Newton's Laws of Motion Resonance

NLM

1.

Which of Newton's laws of motion is involved in rocket propulsion ? Ans. Newton's third law of motion.

2.

Action and reaction are equal in magnitude and opposite in direction. Then, why do they not balance each other ? Ans. Action and reaction act on different bodies.

3.

A body is acted upon by a number of external forces. Can it remain at rest ? Ans. If the vector sum of all the external forces is zero, then the body will remain at rest.

4.

Identify the contact surface :

(a)

(b)

(c)

(d)

(e)

//////////////////////////////////////////////////////////////

(f)

(g) 5.

6.

7.

///////////////////////////////////////////////////

Find the velocity and acceleration of block A.

Ans. VA = 2 m/s (), aA = 4 m/s2 () Find the velocity of block B.

Ans. VB = 5 m/s (  )

Find all the normal reactions and the accelerations : (a) Ans.

(b) (a) N = 100 N, a = 2 m/s2 (c) N = 20 N, a = 2 m/s2|

RESONANCE

(c)

(d)

(b) N = 50 N, 150 N, a = 0 (d) N = 50 N, 150 N, a = 2 m/s2

Page # 1

8.

9.

Draw the FBD for the following systems : (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

A body of mass m is suspended by two strings making angles  and  with the horizontal. Find the tension in the strings. See figure

(A) T1 

mg cos   T2 sin(  )

(C*) T1  10.

mg cos  mg cos  , T2  sin(  ) sin(  )

12.

13. 14.

mg sin  sin(  )

(D) none of these

A small block B is placed on block A of mass 1 kg and length 20 cm. If initially the block is placed at the right end of block A. A constant horizontal force of 10 N is applied on the block A. All the surfaces are assumed frictionless. Find the time in which B separates from A. [RRD PEARSON\Q.17/Pg-78]

(A*) 0.2 s 11.

(B) T1  T2 

(B) 0.32 s

(C) 0.39 s

(D) 0.45 s

A toy train consists of three identical compartment A, B and C. It is being pulled by a constant force F along C. The ratio of the tensions in the string connecting AB and BC is (A) 2 : 1 (B) 1 : 3 (C) 1 : 1 (D*) 1 : 2 [RRD PEARSON\Q.1/Pg-84] Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5N on the heavier mass, the force on the lighter mass will be (A*) 2N (B) 4 N (C) 5 N (D) None of these [RRD PEARSON\Q.4/Pg-85] Two masses m1 and m2 are attached to a string which pass over a frictionless fixed pulley. Given that m1 = 10 kg and m2 = 6 kg and g = 10 ms–2, What is the acceleration of the masses? (A*) 2.5 ms–2 (B) 5 ms–2 (C) 20 ms–2 (D) 40 ms–2 Two masses m1 and m2 are attached to the ends of a string which passes over a pulley attached to the top of an inclined plane. The angle of inclination of the plane in . Take g = 10 ms–2

RESONANCE

Page # 2

If m1 = 10kg, m2 = 5kg,  = 30º, what is the acceleration of mass m2? [RRD PEARSON\Q.30/Pg-86] (A*) zero (B) (2/3) ms–2 (C) 5 ms–2 (D) 10 ms–2 15.

A body is placed over an inclined plane of angle . The angle between normal reaction and the weight of the body is (A) 

16.

(B)

 – 2

(C*)  – 

Find out the mass of block B to keep the system at rest :

(a)

(D) 0

[RRD PEARSON\Q.35/Pg-87] [BUT DPP#13/Q.11]

(b)

Ans. 10 kg 17.

18.

Find out the accelerations of the blocks and tensions in the strings.

[BUT DPP#13/Q.13]

Ans. zero

Find the velocity of A.

[Q.12/Build-up-DPP-9] Ans.. 19.

VA = 24 m/s ()

Find the acceleration of B.

[Q.13/Build-up-DPP-9] Ans.

aB = 2m/s2 (

RESONANCE

)

Page # 3

20.

Find the velocity of A.

[Q.15/Build-up-DPP-9] Ans. 21.

VA = V cos  ()

Find the acceleration of wedge A

[Q.2/Build-up-DPP-10]

22.

Ans. aA = b tan

Find the acceleration of wedge A.

[Q.3/Build-up-DPP-10]

23.

Find the acceleration of B.

[Q.4/Build-up-DPP-10]

a cos 1 Ans. aB = cos  2

For the following questions assume. 24.

Find the acceleration of A w.r.t. ground.

[Q.8/Build-up-DPP-10] Ans. – b ˆi – 4b ˆj 25.

Find the acceleration of C w.r.t. ground

RESONANCE

Page # 4

[Q.9/Build-up-DPP-10]

26.

Ans. a ˆi – 2 (a + b) ˆj

Find the velocity of B w.r.t. ground.

[Q.10/Build-up-DPP-10]

Ans. – 4 ˆi + 8 ˆj 27.

Find the acceleration of B w.r.t ground.

[Q.6/Build-up-DPP-10] Ans. a (1 – cos) ˆi + a sin ˆj 28.

Sol.

29.

Sol.

Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of 10 m/s2. In this box if 1 kg body is suspended with a light string, the tension in the string in equilibrium position. (w.r.t. experimenter) will be. (Take g = 10 m/s2) [Made PKS, 2005] (A) 10 N (B*) 10 2 N (C) 20 N (D) zero Acceleration of box = 10 m/s 2 Inside the box forces acting on bob are shown in the figure

T=

(mg)2  (ma)2 = 10 2 N

The block of mass ‘m’ is being pulled by a horizontal force F = 2 mg applied to a string as shown in figure (where ‘g’ is acceleration due to gravity). The pulley is massless and is fixed at the edge of immovable table. The force exerted by supporting table to the pulley is [Made MPS-2005]

(A) mg

(B) 2mg

The F.B.D. of pulley is as shown

(C)

3 mg

(D*) 2 2 mg

   Let T1 , T2 and FS be the forces exerted by the horizontal string, vertical string and the support on the

RESONANCE

Page # 5

massless pulley respectively. Then    T1 + T2 + FS = 0    or | FS | = | T1 + T2 | = 2 2 mg   (  Tension in each string is | T1 | = | T2 | = 2 mg) 30.

If the resultant of three forces F1 = pi + 3j - k, F2 = -5i + j + 2k and F3 = 6i - k acting on a particle has magnitude equal to 5 units, then the value(s) of p is (are) (A) -6 (B*) -4 (C*) 2 (D) 4 [1994]

31.

Select the incorrect statement : (A)

(B*)

(C*) Sol.

(D)

A powerful man pushes a wall and the wall gets deformed. The magnitude of force exerted by

the wall on man will be equal to the magnitude of force exerted by man on the wall.

Ten person are pulling horizontally an object on a smooth horizontal surface in different

directions. The resultant acceleration of the object is zero. Then the pull of each man is equal to the pull of the other nine men.

A massless object cannot exert force on any other object.

All contact forces are electromagnetic in nature.

(B) & (C)

The direction of pull of each man is in opposite direction to that of net pull by nine other people. A massless body can exert force on other bodies if other bodies exert force on it 

32.

Sol.

B & C are correct choices. marks shall be awarded for either correct choice.

Two springs are in a series combination and are attached to a block of mass ‘m’ which is in equilibrium. The spring constants and the extensions in the springs are as shown in the figure. Then the force exerted by the spring on the block is :

k1 k 2 (A*) k  k (x 1 + x 2) 1 2

(B) k 1x 1 + k 2x 2

(C*) k 1x1

(A, C) Tension in both springs are same i.e. k 1 x 2 = k 2 x 2 = force exerted by lower spring on the block.

RESONANCE

(D) None of these

Page # 6

Comprehensions # 1 Two blocks (as shown in figure) are connected by a heavy uniform rope of mass 4 kg. An upward force of 300 N is applied on upper block. (g = 10 m/sec2) [D TEST 2007]

33.

The acceleration of the system is equal to ; (A) 20 m/sec2 upward (B*) 10 m/sec2 upward (C) 10 m/sec 2, downward (D) 30 m/sec 2, downward

34.

The tension at the top of the heavy rope is equal to (A) 90 N (B) 60 N (C*) 180 N

(D) 240 N

35.

The tension at the mid-point of the rope is equal to (A) 240 N (B) 180 N (C) 70 N

(D*) 140 N

Sol.

T 2 – (5 + 2)g = (5 + 2) a T 2 = 140 N

Match the following : 36. (p) force between the earth and the falling stone (i) Gravitational force (q) The pressing force between one block and another block (ii) Electromagnetic force (r) The stretching force developed in a spring. (iii) Strong nuclear force (s) The force between the proton and the neutron in a nuclear (iv) Weak nuclear force Ans. (p-1), (q-ii), (r-iii), (s-iii) 37.

The expression given below have their usual meaning match them with column on right side u v  t 1. S =   2 

2. Internal force of a system always add upto zero 3. To stop a car in minimum distance break is applied such that car does not slip Ans. 1-b, 2-d, 3-e, 4-g

RESONANCE

(a) variable acceleration

(b) Constant acceleration (c) Newton’s second law (d) Newton’s third law (e) Bodies which can deform (f) Bonding between the molecules (g) s > k

Page # 7

38.

Assertion : The pressing force between two blocks is an example of electromagnetic interaction force. Reason : At microscopic level, all bodies are made of charged constituents (Nuclei and electron.) so any mechanical contact causes mutual forces between there charges. [Made A.K. 2007] (A*) If both Assertion and Reason are true and the Reason is correct explanation of Assertion. (B) If both Assertion and Reason and true but Reason is not a correct explanation of Assertion (C) If Assertion is true but Reason is false. (D) If both Assertion and Reason are false.

