 2 ⎠ ⎝ m= ⎛ d ⎞ ⎜⎜1– ⎟⎟ ⎝ d1 ⎠ 244. Answer (4) h

d τ = ∫ x ρ g b (h − x )dx 0

⎡ h3 h3 ⎤ = ρg b ⎢ 2 − 3 ⎥ ⎣ ⎦ =

1 ρ g b h3 6

245. Answer (3) Pressure at P = P0 + (H – h)ρg ∴ gauge pressure i.e., excess of pressure is (H – h)ρg 246. Answer (2) Weight of block = weight of displaced water + weight of displaced mercury 7.8 × 10 × 100 = 1 × h × 100 + 13.6 × 100 × (10 – h) ⇒ h = 4.6 cm Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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53

247. Answer (4) 4 3 1 4 3 πr dg = 2πrT + πr ρg 3 2 3

4 3 ⎛ 1 ⎞ πr g ⎜ d − ρ ⎟ 3 2 ⎠ ⎝ =T 2 πr T =

1 2 r g (2d − ρ ) 3

248. Answer (1) From equilibrium if block is pushed downward by x then, ⎛h ⎞ ⎛h ⎞ ma = ⎜ + x ⎟d 2 Ag + ⎜ – x ⎟d1Ag – hAgd ⎝2 ⎠ ⎝2 ⎠

d

d1 d2

(in eq. dhAg =

a=

h h d 2 Ag + d1Ag ) 2 2

–(d1 – d 2 )Agx , a ∝ –x dAh

This body will execute SHM. 249. Answer (3) After putting the block, the height of liquid level increases. Pressure at A will be

P = P0 + hρg +

P0 M

Mg A

250. Answer (1) 1

1

R = (n ) 3 r ⇒ R = (125) 3 r ⇒ R = 5r Energy released = σ[125 × 4πr 2 – 4π(5r)2] = σ4πr 2[125 – 25] = 400 πσr 2 Rise in temperature =

=

400 πσr 2 4 π(5r )3 × ρ × s × J 3

12 σ 5 ρrJs

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251. Answer (2) 2 gh . Rate of flow of liquid is a 2 gh

Velocity of efflux

Let x is the height of liquid surface at an instant and decrease in level is ∴ –Adx = a 2gx dt H2

A

∫

a 2gx

H1

A

t

dx = –

∫

dt

0

H

A 2 ( H1 – H 2 ) = t a g

a

252. Answer (1) Weight of liquid held between plates = Force due to surface tension ρg( dh) = 2 σ 2σ ρ gd

h=

253. Answer (3) Velocity of efflux, V = 2gh Mass flow rate = ρaV ∴ Force on vessel = ρaV2 = 2ρagh μMg = 2ρagh

μM 2ρh 254. Answer (2) ⇒ a=

Before melting of ice entire weight of metal was being balanced out. Now, after sinking, entire weight of metal is not balanced out. So, volume of displaced water is more in the former case. So, water-level will decrease. 255. Answer (3) k =

ρ=

P ΔV ΔP ΔP ⇒ =– =– ΔV V k k – V

m V

Δρ ΔV P =– = ρ V k Pρ k 256. Answer (1) Δρ =

In air mg 0 4A In water a

=

w

=

mg 0 4A

⎛ σ⎞ ⎜⎜1– ⎟⎟ ⎝ ρ⎠

Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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= 1–

a

σ σ ⇒ = 1– ρ ρ

Relative density =

ρ = σ

=

w

Machanics a

–

a

55

w

a a a –

w

257. Answer (3) Tension in rod at point P is w 1 + ∴ stress = w1 +

=

3w 4

w

T A

3 /ρ

3w 4

W1

S

258. Answer (3) π⎞ ⎛ x = 0.3 sin ⎜10 πt + ⎟ 6 ⎝ ⎠

⇒ ω = 10π = T =

2π T

2 1 = s 10 5

Initial phase of body is =

π 6

π⎞ ⎛ ⎜π – ⎟ 6⎠ Time required to reach mean position = ⎝ ω

=

5π 6 ω

5π = 6 10π

=

1 s 12

259. Answer (1)

k

T1 = 2π

m k

T2 = 2π

4m k

m

m

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k

T3 = 2π

m 4k

m T1 : T2 : T3 = 1 : 2 :