39.

A monkey of mass 20 kg is holding a vertical rope. The rope can break when a mass of 25 kg is suspended from it. What is the maximum acceleration with which the monkey can climb up along the rope? [Pearson/Page No. 85/Q.No. 8] (A) 7 ms–2 (B) 10 ms –2 (C) 5 ms –2 (D*) 2.5 ms –2

40.

A cricket player catches a ball of mass 100g moving with a velocity of 25 m/s. If the ball comes to rest in 0.1s after being caught, then the force of the blow exerted on the hand of the player is : [Pearson/Page No. 87/Q.No. 41] (A) 4N (B) 40 N (C*) 25 N (D) 250 N

41.

Find the extension in the spring if : (i) a = g/2 upwards (ii) a = g/2 downwards

Ans. (i) 42.

(iii) a = 0

3 mg 1 mg mg (ii) (iii) 2 k 2 k k

Find out the reading of the weighing machine in the following cases.

Ans. 10 3 N or

3 kg

43.

A block of metal is lying on the floor of a bus. The maximum acceleration which can be given to the bus so that the block may remain at rest, will be (A) g2 (B) 2 g (C*) g (D) /g [Q.14/Objective Physics/Pg.-85]

44.

A block of mass 4kg is suspended through two light spring balances A and B. Then A and B will read respectively : [GRB/obj.physics/pg.148/Q.251]

(A*) The reading of A is 4kg (C*) The reading of B is 4kg

RESONANCE

(B) The reading of A is 0 kg (D) The reading of B is 2kg

Page # 8

Comprehension 45. Mr. Abdulla wishes to move a heavy block of mass 10 kg by means of an engine through a string such that reading of the spring balance is 5 kg. The acceleration of the block will be :

(A*) 4m/s2 46.

(B) 2m/s 2

(C) 2.6m/s2

(D) none of these

If there is a weighing machine on which the 10 kg block is placed and two birds 1 and 2 as shown in figure, the weighing machine reads 7.5 kg and the friction coefficient between the block A and weighing machine is  = 0.6. Mr. Abdulla himself applies a force of 5N as shown. Then the acceleration of the block will be : (Bird '2' is of mass 2 kg)

(A) 3.5 m/s2

(B) 4m/s 2

(C*) 0 m/s 2

(D) 1 m/s 2

47.

What will be the acceleration of the block, if Abdulla makes the bird 1 flew away ? (A) 0 m/s2 (B) 1 m/s2 (C) 0.5 m/s2 (D*) 0.3 m/s 2

48.

What will be the mass of the bird 1 ? (A) 600 gm (B) 1 kg

49.

Find the acceleration of B.

(C*) 500 gm

(D*) can not be determined

[Q.11/Build-up-DPP-9] Ans. 50.

aB = 2 m/s2 ()

If block A is released from rest, then find the extension in spring and accelerations of the blocks of mass m just after release.

(a) Ans. 51.

(b) (a) aA = g ; aB = 0 ; x = mg/K ; (b) aA = g ; x = 0

Find out the accelerations of the block B in the following systems :

[Q.9/Build-up-DPP-13]

RESONANCE

Page # 9

Ans. aB = 2g/3 52.

A bob is suspended with the help of a thread whose breaking load is twice the weight of the bob. Taking g = 10 ms–2, what is the minimum time in which the bob can be raised by 10 m? (A*) 2 s (B) 2 2 s (C) 1/ 2 s (D) 1 s

53.

Find out the time taken by the block A of mass ‘m’ to reach pt. B from pt. A. The length of inclined plane is .

Ans. 54.

t=

8 3g

System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the velocity of the block will be: [ Made 2004]

(A*)

3u 4

(B)

3u 2

(C)

u 4

(D) none of these

55.

The velocity time graph of a lift moving downwards is a straight line inclined to the time axis at 45°. If mass of the lift is m kg. What is the effective weight (in newton) of the lift? Take g = 10 ms –2 (A) m (B*) 9m (C) 10 m (D) None of these

56.

Find out all the normal reactions acting in the system.

Ans.

N1 = m (g cos  + a sin)

Ng = N1 cos + Mg

57.

A book is lying on the table. What is the angle between the action of the book on the table and the reaction of the table on the book? [Q.20/Objective Physics/Pg. 86] (A*) 180° (B) 90° (C) 45° (D) 0°

58.

When we kick a stone, we get hurt. Due to which one of the following properties does it happens? (A) velocity (B) momentum (C) inertia (D*) reaction [Q.40/Objective Physics/Pg. 87]

59.

Find the velocity of A with respect to pulley P.

Ans.

VAP = 8 m/s ()

RESONANCE

Page # 10

60.

In the figure M = 2m. The coefficient of friction at all surfaces is . String is light and inextensible and all the pulleys are smooth. D.C. Pandey_pg.220_30

M

61.

m

The free body diagram for block M will be as follows (True/False).

T

T N1

N1

Mg

T

a

Ans.

False (normal reaction and friction due to the small block is not shown).

62.

The free body diagram for block m will be as follows (True/False).

T

N2

a

N2 mg

Ans.

False (acceleration will be 2a).

63.

Figure shows a small block A of mass m kept at the left end of a plank B of mass M = 2m and length . The system can slide on a horizontal road. The system is started towards right with the initial velocity v. The friction coefficients between the road and the plank is 1/2 and that between the plank and the block is 1/4. Find D.C. Pandey_pg.219_29

A

B

(a) the time elapsed before the block separate from the plank. (b) displacement of block and plank relative to ground till that moment.

 Ans. (a) t = 4 3g 64.

 2  5 (b) SA = 4 V 3g  3  , SB = 4 V 3g  3  ,

The diagram shows a cord supporting a picture. Which of the graphs shown in figure correctly represents the relationship between the tension T in the cord and the angle  ? [JP DPP-24 _Q.2 ]

RESONANCE

Page # 11

(A)

65.

(C)

(D)

In the figure is shown the top view of two horizontal forces pulling a box along the floor. (a) (b)

66.

(B)

A batch DPP 46_19.7.05

How much work does each force do as the box is displaced 70 cm along the vertical line? What is the total work done by the two forces in pulling the box this distance?

[ Ans.: (a) w 1 = 29.7 J, w 2 = 51.89 J (b) 81.2 J]

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length, the line OA will [3 marks] USS_Pg.296_22

F

(A*) Shift towards F-axis (C) Remain as it is

O

A x

(B) Shift towards X-axis (D) Become double in length

67.

In the system of pulleys , the value of M such that 14 kg block remains at rest is :

68.

(A) 28 kg (B) 35 kg (C*) 24 kg (D) 42 kg Two blocks of masses m 1 = 5 kg and m 2 = 2 kg hang on either side of a frictionless cylinder as shown in the figure . If the system starts at rest , what is the speed of m 1 after it has fallen 40 cm ?

DPP 47_ACJ_05-06

DPP 48_ACJ_05-06

[ Ans. :

24 m/s ] 7

RESONANCE

Page # 12

69.

A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass . Constant forces F1 = 20 N and F2 = 8

N start to act at time t = 0 when the distance of block B from pulley is 50 cm . Time when block B reaches the pulley is _______ .

DPP 48_ACJ_05-06

[ Ans. : 0.5 sec ] 70.

71.

A ladder that is 3.0 m long and weighs 200 N has its center of gravity 120 cm from the foot. At its top end is 50 N weight. Compute the work required to raise the ladder from a horizontal position on the ground to a vertical position. DPP 49_ACJ_05-06 [ Ans : 390 J ]

Two blocks with masses m 1 = 4 kg and m 2 = 5 kg are connected by a light rope and slide on a frictionless wedge as shown in the figure . Given that it starts at rest , what is the speed of m 2 after it has moved 40 cm along the incline ? DPP 49_ACJ_05-06

[ Ans. : 1.19 = 8/3 5 ] 72.

(i)

(ii)

(iii)

(iv)

(v)

In all the given cases blocks are at rest, are in contact and the forces are applied as shown. All the surfaces are smooth. Then in which of the following cases, normal reaction between the two blocks is zero : (A) (i) , (iv) (B) (ii) , (iii) (C) (iii) (D*) (v) 73.

 A block of 6 kg is put between two smooth walls. If F = 50 2 is also applied as shown in figure, then [5 Marks]

[Made M. Pathak]

P&Q_DPP 44_8

(A*) Interaction force on the block due to walls = 50 ˆi  110 ˆj (B) Interaction force on the walls due to the block = 50 ˆi  110 ˆj  (C) If F were reversed, now interactron force on the block due to wall =  50 ˆi  110 ˆj

 50 ˆ i m/sec 2. (D*) If F were reversed, now the acceleration of the block = 6

RESONANCE

Page # 13

Sol. Ny = 110 Nx = 50  Nblock  50 ˆi  110ˆj So A is correct  N wall will be opposite of Nblock =  50 ˆi  110 ˆj So B is incorrct.  If F were reversed



it will loose contact with verticle wall and Ny = 10 ˆj So option C is incorrect.

 Net force = 50 ˆi  6 a So D is correct. 74.



 50 ˆ a i 6

A block of mass m is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2), (3) and (4) as shown. If the normal reactions in situation (1), (2), (3) and (4) are N1, N2, N3 and N4 respectively and acceleration with which the block slides on the wedge in situations are b1, b2, b3 and b4 respectively then : [5 Marks] P&Q_DPP 44_9

(A*) N3 > N1 > N2 > N4 (B) N4 > N3 > N1 > N2

Sol.