1 = 2:4:1 2

260. Answer (4) Let particle starts motion from mean position and equation is y = a sinωt p = a sinω

...(i)

q + p = a sin2ω

...(ii)

2p 2 By (i) & (ii) we get amplitude a = 3p − q 261. Answer (2)

τ due to mg about C = mgx = mgr sinθ For small ‘θ’ τ = mgr (θ) ω2 =

mgr mgr g = = I mr 2 r

1

= 20

d 2y dx 2

r x

⎛ dy ⎞ = 0 at origin ⎟ ⎜∵ ⎝ dx ⎠

20 g

T = 2π

ω =

y θ

r , r = radius of curvature g

T = 2π

r =

C

2π = T

g 20

262. Answer (3) At equilibrium, 2T cos30 = mg and 2T sin30 = K(0.5 l) ⇒ K =

2 mg 3l

When the block goes down by a small displacement x, length of the spring decreases by Δx0 = the restoring force is F = –2ΔT cos30°. Also 2ΔT sin30° = K × 3 x .

3 x . Now,

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57

⇒ Frest = − 3 ΔT = − 3K 3 x = –3Kx ∴

T = 2π

l 3 = 6g

m l = 2π 3K 2 3g

263. Answer (2) T2 = 2T1 Suppose after time t, both pendulums are in phase again 2πt 2πt − = 2π T1 T2

⇒

t t − =1 T1 2T1

t =1 2T1

t = 2T1 264. Answer (2) Time required to 0 to

Time required to

A T = = T1 2 12

T A to A = = T2 6 2

T 6 1 T1 12 = = = T 12 2 T2 6 265. Answer (4) Let equation of particle is p = a sinωt x = a sinωt Y = a sin(ω)(t + 1) z = a sinω(t + 2) ⇒

x+z 2y

= cos ω

266. Answer (1)

v 12 = ω2 (a 2 – x12 )

...(i)

v 22 = ω2 (a 2 – x 22 )

...(ii)

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267. Answer (1) At mean position, potential energy is minimum. 268. Answer (3) 2 T = mgR sinθ ~ – mgRθ = Iω θ

ω2 =

mgR I

=

mgR 2

m 12

12 × g.R

=

2

= 12 × 3 = 36

N mg θ

2π π = ω=6 ⇒ T = 3 ω

R = 30 cm

269. Answer (4) 4v2 = 25 – x2 Standard Equation is v2 = ω2 (a2 – x2)

x2 25 – 4 4 5 aω = 2 1 ω = 2

v2 =

→

2π 7

→ T

=

...(i) ...(ii)

1 2

= 4π

270. Answer (1) Acceleration = Aω2 sinωt T /2

Avg. acceleration = Aω2

∫ sin ωt 0

T /2

=

2 Aω 2 π

271. Answer (4) Oscillation of first particle, x1 = 8 + 3 sinωt, First particle oscillates between 5 and 11. Second particle oscillates between –4 and 4. Collision is not possible between x1 and x2 . 272. Answer (2) The separation is x = x1 – x2 = 8 + 3 sinωt – 4 cosωt. For x to be minimum,

dx =0 dt

273. Answer (2) Time period of compound pendulum

T = 2π I=

I Mgd

ml 2 ml 2 2ml 2 + = 3 3 3

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59

M=2m d=

l l cos 60 = 2 4

Putting these values, we get T = 4π

l 3g

274. Answer (2) Δ φ= ω1T – ω2T =

2π 2π × 4 ⋅T − × T = 72 T 5T

275. Answer (1) Elongation in spring in equilibrium = After removal mean position = Hence amplitude is

(m + M ) g k

k

mg k

m M

Mg k

276. Answer (2) Time period of particle is 4 s. In 3 s, particle will complete

3 th oscillation. 4

⇒ Distance travelled = 3a 277. Answer (3)

I mgd

Time period of physical pendulum T = 2π 278. Answer (4) Time period of physical pendulum is

T = 2π

I . I = ml2, d = l ⇒ mgd

2π

l = 1.88 s g

279. Answer (2)

I mgd

Time period of physical pendulum T = 2π 280. Answer (3)

k2 l0–2x

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KE =

1 1 mr 2 v 2 3 mv 2 + = mv 2 2 2 2 2 r 4

PE =

x2 1 1 k1x 2 + k 2 (2 x )2 = (k1 + 4k 2 ) 2 2 2

For S.H.M., → mechanical energy is constant M.E. = KE + PE =

x2 3 mv 2 + (k1 + 4k 2 ) ...(i) 4 2

→ Differentiate (i) w.r.t. time to get

d 2x dt

2

⎛ k + 4k 2 ⎞ = −⎜ 1 ⎟x m ⎠ ⎝

280a. Answer (3)

(IIT-JEE 2009)

Restoring torque ⎛ l ⎞ l τ = − ⎜⎝ k θ⎟⎠ × 2 2 2

= –

kx klθ 2

θ

kl 2 θ 2

θ

⇒

= –

6k θ = – ω2θ m

⇒

ω=

6k m

kx = klθ 2

– kl 2 θ 2 ml 2 12

τ α= = I

= 2πf

281. Answer (1) At mean position, KE is maximum and PE is minimum (KE)max + Umin = 9J (KE)max + 5 = 9 KEmax = 4 =

1 ma 2 ω2 2

1 × 2 × (0.01)2 ω2 2 2π 2 ω = = 200 = T 0.01 4 =

T =

π s 100

282. Answer (4) KE =

ω2 1 1 2 mv 2 + ⋅ mR 2 ⋅ 2 2 2 5 R

PE = mgR (1 – cosθ) (∵ h = R – R cosθ)

R

R cosθ h

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KE + PE =

Machanics

MR 2 ω2 1 mv 2 + 2 5

+ mgR (1 – cos θ)

61

...(i)

⎛θ⎞ θ Differentiate equation w.r.t. time, take sin⎜ ⎟ ≈ ⎝2⎠ 2

283. Answer (1)

⎛ Time period = ⎜⎜ 2π ⎝

m k

⎞ 1 ⎛ ⎟ + ⎜ 2π ⎟ 2 ⎜ ⎠ ⎝

m k

m 2k

= π

m + π k

= π

m (1 + 0.71) = 5.36 k

⎞ 1 ⎟ ⎟2 ⎠

1 2 m k

284. Answer (4)

v1 = ω A12 − ( α )2

2

⇒ A1 =

α2 +

v12 g

285. Answer (3) Reference direction of ω is anticlockwise

ω

t=0

α ω

⎛ α⎞ φ1 = π − sin ⎜ ⎟ ⎝ A1 ⎠

286. Answer (3) Tension in each spring is same So, k1x1 = k2x2 = k3x3

U1 =

1 k1x12 2

U2 =

1 k 2 x 22 2

U3 =

1 k 3 x 32 2

⇒ x1 : x2 : x3 = 3 : 1.5 : 1

⇒ U1 : U2 : U3 = x1 : x2 : x3 = 3 :

3 :1 2

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287. Answer (1) T = 20 N

f = −T ⋅ ⇒ k=

4y l

4T l

Time period = 2π

m ml = 2π k 4T 0.1× 0.2 = 0.03π 4 × 20

= 2π 288. Answer (4)