(C*) b2 > b3 > b4 > b1

(D) b2 > b3 > b1 > b4

(1) Balancing forces perpendicular to incline N = mg cos37° + ma sin37° N1 =

RESONANCE

4 3 mg + ma 5 5

Page # 14

and along incline mg sin37° – ma cos37° = mb1



b1 =

3 4 g– a 5 5

(2) Similarly for this case get N2 = N2 =

4 3 mg – ma 5 5

4 3 mg – ma 5 5

and b2 =

3 4 g+ a 5 5

(3)

Similarly for this case get N3 =

4 4 3 3 mg + ma and b3 = g + a 5 5 5 5

Similarly for this case get N 4 =

4 4 mg – ma 5 5

(d)

and b4 = 75.

3 3 g– a 5 5

Find the tensions in the strings (1), (2) and (3) and the acceleration of the mass ‘m’ just after : [S-05-06/DPP 39/Q.10]

(a) string (1) is cut Ans.

(b) string (2) is cut

(a) T1 = 0 ; T2 = mg ; T3 = 2 mg ; a = g (b) T1 = mg ; T2 = 0 ; a = 0 (c) T1 = mg ; T2 = 0 ; T3 = 0 ; a = 0

RESONANCE

(c) string (3) is cut

Page # 15

76.

A body of mass 10 kg is on a rough inclined plane of inclination = sin 1 (3/5) with the horizontal. When a force of 30 N is applied on the block parallel to and upward the plane, the total reaction by the plane on the block is nearly along : [S-05-06/DPP 44/Q.5]

(A*) OA 77.

(B) OB

(C) OC

(D) OD

Three equal balls 1, 2, 3 are suspended on springs on below the other as shown in the figure. OA is a weightless thread. (a) If the string is cut, the system starts falling. Find the acceleration of all the balls at the initial instant.

(b) Find the initial accelerations of all the balls if we cut the spring BC which is supporting ball 3 instead of cutting the thread. [Bank/Physics_Bank/Jaipur/DPP(A)/Q.10] Ans. [(a) 3g, 0, 0, (b) 0, g, g] 78.

System is shown in the figure. Assume that cylinder remains in contact with the two wedges.The velocity of cylinder is A-Batch_DPP-58_05-06_3

(A)

19  4 3

u m/s 2

(B)

13 u m/s 2

(C)

3 u m/s

(D*)

7 u m/s

79.

In the figure shown the velocity of different blocks is shown. The velocity of C is : (A) 6 m/s JP DPP_51_9 (B*) 4 m/s (C) 0 m/s (D) none of these

80.

Two blocks A (5 kg) and B (3 kg) resting on a smooth horizontal plane are connected by a spring of stiffness 294 N/m. A horizontal force F of magnitude 3 × 9.8 N acts on A as shown. At the instant B has an acceleration of 4.9 m/s2 towards left : JP DPP_51_10

(A) the acceleration of A is 3 × 1.96 m/s2 (C) the extension in the spring is 0.1 m

RESONANCE

(B*) the acceleration of A is 3 × 0.98 m/s2 (D) the extension in the spring is 0.2 m

Page # 16

81.

82.

83.

A ball of mass m is attached to the lower end of a light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal length, and comes to rest again after descending through a distance x : JP DPP_52_1 (A) x = mg/k (B*) x = 2mg/k (C*) The ball will have no acceleration at the position where it has descended through x/2 (D*) The ball will have an upward acceleration equal to g at its lowermost position. Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring. JP DPP_52_2 Ans. [2 mg/k]

In the figure shown neglecting friction and mass of pulleys, what is the acceleration of mass B? fp=kesan'kkZ,vuqlkj?k"kZ.k,oaiqyhdsnzO;ekudksux.;ekusa]nzO;ekuBdkRoj.kD;kgksxk? [ 3.46_NLM ]

(A) 84.

g 3

(B)

5g 2

(C)

2g 2

(D*)

2g 5

Two blocks A and B of equal mass are connected by a light inextensible string passing over two light smooth pulleys fixed to the blocks. The parts of the string not in contact with the pulleys are horizontal. A horizontal force F is applied to the block A as shown. Then: M.Bank_NLM_3.57 nksxqVdsArFkkBftudsnzO;ekulekugS],dgYdhvrU;jLlh}kjk¼jLlhtksxqVdksalsca/ksnksfpduhiqyhesa lsxqtjrhgSA½tqM+sgq,gSaAjLlhdkogHkkxtksiqyhlstqM+kughagS]{kSfrtgSA,d{kSfrtcyFxqVdsAijfp=kkuqlkj yxk;ktkrkgS]rks-

(A) the acceleration of A will be equal to that of B (B*) the acceleration of A will be greater than that of B (C) the acceleration of A will be less than that of B (D) none of the above is correct. (A)Adk Roj.k] B ds Roj.k ds cjkcj gksxk (B*)Adk Roj.k] B ds Roj.k ls vf/kd gksxk (C)Adk Roj.k] B ds Roj.k ls de gksxk (D)buesalsdksbZugha 85.

A weightless inextensible rope on a stationary wedge forming angle  with the horizontal. One end of the rope is fixed to the wall at point A. A small load is attached to the rope at point B. The wedge starts moving to the right with a constant acceleration. Determine the acceleration a1 of the load when it is still on the wedge. M.Bank_NLM_8.12 ,dgYdhvrU;jLlhfLFkjostij{kSfrtlsdks.kcukrsgq,j[khgSAjLlhdk,dfljkfcUnqAijfLFkjgSAfcUnq B ij,dNksVklknzO;ekutksM+j[kkgSAostnk;havksjfu;rRoj.klsxfrdjuk'kq:djnsrkgSAB ijtqM+snzO;ekudk

RESONANCE

Page # 17

Roj.k a1 Kkr dhft,] tcfd ;gvHkhHkhost ijgks-

M.Bank_NLM_8.12

[ Ans: 2 a sin (/2) ] 86.

In the figure shown the blocks A & C are pulled down with constant velocities u . Acceleration of block B is : [ Made 2003] fn[kk;sx;sfp=kesaArFkkC CykWdksadksfu;rosxuls[khapktkrkgSACykWdBdkRoj.kgksxk : [ Made 2003] M.Bank_NLM_8.13

(A)

87.

(C)

u2 tan2  sec  b u2 sec 2  tan  b

(D) zero ‘'kwU;

(B*) m g

(C) 2 m g

(D) none of these

In the figure shown, find out the value of  [ assume string to be tight ] [ Made 2004] fn[kk;sx;sfp=kesadkekucrkb;s[jLlhdksdlkgqvkekus] [ Made 2004] M.Bank_NLM_8.25

(A*) tan1 89.

u2 tan3  b

In the figure shown the car moves down with a constant acceleration g. A bob of mass m is attached to a string whose other end is tied to the ceiling of the car. If the bob remains stationary relative to the car. Then tension in the string is: M.Bank_NLM_5.20

(A) m g/2 88.

(B*)

3 4

(B) tan1

4 3

(C) tan1

3 8

(D)noneofthesebuesalsdksbZugha

A block of weight 9.8N is placed on a table. The table surface exerts an upward force of 10 N on the block. Assume g = 9.8 m/s 2. [M.Bank_NLM_4.3] 9.8N Hkkj okyk ,d xqVdk est ij j[kk gqvk gSA est dh lrg xqVds ij Åij dh vksj 10 N dk cy yxkrh gSA ekfu, g = 9.8 m/s2 (A*) The block exerts a force of 10N on the table xqVdk est ij 10N dk cy yxkrk gSA (B) The block exerts a force of 19.8N on the table xqVdk est ij19.8N dk cy yxkrk gSA (C) The block exerts a force of 9.8N on the table xqVdk est ij 9.8N dk cy yxkrk gSA (D*) The block has an upward acceleration. xqVdsdkÅijdhvksjRoj.kgksxkA

RESONANCE

Page # 18

90.

91. 92.

Two weights W 1 & W 2 in equillibrium and at rest, are suspended as shown in figure. Then the ratio W 1/ W 2 is about : [M.Bank_NLM_3.57] nksHkkjW1 rFkkW2 fp=kkuqlkjyVdsgq,gSrFkklkE;koLFkkvkSjfojkeesagSrksW1/W2 dkvuqikrgksxk:

(A*) 5/4

(B) 4/5

(C) 8/5

(D) none of these buesalsdksbZugha

In the previous question the tension in the string attached to the wall is:

iwoZç'uesanhokjlstqM+hjLlhesarukogksxk& (A*) 0.6 W 1

(B) 0.8 W 1

(C) W 1

[M.Bank_NLM_3.58]

(D) none of these buesalsdksbZugha

Two smooth spheres each of radius 5 cm and weight W rest one on the other inside a fixed smooth cylinder of radius 8 cm. The reactions between the spheres and the vertical side of the cylinder are : [M. Ban k (0 7-

08)_NLM_4.2]

8 lseh0f=kT;kdsfpdusfLFkjcsyudsvUnjnksfpdusxksysçR;sdf=kT;k5lseh0rFkkW HkkjokysfLFkjj[ksgSA

xksyksadhm/oZnhokjksadslkFkçfrfØ;k,saKkrdjks\

93.