At mean position, F = – 2x + 4 = –2(x – 2) F = 0 is at x = 2 and motion is SHM. 289. Answer (4) F =

dU 2p q =− 3 + 2 =0 dx x x

⇒

x=

Equilibrium position of particle at x =

ω=

2p q

⎛2p⎞ f '⎜ ⎟ ⎝ q ⎠ m

⎛ 2 p ⎞ dFR f '⎜ ⎟= ⎝ q ⎠ dx

=

2p q

x=

2p 4

q4 8 p3

ω=

q4 8 mp3

290. Answer (1) h is the height of cube inside water h× A 0.9 = 10 × A 1

⇒ h = 9 cm

T = 2π

h 9 = 2π ≈ 0.09486 g 1000 = 2 × 0.3 s

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63

Machanics

291. Answer (1)

Ml 2 I + Md 2 , solve for d. . I= 12 mgd

For physical pendulum T = 2π 292. Answer (3)

KE =

1 1 m(2v )2 + mv 2 2 2

PE =

1 k (2 x ) 2 2

Total energy =

=

2v m

m

4 ×1 2 1 kx + (4mv 2 + mv 2 ) 2 2

5mv 2

2

+

4kx 2

2x

x

v

θ

2

...(i)

Differentiate (i) w.r.t. time dv + 4kx.v = 0 dt

5mv

dv 4k .x = dt – 5m ω2 =

4k 5m

293. Answer (2) At equilibrium, mBgR sinθ = mAgR ⇒ θ=

π 6

For small displacement, mAgR – mBgR sin(θ + φ) =

I=

4kg, R = 1

Id 2 φ

mB

dt 2

MR 2 + M AR 2 + M BR 2 2

Take sin φ ≈ φ and cosφ ≈ 1,

d 2φ dt 2

1 kg (mA)

g 2k θ

= −2 3 φ

294. Answer (1)

k

⇒ y

z

x

m 2x = y + z

...(i)

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At equilibrium T = mg

...(ii)

T = ky 2

...(iii)

T = kz 2

...(iv)

T = k (y + 2) T = k(2x) ⇒ ΔT = 2kΔx ⇒ mω2ax = 2kax 2k m

ω2 =

295. Answer (4) 1 1 MR 2 Mv 2 + 2 2 2

Sum of KE and PE =

=

v2 R2

1 1 1 2 Mv 2 + Mv 2 + kx = constant 2 4 2

3 1 2 Mv 2 + kx = constant 4 2

+

1 2 kx 2

M,R x

...(i)

Differentiate (i) w.r.t. time 3 dv 1 M ⋅2 v + k 2 xv = 0 4 dt 2

3 kx Ma + =0 4 2 a =

2k x = ω2 x 3M

ω2 =

3M 2k

T = 2π

295(a).

3M 2k

Answer (4)

a=−

IIT-JEE 2008

2kx I M+ 2 r

with a = –ω 2x ⇒ ω=

2k M+

I

=

4k 3M

r2

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65

296. Answer (1)

FR = −2T sin θ = −

2T x x2 +

FR = −

4T

2

/2 4

θ

θ

x

x

47 m

ω=

T = 2π

m = 3.14 4T

0.04 × 1 = 0.2 s 10

297. Answer (2) Displacement of particles 1, 2, is

A 2

To get separation between two particles having displacements of amplitudes

A we have 2

A = A sin (Kx – ωt) 2

Kx =

y

π 5π , (at t = 0) 6 6

1

2

λ x2 – x1 = 3

x

∴ separation between 1 & 2 is ⇒

λ 3

3

4

λ = 4cm or λ = 12 cm 3

For two particles having same displacement equal to amplitude, distance λ = 12 cm 298. Answer (4)

V =6m ν Δφ Δx = 2π λ

λ=

⇒ Δφ = 2π ×

1 π = 6 3

S1 is lagging in path, so in phase as well 299. Answer (3) In an open organ pipe, both ends are free ends, hence both are displacement antinodes & hence pressure nodes. 300. Answer (1) 4 y(x, t) = ( x + 2t )2

dy 16 vp = dt = − ( x + 2t )3 vp (x = 2, t = 2) = −

2 m/s . 27

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301. Answer (4) f1 = 3.