(A) W/4 & 3W/4

(B) W/4 & W/4

(C*) 3W/4 & 3W/4

(D) W & W

Two blocks A and B of masses 4 kg and 12 kg respectively are placed on a smooth plane surface. A force F of 16 N is applied on A as shown. The force of contact between A & B is: 4 fdxzk0 rFkk 12 fdxzk0 nzO;eku dsnksCykWdØe'k% ArFkkB fpdus {kSfrtlrgij j[ks gq, gSaA ,d 16 N dkcy F, Aij fp=kkuqlkj yxk;k tkrk gSAArFkk B ds e/; yxus okys lEidZ cy dk eku gksxk & [M.Bank(07-08)_N.LM_4.7]

(A) 4 N 94.

(B) 8 N

(C*) 12 N

(D) 16 N

The block of mass ‘m’ equal to 100 kg is being pulled by a horizontal force F = 2 mg applied to a string as shown in figure (Take g = 10 m/s2). The pulley is massless and is fixed at the edge of an immovable table. What is the value of force exerted by the supporting table on the pulley (in Newton) ‘m’xqVdsdknzO;eku100kggSvkSjbls{kSfrtcyF=2mgtksfdfp=kkuqlkjyxrkgS]ls[khapktkrkgSA(g=10m/ s2ysa)AiqyhnzO;ekughugSrFkkfLFkjestdsfdukjsijc¡/khgqbZgSAest}kjkiqyhijyxk;sx,cydkeku(U;wVuesa)

gksxkA

Sol.

Made_MPS_2005

[M.Bank(07-08)_NLM_3.56]

The F.B.D. of pulley is as shown

RESONANCE

Page # 19

   Let T1 , T2 and FS be the forces exerted by the horizontal string, vertical string and the support on the massless pulley respectively. Then    T1 + T2 + FS = 0    or | FS | = | T1 + T2 | = 2 2 mg   ( Tension in each string is | T1 | = | T2 | = 2 mg) Ans. 2828 95.

A horizontal force of magnitude F is applied on a body of mass m as shown in figure. The magnitude of net normal reaction on the block is : ,dm nzO;ekudhoLrqijF cyfp=kkuqlkjyxk;ktkrkgSAxqVdsijifj.kkehvfHkyEcçfrfØ;kcydkifjek.kgksxk : [M.Bank(07-08)_NLM_4.9]

96.

(A*) 10 2 N

(B)

10

2

N

(C) 10 N

(D) none of these buesalsdksbZugha

Consider the system as shown in the figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (g = 10 m/s 2).

fp=kesan'kkZ;svuqlkjfLFkfrdksekfu,AiqyhrFkkjLlhnzO;ekughugSrFkklHkhlrg?k"kZ.kghugSAjLlhesarukogksxkA

97.

(g = 10 m/s2).

[M.Bank(07-08)_NLM_3.30]

(A) 0 N

(B) 1 N

(C) 2 N

(D*) 5 N

A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s 2, then: [M.Bank(07-08)_NLM_3.45] 2 (fp=kkuqlkj,disUVjLo;acyyxkdjviusvkidksrFkkck¡Dldks5 m/s dsRoj.kls,dnzO;ekughujLlhoiqyhO;oLFkk }kjkfp=kkuqlkjÅijmBkrkgSA;fnisUVjdknzO;eku100 kg rFkkck¡Dldk50 kg gSrFkkjLlhdknzO;ekuux.;gS (;fn g = 10 m/s2 ) rks

tension in the rope is 1125 N (jLlh esa ruko 1125 N) tension in the rope is 2250 N (jLlh esa ruko 2250 N) force of contact between the painter and the floor is 375 N (isUVj rFkk Q'kZ ds chp lEidZ cy 375 N gS) (D) none of these (buesalsdksbZugha) For painter ; R + T – mg = ma R + T = m(g + a) ............(1) For the system ; (A*) (B) (C*)

Sol.

RESONANCE

Page # 20

2T – (m + M)g = (m + M)a 2T = (m + M) (g + a) where ; m = 100 kg M = 50 kg a = 5 m/sec2 

and ; 98.

150  15 2 R = 375 N. T=

=

..............(2)

1125 N

Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on the 2blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is : (A and B are smooth) nks xqVds ‘A’o ‘B’, izR;sd dk nzO;eku‘m’,d fpduh {kSfrt lrg ij j[ks gSaA nks {kSfrt cyF o 2F Øe'k% xqVdsAo ‘B’ijfp=kkuqlkjvkjksfirgSAxqVdkA, xqVdsB ijughafQlyrkgSrksnksxqVdksadse/;dk;ZjrvfHkyEcizfrfØ;kcy

gSA ¼A rFkkB dh laifdZr lrg ?k"kZ.kghu gS½

(A) F Sol.

(B) F/2

nksnzO;ekufudk;dkRoj.kgSa=

99.

(C)

Acceleration of two mass system is a =

FBD of block A

[Q.33/RK_BM/Constrained Motion] [M.Bank(07-08)_NLM_8.29]

F 2m

F

(D*) 3F

3

F leftward 2m

ck;havksj

N F

xqVds Adk eqDr oLrq js[kkfp=k

60°

30°

mF N cos 60° – F = ma = solving N = 3 F 2m In the shown mass pulley system, pulleys and string are massless. The one end of the string is pulled by the force F = mg. The acceleration of the block will be [M.Bank(07-08)_NLM_3.51] fp=kesan'kkZ;sf?kjuhnzO;ekufudk;esaf?kjuhrFkkjLlhnzO;ekughugSAjLlhdk,dfljkcyF=mg}kjk[khapk

tkrkgSAxqVdsdkRoj.kgksxk&

(A) g/2 100.

Sol.

(B) 0

(C*) g

(D) 3g

The system starts from rest and A attains a velocity of 5 m/s after it has moved 5 m towards right. Assuming the arrangement to be frictionless every where and pulley & strings to be light, the value of the constant force F applied on A is : [Made 2006, VSS, GRSTUX] [M.Bank(07-08)_NLM_3.54] fudk;fojkelsizkjEHkgksrkgSrFkkAnk;havksj5m pyusdsi'pkr~5m/s dkosxizkIrdjrkgSAO;oLFkkdkslHkhtxg ?k"kZ.kjfgrof?kjfu;ksaoMksfj;ksadksgYdhekursgq,Aijvkjksfirfu;rcyFdkekugS-

(A) 50 N (B)

RESONANCE

(B*) 75 N

(C) 100 N

(D) 96 N Page # 21

a=

v 2 25  = 2.5 m/s2 2s 10

For 6 kg : – F – 2T = 6a For 2 kg : – T – 2g = 2 (2 a) From (1) & (2) F = 75 N 101.

In the figure shown, find out the value of  [ assume string to be tight ] fn[kk;sx;sfp=kesadkekucrkb;s[jLlhdksdlkgqvkekus] [ Made 2004]

(A*) tan1 102.

ugha

3 4

(B) tan1

4 3

(C) tan1

3 8

[M.Bank(07-08)_NLM_8.23]

(D) none of these buesa ls dksbZ

A system is as shown in the figure. All speeds shown are with respect to ground. Then the speed of Block B with respect to ground is : [M.Bank(07-08)_NLM_8.27] fp=kesa,dfudk;iznf'kZrgSAlHkhpky/kjkrydslkis{kgSaArks/kjkrydslkis{kCykWdBdhpkygksxhA

[Q.31/RK_BM/Constrained Motion]

Sol.

(A) 5 m/s (B)

(C) 15 m/s

(D) 7.5 m/s

 1 + 2  2 = constant

d 1 2 d 2 + =0 dt dt ( 5 + 5) + 2 (5 + vB) = 0 or

 103.

(B*) 10 m/s

vB = 10 m/s

Three blocks are connected by strings as shown in figure and pulled by a force F = 60 N. If m A = 10 kg, m B = 20 kg and m C = 30 kg, then : rhu CykWdksa dks fp=kkuqlkj jLlh ls tksM+dj F = 60 N cy ls [khapk tkrk gSA ;fn mA = 10 kg, mB = 20 kg rFkk mC = 30 kg gks rks: [M.Bank_NLM_3.34]

RESONANCE

Page # 22

(A) acceleration of the system is 2 m/s 2 (B*) T 1 = 10 N (C*) T 2 = 30 N 104.

fudk;dkRoj.k2m/s2gksxkA

(D) T 1 = 20 N & T 2 = 40 N

A cart of mass 0.5 kg is placed on a smooth surface and is connected by a string to a block of mass 0.2 kg. At the initial moment the cart moves to the left along a horizontal plane at a speed of 7 m/s. (Use g = 9.8 m/ s2) nzO;eku0.5fdxzk-dh,dxkM+h]0.2fdxzk-nzO;ekudsfi.MlsjLlh}kjktqM+hgS]izkjEHkesaxkM+hfpdus{kSfrtlery

ijck;ahavksj7eh-/ls-. dhpkylsxfrekugS(g=9.8eh-/ls2 ysa) [M.Bank_NLM_3.47]

0.5 kg

[Modified AKS_2007]

0.2 kg 2g towards right. 7 (B*) The cart comes to momentary rest after 2.5 s. (C*) The distance travelled by the cart in the first 5s is 17.5 m. (D) The velocity of the cart after 5s will be same as initial velocity. (A*) The acceleration of the cart is

(A*) xkM+hdkRoj.k

2g nk;harjQgksxkA 7

(B*)xkM+h2.5lsd.M i'pkr~{kf.kd:dsxhA

(C*)igys5 lsd.MesaxkM+h}kjkr;dhxbZnwjh17.5eh-gSA (D)5lsd.Mi'pkr~xkM+hdkosxizkjfEHkdosxdslekugksxkA Sol.



0.2 g = 0.7 a  a =

2g m/s 2 7

For the case, it comes to rest when V = 0

ml fLFkfr ds fy, tc ;g fLFkj gksrk gS V = 0  2g  t 0 = 7 +   7 

49  t = 2g = 2.5 s

Distance travelled till it comes to rest

7 m/s

a

T T

T = 0.5 a

fLFkjvoLFkkvkusrdpyhxbZnwjh

a 0.2g

0.2 - T = 0.2 a

 2g  s 0 = 72 + 2    7 

S = 8.75 m So in next 2.5s, it covers 8.75 m towards right. Total distance = 2 x 8.75 = 17.5 m After 5s, it speed will be same as that of initial (7 m/s) but direction will be reversed.

105.

vr%vxys2.5 ls-esa;gnk;havksj8.75 eh-nwjhr;djsxkA dqy nwjh = 2 x 8.75 = 17.5 m 5 ls-i'pkr~bldhpky]izkjfEHkdpky(7 eh-/ls-)dscjkcjgksxhijUrqfn'kkfoijhrgksxhA

A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass. Constant forces F 1 = 20 N and F 2 = 8 N start to act at time t = 0 when the distance of block B from pulley is 50 cm.Time when block B reaches the pulley is _______. (Assume that friction is absent every where. Pulley and string are light.

,d 1 fdxzk- nzO;eku dk CykWd ‘B’ leku nzO;eku ds CykWd ‘A’ ij fLFkj j[kk gS t = 0 ij cy F1 = 20 N rFkk

RESONANCE

Page # 23

F2 = 8 N yxk;stkrsgSblle;CykWdB dhiqyhlsnwjh50 cm gSrksCykWdB fdrusle;esaiqyhrdigqapsxk_____(;g ekfu,fdlHkhtxg?k"kZ.kvuqifLFkrgSA¼iqyhrFkkjLlhgYdhgS) [M.Bank(07-08)_NLM_3.2]

[ Ans: 0.5 sec. ] 106.

In the figure if blocks A and B will move with same acceleration due to external agent, there is no friction between A and B, then the magnitude of interaction force between the two blocks will be () : fp=kesa;fnCykWdArFkkB fdlhckácydsdkj.k,dlekuRoj.klsxfrdjsarFkkArFkkB dse/;dksbZ?k"kZ.kcy

ughagksrksnksuksaCykWdksdse/;vfHkyEcçfrfØ;kcydkekuD;kgksxkA [M.Bank_NLM_4.10]

Sol.

(A*) 2 mg/cos  N cos 

(B) 2 mg cos  (C) mg cos  N = 2 mg/cos . Ncos

N a

Nsin

(D) none of these



2m A



a

2mg

107. A 2 kg toy car can move along an x axis. Graph shows force F x, acting on the car which begins at rest at time t = 0. The velocity of the particle at t = 10 s is :

,d2 kg dh f[kykSuk dkj x v{k ds vuqfn'k xfr dj ldrh gSA xzkQ cy Fx, dks çnf'kZr djrk gS] tks le; t = 0 ij fojke ij dkj ij yxuk çkjEk gksrk gSA t = 10 s ij d.k dk osx gS % [Made A.K.S. sir] [M.Bank_NLM_7.35]

Fx(N) 4

0 -2

Sol.

(A) – i m/s

 dp

= pf – pi

=

 F dt

pi = 0 Net Area = 16 – 2 – 1

4

(B) – 1.5 i m/s

8

9 10

= Area under the curve.

t(s)

11

(C*) 6.5 i m/s

(D) 13 i m/s

= 13 N-s

13 = 6.5 i m/s 2 [As momentum is positive, particle is moving along positive x axis.] = Vf =

RESONANCE

Page # 24

108.

Sol.

109.

Sol.

110.

Three rigid rods are joined to form an equilateral triangle ABC of side 1m. Three particles carrying charges 20  C each are attached to the vertices of the triangle. The whole system is at rest always in an inertial frame.The resultant force on the charged particle at A has the magnitude. rhulqn` N2 > N4 (B) N4 > N3 > N1 > N2 (A), (C)

(C*) b2 > b3 > b4 > b1

a

(1)

mgsin37°

N

(D) b2 > b3 > b1 > b4 macos37° ma (t

M+Roh; cy )

masin37°

mg mgcos37°

Balancing forces perpendicular to incline N = mg cos37° + ma sin37° ur ry ds yEcor~ cyksa dks larqfyr djus ij N = mg cos37° + ma sin37° N1 =

4 3 mg + ma 5 5

b1 =

3 4 g– a 5 5

and along incline mg sin 37° – ma cos 37° = mb1 ,oa ur ry ds vuqfn'k mg sin 37° – ma cos 37° = mb1

(2)

(t

masin37° N

M+Roh; cy ) ma

macos37° mgsin37°

a mg mgcos37°

Similarly for this case get N 2 =

4 3 mg – ma 5 5

4 5

3 5

blh izdkj bl fLFkfr ds fy, ge ikrs gSa N2 = mg – ma and

vkSj b2 = N2 =

RESONANCE

3 4 g+ a 5 5

4 3 mg – ma 5 5 Page # 35

(3)

Similarly for this case get N3 =

4 4 mg + ma 5 5 4 5

4 5

blh izdkj bl fLFkfr ds fy, ge ikrs gSaN3 = mg + ma and vkSj b3 =

(d)

3 3 g+ a 5 5

Similarly for this case get N4 =

4 4 mg – ma 5 5 4 5

4 5

blh izdkj bl fLFkfr ds fy, ge ikrs gSa N4 = mg – ma and vkSj b4 = 131.

3 3 g– a 5 5

The force acting on a body moving along x axis varies with the position of the particle as shown in the fig. The body is in stable equilibrium at Bank new _WPE_34 [M.Bank(07-08)_WPE_6.3] x v{kdsvuqfn'kxfr'khy,doLrqijdk;Zjrcyd.kdhfLFkfrdslkFkfp=kkuqlkjifjofrZrgksrkgSAoLrqfdlfLFkfr

ijLFkk;hlkE;koLFkkesagSA

Sol.

132.

Sol.

(A) x = x1 (B*) x = x 2 (C) both x 1 and x 2 (D) neither x 1 nor x2 (A) x = x1 (B*) x = x 2 (C)x1 rFkkx2nksauks (D)x1 rFkkx2nksauksugha Body will be in equilibrium at both x1 & x2 as at these points force will be zero. At x2 on increasing x force becomes -ve & on decreasing x force becomes + ve. so force & displacement have opposite signs. so it a pt. of stable eq. S1 S2 S3 S4

: : : :

S1 S2 S3 S4

: : : :

Newton's third law depends on Newton's second law. Newton's first law can be derived from Newton's second law. All the three Newton's laws are independent of each other. A stationary body is kept stationary on ground. Then the gravitational force exerted by earth on block and normal reaction exerted by block on earth is an example of action reaction pair illustrating Newton's third law.

U;wVu dk rhljk fu;e] U;wVu ds f}rh; fu;e ij fuHkZj djrk gSA U;wVu ds f}rh; fu;e ls] U;wVu ds izFke fu;e dks O;qRiUu fd;k tk ldrk gSA U;wVudslHkhrhuksafu;e,dnwljslsLorU=kgSA ,doLrqdkstehuijfojkeesaj[kktkrkgSrcCykWdiji`Foh}kjkyxk;kx;kxq:Rokd"kZ.kcyrFkkCykWd }kjk i`Foh ijyxk;kx;k vfHkyEccy]U;wVu ds rhljs fu;e esafØ;k o izfrfØ;k ;qXe dks crkrs gSA

(A*) F T F F (B) F F T T (C) T T T F S1 : Newton’s third law is independent of first law. Hence S1 is false.

(D) T F F F

S2 : Newton’s first law is a special case of second law when a = 0. S2 is true. S3 : Newton’s first law is derived from second law. S3 is false. S4 : The normal reaction by earth on block and by block on earth form action and reaction pair. S4 is false. S1 : U;wVu dk rhljk fu;e] igys fu;e ls Lora=k gSA vr% S1 vlR; gSA S2 :U;wVu dk izFke fu;e]U;wVu dsnwljs fu;edk ghfo'ks"kfLFkfr gSA tc a=0.S2 lR; gSA S3 : U;wVu dk izFke fu;e dks U;wVu ds f}rh; fu;e ls O;qRiUu fd;k tk ldrk gSA vr% S3 vlR; gSA S4 :CykWd ij i`Foh }kjk yxk;k x;k vfHkyEc cy o CykWd }kjk i`Foh ij yxk;k x;k vfHkyEc cy fØ;k izfrfØ;k ;qXe cukrsgSaAvr%S4 vlR;gSA

RESONANCE

Page # 36

133.

A particle is projected with a velocity 20 m/s from bottom of inclined smooth, fixed plane of inclination 30º and height 10 m. Find out the maximum height (from ground) reached by the particle. (g = 10 m/s2) ,dCykWd dks = 30° dsfpdusurrydsfuEurefcUnqls20 m/s dsosx lsiz{ksfir fd;k tkrk gSA urry dh Å¡pkbZ 10 m gSA CykWd urry ij vf/kdre fdruh Å¡pkbZ ¼tehu ls½ ij igq¡psxkA (g = 10 m/s2)[M.Bank(07-08)_WPE_5.42]

[Ans.: 12.5 m] 134.