V 4L

V 6V V 5L f2 = = = 5L L′ 6

f1 3 5 5 × = = f2 4 6 8

302. Answer (4) 1 Hz 5

f=

λ = 2 × 10 = 20. As v = f λ,

10 m

⇒ v = 4 m/s 303. Answer (3)

λ' =

Wave speed relative to observer f

λ' =

v + v ω 340 = × 33 = 34 m . f 330

304. Answer (3) For wave y = A sin(ωt – kx), v0 = Aω At y = ∴

A 1 , sin(ωt – kx) = 2 2

dy 3 = Aω cos(ωt – kx) = v 0 . dt 2

305. Answer (1) As source approaches detector, frequency received by detector increases. So energy received by detector per unit time increases. 306. Answer (3) fA ~ fB = beat frequency fA = fB ± beat

(i)

fB = fA ± beat

(ii)

If B is loaded fB decreases and fA remains same ∴ from (i) fB = 208 or 216. If fB = 208, then on loading, fB decreases and beat frequency increases so fB = 216 Hz. 307. Answer (4) y1 = (5 × 10–5) sin(100 πt) ; at x = 0 y2 = (5 × 10–5) sin(100 πt+ φ) ; at x = – 3 m where,

φ Δx = 2π λ

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67

⇒ φ = π y2 – y1 = (10 × 10–5 m) sin

φ φ⎞ ⎛ . cos ⎜100 πt + ⎟ 2 2 ⎝ ⎠

Maximum difference in displacements = (10 × 10–5 m) sin

φ = 10–4 m 2

308. Answer (3) u =

I ; I = Intensity = constant v

I = 2π2ν2A2ρv

I = 2π2ν2A2ρ v

⇒u=

i.e. u ∝ ρ 309. Answer (3) There is no power transmission in a standing wave. 310. Answer (2)

1 mω2 A 2 = Energy density 2 V 1 ρ ω 2 A 2 = 0.16 π2 2 Putting values of ρ and A, α = 400 π ⇒ Frequency = 200 Hz 311. Answer (3)

v =

ω = 2 m/s k

v=

T μ

2=

m ×10 μ

μ =

m = ρ× A l

m⎞ ⎛ ⎜∵ ρ = ⎟ Al ⎝ ⎠

⇒ m = 0.004 kg. 312. Answer (1) Velocity of motorist = 36 km/hr = 10 m/s

A 5m

AB = ( AF )2 + (BF )2 = 52 + x 2

F

B

C ⇒

AB + BC = 2 5 + x 2

2

⇒ 330 = 2 52 + x 2 Solving we get x = 164.9

165

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313. Answer (2) f is constant till car approaches traffic signal, given by f =

v f v − vs 0

but

f > f0.

f is constant, while car moves away given by f =

v f v + vs 0

f < f0.

314. Answer (1) ΔP is maximum at x, where y is minimum ⇒

So x =

dy =0 dx 34 = 0.34 . 100

315. Answer (4) ⎛ v ⎞ ⎟f ∴ fmax = ⎜ ⎝ v – ωR ⎠

fmin

⎛ v ⎞ ⎟f = ⎜ ⎝ v + ωR ⎠

20 cm s

v = ω R = 200

fmax v + ωR 53 = = . fmin v – ωR 13

316. Answer (1) I Loudness = 10 log I dB 0

L = 0, when I = I0 317. Answer (1) Maximum pressure amplitude of wave can be at most equal to atmospheric pressure

I=

Δp02 (1.01× 105 )2 = = 1.18 × 107 W/m2 2ev 2 × 1.3 × 332

L = 10 log

I I0

10log

107 ≈ 190 dB 10 −12

318. Answer (3) f =

v λ

f1 – f2 =

10 3

v v 10 − = 1 1.01 3

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69

319. Answer (3) V1 V2 = 4L 2L ⇒ V1 = 2V2 = 660 m/s

This is speed of sound in gas at 30°C. If V is speed of sound in gas at 7°C. V = V1