A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is x2 = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is : [M.Bank(07-08)_CM_1.8]

fp=kkuqlkjijoy;dhlehdj.kx2=4aygSAbldhÅ/okZ/kjv{krFkk'kh"kZewyfcUnqijgS]blijoy;rkjijmnzO;eku dh eudkfLFkrgSA rkj dk Ýse fLFkj gS rFkkeudk ¼eudk½fcuk ?k"kZ.k ds ijoy; ij fQly ldrk gSA rkj Ýse ij y =4afcUnqlseudkfLFkjkoLFkklsNksM+ktkrkgSAtceudky=afLFkfrijigq¡prkgSrksbldkLi'kZjs[kh;Roj.kgS: BM_CM_161

[MB_Q. 1.8]

[Q.161/RK_BM/Circular Motion] [Made MPS, 2005]

Sol.

g 3g (B) 2 2 x2 = 4ay Differentiating w.r.t. y, we get yds lkis{k vodyu (A)

(C*)

g

(D)

2

g

5

dy x = dx 2a

dy =1  hence vr%  = 45° dx the component of weight along tangential direction is mg sin . HkkjdkLi'kZjs[kh;fn'kkesa?kVd mgsin.



At (2a, a),

hence tangential acceleration is g sin  =

vr% Li'kZ js[kh; Roj.k gsin= 135.

g

g

2

2

A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity

3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60º with the ground is: [ Made 2004]

[M.Bank_NLM_8.16]

,dNM+ABfp=kesaiznf'kZrgSANM+dkAfljkHkwfeijfLFkj¼c¡/kk½gSrFkkCykWdnk¡;hvkSj 3 m/sosxlsxfrdjjgk gSAtcNM+/kjkryls60º dkdks.kcuk;srksNM+dsfljsBdkosxgksxk&

RESONANCE

Page # 37

[Made 2004]

Sol.

(A) 3 m/s (B*) 2 m/s Let AB =  , B = (x , y)  v B = vx ˆi + vy ˆj  vB =

3 ˆi + v y ˆj

x2 + y2 = 2 

136.

Sol.

2x vx = 2y vy = 0



(D) 3 m/s

(i)

3 +

3 + (tan600) vy = 0  vy = – 1 Hence from (i)  v B = 3 ˆi – ˆj Hence vB = 2 m/s

y v =0 x y

A body with mass 2kg moves in x direction in the presence of a force which is described by the potential energy-displacement graph. If the body is released from rest at x=2m, then its speed when it crosses x = 5m is : ,d2kgnzO;ekudkfi.M,dcytksfdfLFkfrtÅtkZ&foLFkkiuoØ}kjkçnf'kZr gS fd mifLFkfresaxfn'kk esaxfr dj jgk gSA ;fn bls fcUnq x=2m,ls NksM+k tk;s rks x=5mfcUnq dks ikj djrs le; bldh pky gS& [M_Bank(07-08)_WPE_7.7]

(A) zero (B) 1 ms1 (C*) 2 ms1 Loss in potential energy = gain in kinetic energy 6–2=

Alternate : Sol.



(C) 2 3 m/s

(D) 3 ms 1

1 . 2 . (v2) 2 v = 2 m/s.

du =–F dx Since, Slope is constant and negative from x = 1 to x = 3.5 m, the force is accelerating and constant (constant acceleration case)

Slope of U – X graph is

Acceleration (x = 1 to 3.5) = F/m = tan/m = Using ; v2 = u2 + 2aS from x = 2 to x = 3.5 m

(10 / 2.5) = 2m/s2 2

3 v2 = 2(2)   =6 2 From x = 3.5 to x = 4.5 :

&

(2 / 1) = – 1m/s2 2 v22 = v12 – 2aS2 a2 =

RESONANCE

Page # 38

= v22 = 4  v2 = 2m/s After which it will move with constant speed. U – X xzkQ

dk F2 . Page # 54

173.

A cart of mass 0.5 kg is placed on a smooth surface and is connected by a string to a block of mass 0.2 kg. At the initial moment the cart moves to the left along a horizontal plane at a speed of 7 m/s. (Use g = 9.8 m/ s2) ?k"kZ.kjfgrlrgijj[khnzO;eku0.5fdxzk-dh,dxkM+h]0.2fdxzk-nzO;ekudsfi.MlsjLlh}kjktqM+hgSAizkjEHkesaxkM+h

{kSfrtryijck;ahvksj7eh-/ls-. dhpkyls xfrekugS(g=9.8eh-/ls2 ysa)

[M.Bank(07-08)_NLM_3.47]

0.5 kg

[Modified AKS_2007] 0.2 kg

2g towards right. 7 (B*) The cart comes to momentary rest after 2.5 s. (C*) The distance travelled by the cart in the first 5s is 17.5 m. (D) The velocity of the cart after 5s will be same as initial velocity. (A*) The acceleration of the cart is

(A*) xkM+hdkRoj.k

2g nk;harjQgksxkA 7

(B*)xkM+h2.5lsd.M i'pkr~{kf.kd:dsxhA

(C*)igys5 lsd.MesaxkM+h}kjkr;dhxbZnwjh17.5eh-gSA (D)5lsd.Mi'pkr~xkM+hdkosxizkjfEHkdosxdslekugksxkA Sol.



0.2 g = 0.7 a  a =

2g m/s 2 7

For the case, it comes to rest when V = 0

ml fLFkfr ds fy, tc ;g fLFkj gksrk gS V = 0  2g  t 0 = 7 +   7 

49  t = 2g = 2.5 s

Distance travelled till it comes to rest

7 m/s

a

T T

T = 0.5 a

fLFkjvoLFkkvkusrdpyhxbZnwjh

a 0.2g

0.2 - T = 0.2 a

 2g  s 0 = 72 + 2    7 

S = 8.75 m So in next 2.5s, it covers 8.75 m towards right. Total distance = 2 x 8.75 = 17.5 m After 5s, it speed will be same as that of initial (7 m/s) but direction will be reversed.

vr%vxys2.5 ls-esa;gnk;havksj8.75 eh-nwjhr;djsxkA dqy nwjh = 2 x 8.75 = 17.5 m 5 ls-i'pkr~bldhpky]izkjfEHkdpky(7 eh-/ls-)dscjkcjgksxhijUrqfn'kkfoijhrgksxhA 174.

Sol.

175.

A uniform thick string of length 5 m is resting on a horizontal frictionless surface. It is pulled by a horizontal force of 5 N from one end. The tension in the string at 1m from the force applied is: ,dlekueksVhjLlhdhyEckbZ5m gStks?k"kZ.kjfgrlrgijj[khgSA;g5N ds{kSfrtcyls,dfljsls[khaphtkrh gSrkscyyxkusokysfcUnqls1m dhnwjhijjLlhesarukogksxk: [ M . B a n k ( 0 7 08)_NLM_3.11] (A) zero (B) 5 N (C*) 4 N (D) 1 N a = 5/M a T M M 5 5N .a = . 5–T= 5 5 M 4M/5 M/5 T = 4 N. In the system shown in the figure the friction and mass of rope is negligible, then acceleration of 2 m is: Page # 55

RESONANCE

uhpsfn[kk;sx;sfudk;esajLlhnzO;ekujfgrrFkk?k"kZ.kjfgrgSArks2m nzO;ekudkRoj.kgksxk&

[M.Bank(07-08)_NLM_3.36]

(A)

g 5

(B)

2g 5

(C*) 0

(D)

5g M 5 . 2 5 M

P1 P2

Sol.

a 2m  aP2 =

am  0 2

P2 mg – T = ma

2T

T

a 2 From eq. (i) & (ii) a=0 Ans. 2T – 2mg = 2m.

176.

T

m

T

a

mg

2m

2T

a/2

2mg

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

A weight W is supported by two strings inclined at 60º and 30º to the vertical. The tensions in the strings are T 1 & T 2 as shown. If these tensions are to be determined in terms of W using a triangle of forces, which of these triangles should you draw ? ,dHkkjW nksjfLl;ksatksÅ/oZls60ºrFkk30ºij>qdhgSijlarqfyrgSAjLlh;ksaesarukofp=kkuqlkjT1 rFkkT2 gSA;s rukoksaW ds:iesacyksadkf=kHkqtcukdjKkrdjusgSArks rqefuEuesalsdkSulkf=kHkqt[khapkxks?[M.Bank(0708)_NLM_3.13]

(A)

177.

(B)

(C)

(D)

(E*)

Figure shows two pulley arrangements for lifting a mass m. In figure (a) the mass is lifted by attaching a mass 2 m while in figure (b) the mass is lifted by pulling the other end with a downward force F=2 mg, If f a & f b are the accelerations of the masses in figures (a) and (b) then (Assume string is massless and pulley is ideal)

RESONANCE

Page # 56

fp=kesanksiqyhO;oLFkkn'kkZ;hx;hgStksm nzO;ekumBkrhgSA(a)esanzO;ekudks2m nzO;ekudklgk;rklsmBkrs gStcfd(b) esam nzO;ekudksnwljsfljsijF=2mg cyuhpsdhvksjyxkdjmBkrsgSA;fnfa ofb nksuksanzO;ekuksads Roj.kgSarks¼ekuyksjLlhnzO;ekujfgrgS,oaf?kjuhvkn'kZgSA½ [M_Bank(06-07)\Q.1.10\NLM]

(A) f a = f b Sol.

 2m  m  g g = fa =  3  2m  m 

(B) f a = f b/2

(a)

(b)

(C*) f a = f b/3

(D) f a = 2 f b

2mg  mg =g m So, f a = f b/3. fb =

180.

A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s 2, then: [M.Bank(07-08)_NLM_3.45] 2 (fp=kkuqlkj,disUVjLo;acyyxkdjviusvkidksrFkkck¡Dldks5 m/s dsRoj.kls,dnzO;ekughujLlhoiqyhO;oLFkk }kjkfp=kkuqlkjÅijmBkrkgSA;fnisUVjdknzO;eku100 kg rFkkck¡Dldk50 kg gSrFkkjLlhdknzO;ekuux.;gS (;fn g = 10 m/s2 ) rks

tension in the rope is 1125 N (jLlh esa ruko 1125 N) tension in the rope is 2250 N (jLlh esa ruko 2250 N) force of contact between the painter and the floor is 375 N (isUVj rFkk Q'kZ ds chp lEidZ cy 375 N gS) (D) none of these (buesalsdksbZugha) For painter ; R + T – mg = ma R + T = m(g + a) ............(1) For the system ; 2T – (m + M)g = (m + M)a 2T = (m + M) (g + a) ..............(2) where ; m = 100 kg M = 50 kg a = 5 m/sec2 (A*) (B) (C*)

Sol.



and ; 181.

150  15 2 R = 375 N. T=

=

1125 N

In the figure shown the acceleration of A is, a



contact with B)

A

= 15ˆi 15ˆj then the acceleration of B is: (A remains in [M.Bank(07-08)_NLM_1.8]

 n'kkZ;sx,fp=kesaAdkRoj.kgS, aA = 15ˆi 15ˆj rksB dkRoj.kgksxk: (A, B lEidZesajgrkgSA)

RESONANCE

Page # 57

Sol.

(A) 6 ˆi (B)  15 ˆi (D) From wedge constraint

(C)  10 ˆi

(D*)  5 ˆi

aAY

ostca/kuls

(a A )  = (a B )  a aAX cos 53° – aAY cos 37° a = aB cos 53º 37° aB = – 5 m/s  aB  5 ˆi A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity 37°

AX

B

182.

3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60º with the ground is:

,dNM+ABfp=kesaiznf'kZrgSANM+dkAfljkHkwfeijtM+or~(fixed) gSrFkkCykWdnk¡;hvkSj 3 m/sosxlsxfrdj jgk gSA tc NM+ /kjkry ls 60º dk dks.k cuk;s rks NM+ ds fljs B dk osx gksxk& [ Made 2004] [M.Bank(0708)_NLM_8.16]

(A) 3 m/s 60º

Sol.

(D) 3 m/s

V

For regular contact V sin60º =

V = 2 m/s. 183.

(C) 2 3 m/s

(B*) 2 m/s

3

Ans.

In the Figure, the blocks are of equal mass. The pulley is fixed & massless. In the position shown, A is given a speed u and vB= the speed of B. (< 90°) [M.Bank(07-08)_NLM_8.7]

fp=kkuqlkjnksuksaxqVdslekunzO;ekudsgSAiqyhfu;rrFkknzO;ekughugSAfp=kesafn[kkbZfLFkfresaAdkspkyu nh tkrh gSA vB= B dh pkyA (< 90°) P

B



/// /

/// /

/// //

A

u

(A*) B will never lose contact with the ground B dHkhHkhtehulslEidZughaNksM+sxkA (B) The downward acceleration of A is equal in magnitude to the horizontal acceleration of B. A dkuhpsdhvkSjRoj.k]B ds{kSfrtRoj.kdsifjek.kdscjkcjgksxkA (C) vB = u cos  (D*) vB = u/cos 

RESONANCE

Page # 58

P

Sol.

B



/// /

/// /

/// //

A

u

From string constrained motion VBcos = u

u cos  For angle  < 90º Vertical component of tension will be less than mg.  B will never lose contact with the ground.

VB =

184.

In the Figure, the pulley P moves to the right with a constant speed u. The downward speed of A is vA, and the speed of B to the right is vB. fn, x, fp=k esa f?kjuh P nka;h rjQ u fu;r osx ls xfreku gSA Adh uhps dh rjQ pky vA, gS rFkk B dh nka;h rjQ pkyvBgSrks [M.Bank(07-08)_NLM_8.9]

(A) (C) (D*) Sol.

185.

vB = vA (B*) vB = u + vA vB + u = vA the two blocks have accelerations of the same magnitude

nksauksCykWdksadsRoj.kdkifjek.klekugSA

BP + AP = constant

d BP d AP  =0 dt dt u – VB + V A = 0 VB = u + VA Ans. dv B d(u  VA ) = dt dt aB = 0 + aA aB = aA Ans.

The pulley moves up with a velocity of 10 m/sec. Two blocks are tied by a string which passes over a pulley. The velocity V will be _________. Given: vB = 5 m/s  iqyh 10 m/sec osxlsÅijdhvksjxfr'khygSAnksxqVdksadksjLlhtks iqyh ds Åij ls xqtjrh gS dhlgk;rk ls ck¡/kktkrk gSA osxV _________ gksxkAfn;kgSvB = 5 m/s  [Modified_NLM_8.6] [M.Bank(07-08)_NLM_8.5]

[ Ans: 25 m/s Sol.

AP + BP = constant

d AP d BP  0 dt dt 10 – V + 10 + 5 = 0 V = 25 m/s .

RESONANCE

]

P Ans.

V

Page # 59

186.

Sol.

Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on the 2blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is : nks xqVds ‘A’o ‘B’, izR;sd dk nzO;eku‘m’,d fpduh {kSfrt lrg ij j[ks gSaA nks {kSfrt cyF o 2F Øe'k% xqVdsAo ‘B’ijfp=kkuqlkjvkjksfirgSAxqVdkA, xqVdsB ijughafQlyrkgSrksnksxqVdksadse/;dk;ZjrvfHkyEc izfrfØ;kcygSASA [Q.33/RK_BM/Constrained Motion] [M.Bank(07-08)_NLM_8.29]

(A) F

(B) F/2

(D)

(C)

Acceleration of two mass system is a =

nksnzO;ekufudk;dkRoj.kgSa= FBD of block A

F 2m

F leftward 2m

N

60°

30°

mF solving N = 3 F 2m

In the figure shown the velocity of lift is 2 m/s while string is winding on the motor shaft with velocity 2 m/s and block A is moving downwards with a velocity of 2 m/s, then find out the velocity of block B. [ Made 2004] M.Bank_NLM_8.26 fn[kk;s x;s fp=kesa fy¶Vdkosx2m/sgS tcfd jLlh2m/sds osx lseksVj 'kk¡¶V ij fyiV jgh gS rFkk CykWdA, 2m/s osxlsuhpsxfrdjjgkgsS]rksCykWdBdkosxKkrdjks& [S-(06-07)_BEK_DPP-23_Q.5]

(A) 2 m/s 

ugha

VP =

(B) 2 m/s 

V1  V2 2

4  VB 2 VB = 8 m/s . 2=

188.

(D*) 3F

xqVds Adk eqDr oLrq js[kkfp=k

N cos 60° – F = ma =

Sol.

3

ck;havksj

F

187.

F

(C) 4 m/s 

(D*) none of these buesa ls dksbZ

2m/s

VB

B

A 2m/s

Two blocks A & B with mass 4 kg and 6 kg respectively are connected by a stretched spring of negligible mass as in figure. When the two blocks are released simultaneously the initial acceleration of B is 1.5 m/s 2 westward. The acceleration of A is: M.Bank_NLM_1.9

nksfi.MAoBftudsnzO;ekuØe'k%4fdxzko6fdxzkgS,dnzO;ekujfgr[khaphagqbZfLizaxlstqM+sgq,gSAtcnksuksa fi.Mksadks,dlkFkNksM+ktkrkgSrksBdkRoj.kif'pedhrjQ1.5m/s2 gSAAdkRoj.kgSA [S-(06-07)_BEK_DPP-23_Q.6]

RESONANCE

Page # 60

Sol.

(A) 1 m/s2 westward (C) 1 m/s 2 eastward (A) 1 m/s2 if'pe dh rjQ (C) 1 m/s2 iwoZ dh rjQ T = MBaB T = 6 × 1.5 = 9N T = MAaA

9 aA = = 2.25 m/s2 4 189.

Ans.

(B*) 2.25 m/s2 eastward (D) 2.75 m/s 2 westward (B*) 2.25 m/s2 iwoZ dh rjQ (D) 2.75 if'pe dh rjQ a

A

T

T

T

1.5m/s

T

2

B

Match the following : [Made RA sir Batch A 2007] [M.Bank(07-08)_NLM_6.18] Three blocks of masses m 1, m 2 and M are arranged as shown in figure. All the surfaces are frictionless and string is inextensible. Pulleys are light. A constant force F is applied on block of mass m 1. Pulleys and string are light. Part of the string connecting both pulleys is vertical and part of the strings connecting pulleys with masses m 1 and m 2 are horizontal. nzO;ekum1,m2 rFkkMdsrhufi.Mfp=kesan'kkZ;svuqlkjla;ksftrgSaAlHkhlrg?k"kZ.kjfgrrFkkjLlhvfoLrj.kh;

gSAf?kjfu;k¡ojLlhgYdhgSaAnzO;ekum1 dsfi.Mij,dfu;rcyF vkjksfirdjrsgSaAnksuksf?kjuhdkstksM+us okyhjLlhdkHkkxm/oZgSvkSjnzO;ekum1rFkkm2 dksf?kjuh;ksalstksM+usokyhjLlhdsHkkx{kSfrtgSa&

(A) Acceleration of mass m 1 (B) Acceleration of mass m 2

F (Q) m  m 1 2

(D) Tension in the string

m 2F (S) m  m 1 2

(C) Acceleration of mass M

(A) nzO;eku m1 dk Roj.k (B) nzO;eku m2 dk Roj.k (C) nzO;eku M dk Roj.k Sol.