280 300

⇒ V = 638 m/s 320. Answer (2) Let speed of wave be v. For crossing one wave crests to other while travelling in same direction, the surfing speed has to be greater than speed of wave i.e. V < 15 m/s Let wavelength of wave be λm While surfing in same direction λ = (15 - v) × 0.8 While surfing in opposite direction, λ = (15 + v) × 0.8 Equating both, we get v=

15 m/s 7

∴ λ = 10.3 m. 321. Answer (1) ⎛ T2 – T1 ⎞ ⎟x T = T1 + ⎜ l ⎝ ⎠

v= k T dT ⎛ T2 – T1 ⎞ dx ⎛ T2 – T1 ⎞ =⎜ ⎟. ⎟ T = k⎜ dt ⎝ l l ⎠ dt ⎝ ⎠ T2

⇒

∫

t

⎛ T2 – T1 ⎞ ⎟ dt = k⎜ l T ⎝ ⎠0

dT

T1

⇒t=

∫

(

2L

k T1 + T2 322. Answer (4)

)

An open pipe and closed pipe of equal length cannot have same frequency at any harmonic. 323. Answer (1)

V = 330 4L 3 m ⇒L= 4 L 3 ⎛ 1 1⎞ = m=⎜ + ⎟m ⇒ 2 8 ⎝4 8⎠ 3.

λ=

L/2 1 m 8 x=0

x=

V 330 = =1m v 330

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x = 0 is a pressure node and mid-point of organ pipe is at x = ΔP = ΔP0 sin

2π x λ

= ΔP0 sin

2π 1 x 1 8

= ΔP0 sin

π 4

=

1 m. 8

ΔP0 2

324. Answer (1) VAB =

2T μ

VAC =

T μ

VAB = 2. VAC

∴

325. Answer (1) Time taken by pulse to travel a distance x from bottom is

t=2

x g

In this time the distance fallen by particle in its free fall motion is (L − x ) =

1 2 gt 2

L from the bottom 3

⇒

326. Answer (1)

f =

1 T 2L M

f' T' = f T f +6 1.44 T = f T

⇒

f '' = f ''

[As ′′ =

f '' =

f = 30 Hz

+ 0.20 = 1.20 ]

30 = 25 Hz 1.2

Δf = f′′ – f = 25 – 30 = – 5Hz Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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71

327. Answer (4) P

IA =

4π (1)2

; IB =

P 4π ( 2) 2

⇒

IA 4 = IB 1

I ∝ A2 f 2 I A ⎛ AA ⎞ =⎜ ⎟ I B ⎝ AB ⎠

∴

2

AA 2 = AB 1 .

328. Answer (4) μ=

λ y L

mass of length y ⎛λ⎞ m = ⎜ ⎟Y ⎝L⎠

λy m= 2L

dy

2

∴

y

. T = mg + ma

v = T = 2yg μ

L

3g

⇒ u increases with y

dv = 2yg 2g = g a= v dy 2 y t=

⎛ 2L ⎜∵ t = . ⎜ g ⎝

⎞ 2H for uniform accelerati on ⎟ ⎟ g ⎠

329. Answer (4) I ∝ f 2A2 330. Answer (2) For a wave moving towards +x–axis, particles on the positive slope are moving down ward. At the instant, C is moving downwards and wave is travelling left. 331. Answer (3) Option (2) and (4) are obviously wrong as for standing waves two wave should travel in opposite direction for standing wave y = 2[cos(4x – πt) – cos (4x + πt)] = 2 × 2 sin 4x sin πt for node sin4x = 0 As at x = 0 there exists a node. Aakash Educational Services Ltd.-Regd. Office: Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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332. Answer (2) The beats frequency heard are 1, 2, 3, 5, 7 and 8. The combinations 552, 553, 555 and 560 Hz can give all the above frequencies. 333. Answer (3) l = l 1 + l 2 + l 3 + ---. Also

1

=

n , 2f1

2=

n ,−− 2f2

1+ 1 1 = ........... f1 f2 f

334. Answer (2) For n loops,

=

nλ 2

⇒ L = 15 cm.

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