F (P) m 1

(D) jLlhesaruko

(A) Q (b) Q (C) R (D) S

(R) zero

F (P) m 1

F (Q) m  m 1 2

(R)'kwU;

m 2F (S) m  m 1 2

FBD’s

T = m 2a. F – T = m 1a

RESONANCE

Page # 61

F = (m 1 + m 2)a T = m 2a   190.

m 2F  T = m m . 1 2

F a = m m 1 2

F x = 0, aM = 0

Three identical balls 1,2,3 are suspended on springs one below the other as shown in the figure. OA is a weightless thread. M.Bank_NLM_1.2

fp=kesan'kkZ,vuqlkjrhu,dtSlhxsan1,2,3 fLizaxdhlgk;rklsfp=kkuqlkj,dnwljslstqM+hgqbZgSAOA,d nzO;ekughujLlhgSA

(a) If the thread is cut, the system starts falling. Find the acceleration of all the balls at the initial instant

vxjjLlhdksdkVfn;ktkrkgSrksfudk;fxjuk'kq:djnsrkgSAizkjfEHkd{k.kijlHkhxsanksadkRoj.kKkrdhft,A

(b) Find the initial accelerations of all the balls if we cut the spring BC which is supporting ball 3 instead of cutting the thread. Sol.

vxjgejLlhdhtxgxsan3dkstksM+usokyhfLizaxdksdkVnsrsgSrkslHkhxsanksadkizkjfEHkdRoj.kKkrdhft,A [ Ans: (a) 3 g  0, 0, (b) 0, g  g  ]

(a)

T AB = 2mg, T BC = mg TAB A m TAB

mg

B m mg

TBC

TBC Cm mg

For A, 2mg + mg = maA  For B, T AB – mg – T BC = maB  2mg – mg – mg = maB  T BC – mg = mac  (b)

aA = 3g maB = 0 ac = 0.



aB = 0

T AB = 2mg TAB m B m

aB

mg



T AB – mg = maB 2mg – mg = maB aB = g () aA = 0 & aC = g().

RESONANCE

Page # 62

191.

A body of mass 32 kg is suspended by a spring balance from the roof of a verticallyoperating lift and going continuously downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is: [ M . B a n k ( 0 7 08)_NLM_1.11] fLFkjkoLFkklsyxkrkjÅ/okZ/kjuhpsdhvksjtkjgh,dfy¶VdhNrls,dfLçaax}kjk32kgnzO;ekudhoLrqyVdh gq;hgSAfy¶V}kjk20mo50m,nwjhr;djusokys{k.kksaijfLçaxdkikB;kadØe'k%30kgo36kggksrkgSfy¶Vdk

osx%

(A) decreasing at 20 m & increasing at 50 m 20m ij ?kV jgk gS rFkk 50m ij c F2 .

In the figure shown all contact surfaces are smooth. Acceleration of B block will be: fp=kesafn[kkbZfLFkfresalHkhlEidZlrgfpduhgSxqVdsBdkRoj.kgksxk:

[3.52 NLM]

(A*) 1 m/s 2 279.

(B) 2 m/s2

(C) 3 m/s 2

(D) none of these buesa ls dksbZ ugha

In the shown mass pulley system, pulleys and string are massless. The one end of the string is pulled by the force F = mg. The acceleration of the block will be M.Bank_NLM_7.17 fp=kesan'kkZ;sf?kjuhnzO;ekufudk;esaf?kjuhrFkkjLlhnzO;ekughugSAjLlhdk,dfljkcyF=mg}kjk[khapk

tkrkgSAxqVdsdkRoj.kgksxk&

(A) g/2 280.

(B) 0

(C*) g

(D) 3g

A monkey pulls along the ground mid point of a 10 m long light inextensible string connecting two identical objects A & B each of mass 0.3 kg continuously along the perpendicular bisector of line joining the masses. The masses are found to approach each other at a relative acceleration of 5 m/s2 when they are 6 m apart. The constant force applied by monkey is : M.Bank_NLM_7.29 ,d10 m yEch gYdh vrU; jLlh ls nks ,d leku oLrq, ArFkkB izR;sd nzO;eku 0.3 kg ds cU/ks gSA ,d cUnj tehudhfn'kkesaArFkkBdkstksM+usokyhjs[kkdsyEcor~fn'kkesabljLlhdks,dfu;rcylsyxkrkj[khap

RESONANCE

Page # 97

jgk gSa ;s nzO;eku ,d nwljs dh vkSj lkis{k Roj.k 5 m/s2 ls xfr dj jgs gS tc muds chp dh nwjh 6m gSA cUnj }kjk yxk;k x;k fu;r cy dk eku gSA (A) 4 N 281.

(B*) 2 N

(C) 3 N

(D) none

A 15 kg block B is suspended from an inextensible cord attached to a 20 kg cart A. Neglect friction.

[5.16_NLM] (i) (ii)

Draw FBD of cart A & block B indicating all forces acting on them. Determine the acceleration of the cart & block immediately after the system is released from rest. Mass m shown in figure is in equilibrium. If it is stretched further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light. fp=kesafn[kk,vuqlkjnzO;ekuMlkE;koLFkkesagSAvxjblsxnwjhrdvkSj[khapktkrkgSvkSjfQjNksM+fn;ktkrk

282.

gSrksNksM+usdsBhdcknbldkRoj.kD;kgksxk\iqyhrFkkjLlhgYdsgSvkSjiqyhfpduhgSA

[Q.1.34_NLM] [Made 2005, PKS Sir]

(A)

4kx 5m

(B)

2kx 5m

(C*)

4kx m

(D) none of these

Sol.(C)

Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx. 283.

4kx m Three blocks A, B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium and mass of B is M, then: fp=kesan'kkZ,vuqlkjrhuxqVds A, B vkSjC fp=kkuqlkjyVdsgq,gSAArFkkC çR;sddknzO;ekum gSAvxjfudk; lkE;koLFkkesagSvkSjBdknzO;ekuMgSrks: [3.43.NLM] 

acceleration of the block is =

(A) M = 2 m 284.

(B*) M < 2 m

(C) M > 2 m

(D) M  2 m

There is an inclined surface of inclination  = 30º. A smooth groove is cut into it forming angle  with AB. A steel ball is free to slide along the groove, if the ball is released from the point O. The speed when it comes to A is _____. [ g = 10 m/s 2 ] [7.5 _ NLM] urry dk >qdko  = 30º gSA fp=kkuqlkj ABHkqtk ls dks.k ij ,d fpduk [kk¡pk cuk;k tkrk gSA ,d xsan [kk¡ps dsvuqfn'kxfrdjusdsfy,LorU=kgSAxSandksO fcUnqlsNksM+ktkrkgSArksxsandhpkyD;kgksxhtc;gfcUnq A ij igq¡psxh &[g = 10 m/s2 ]

RESONANCE

Page # 98

O

A [ Ans: 40 m / s ] 285.

287.

3m

4m

q

B

In the figure shown neglecting friction and mass of pulleys, what is the acceleration of mass B? fp=kesan'kkZ,vuqlkj?k"kZ.k,oaiqyhdsnzO;ekudksux.;ekusa]nzO;ekuBdkRoj.kD;kgksxk? [ 3.46_NLM ]

(A) 286.

a

g 3

(B)

5g 2

(C)

2g 2

(D*)

2g 5

Two blocks A and B of equal mass are connected by a light inextensible string passing over two light smooth pulleys fixed to the blocks. The parts of the string not in contact with the pulleys are horizontal. A horizontal force F is applied to the block A as shown. Then: M.Bank_NLM_3.57 nksxqVdsArFkkBftudsnzO;ekulekugS],dgYdhvrU;jLlh}kjk¼jLlhtksxqVdksalsca/ksnksfpduhiqyhesa lsxqtjrhgSA½tqM+sgq,gSaAjLlhdkogHkkxtksiqyhlstqM+kughagS]{kSfrtgSA,d{kSfrtcyFxqVdsAijfp=kkuqlkj yxk;ktkrkgS]rks-

(A) the acceleration of A will be equal to that of B (B*) the acceleration of A will be greater than that of B (C) the acceleration of A will be less than that of B (D) none of the above is correct. (A)Adk Roj.k] B ds Roj.k ds cjkcj gksxk (B*)Adk Roj.k] B ds Roj.k ls vf/kd gksxk (C)Adk Roj.k] B ds Roj.k ls de gksxk (D)buesalsdksbZugha

A system in set up as shown in the Figure. At time t = 0, a force P is applied to the block at a constant angle of 370. Magnitude of this force is gradually increased and it is found that the block lifts off from the floor when the extension in the spring is 25cm. The acceleration of the block at this instant is _________. [Friction is absent and the spring was initially in nondeformed state] ,dfudk;fp=kkuqlkjj[kk x;kgSAle;t = 0, ij xqVdsijfu;r dks.k370 ij cyP yxk;k tkrk gSAbl cy dk ifjek.k /khjs&/khjsc