Newton's Laws of Motion Resonance

8/19/2019 Newton's Laws of Motion Resonance

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Page # 1

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. Find the velocity and acceleration of block A.

Ans. V A = 2 m/s (), a

A = 4 m/s2 ()

6. Find the velocity of block B.

Ans. VB = 5 m/s ( )

7. Find all the normal reactions and the accelerations :

(a) (b) (c) (d)

Ans. (a) N = 100 N, a = 2 m/s2 (b) N = 50 N, 150 N, a = 0(c) N = 20 N, a = 2 m/s2| (d) N = 50 N, 150 N, a = 2 m/s2


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Page # 2

8. Draw the FBD for the following systems :

(a) (b) (c) (d)

(e) (f) (g) (h)

(i) (j)

9. A body of mass m is suspended by two strings making angles and with the horizontal. Find the tensionin the strings. See figure

(A) 21T

)sin(

cosmgT

(B) )sin(

sinmgTT

21

(C*))sin(

cosmgT,

)sin(

cosmgT 21

(D) none of these

10. 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 (B) 0.32 s (C) 0.39 s (D) 0.45 s

11. 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]

12. 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


13. Two masses m1 and m

2 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

14. Two masses m1 and m

2 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


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Page # 3

If m1 = 10kg, m

2 = 5kg, = 30º, what is the acceleration of mass m

2? [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 thebody is

(A) (B)2

– (C*) – (D) 0


16. Find out the mass of block B to keep the system at rest : [BUT DPP#13/Q.11]

(a) (b)

Ans. 10 kg

17. Find out the accelerations of the blocks and tensions in the strings. [BUT DPP#13/Q.13]

Ans. zero

18. Find the velocity of A.

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

Ans.. V A = 24 m/s ()

19. Find the acceleration of B.


Ans. aB = 2m/s2 ( )


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Page # 4

20. Find the velocity of A.


Ans. V A = V cos ()

21. Find the acceleration of wedge A


Ans. aA = b tan

22. Find the acceleration of wedge A.




Ans. aB =

2

1

cos

cosa

For the following questions assume.

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


Ans. – b î – 4b ĵ

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


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Page # 5


Ans. a î – 2 (a + b) ĵ

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


Ans. – 4 î + 8 ĵ

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


Ans. a (1 – cos) î + a sin ĵ

28. 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

Sol. Acceleration of box = 10 m/s2Inside the box forces acting on bob are shown in the figure

T = 22 )ma()mg( = 210 N

29. The block of mass ‘m’ is being pulled by a horizontal force F = 2 mg applied to a string as shown infigure (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 (C) 3 mg (D*) 22 mg

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

Let 1T

, 2T

and SF

be the forces exerted by the horizontal string, vertical string and the support on the


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Page # 6

massless pulley respectively. Then

1T

+ 2T

+ SF

= 0

or | SF

| = | 1T

+ 2T

| = 2 2 mg

( Tension in each string is | 1T

| = | 2T

| = 2 mg)

30. If the resultant of three forces F1 = pi + 3 j - k, F

2 = -5i + j + 2k and F

3 = 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) 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.

(B*) 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.

(C*) A massless object cannot exert force on any other object.

(D) All contact forces are electromagnetic in nature.

Sol. (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

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

32. 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 :

(A*)21

21

kk

kk

(x1 + x2) (B) k1x1 + k2x2 (C*) k1x1 (D) None of these

Sol. (A, C)Tension in both springs are same

i.e. k1x

2 = k

2x

2 = force exerted by lower spring on the block.


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Page # 7

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/sec2, downward (D) 30 m/sec2, 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. T2 – (5 + 2)g = (5 + 2) a

T2 = 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

1. S = t2

vu

(a) variable acceleration

(b) Constant acceleration

2. Internal force of a system always add upto zero (c) Newton’s second law

(d) Newton’s third law

(e) Bodies which can deform

3. To stop a car in minimum distance (f) Bonding between the molecules

break is applied such that car does not slip (g) s >

k

Ans. 1-b, 2-d, 3-e, 4-g


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Page # 8

38. Assertion : The pressing force between two blocks is an example of electromagnetic interaction force.

Reason : At microscopic level, al l 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 issuspended 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 (iii) a = 0

Ans. (i)k

mg

2

3 (ii)

k

mg

2

1 (iii)

k

mg

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

Ans. 310 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 wil l read

respectively : [GRB/obj.physics/pg.148/Q.251]

(A*) The reading of A is 4kg (B) The reading of A is 0 kg

(C*) The reading of B is 4kg (D) The reading of B is 2kg


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Page # 9

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 (B) 2m/s2 (C) 2.6m/s2 (D) none of these

46. 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 accelerationof the block will be : (Bird '2' is of mass 2 kg)

(A) 3.5 m/s2 (B) 4m/s2 (C*) 0 m/s2 (D) 1 m/s2

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/s2

48. What will be the mass of the bird 1 ?

(A) 600 gm (B) 1 kg (C*) 500 gm (D*) can not be determined



Ans. aB = 2 m/s2 ()

50. 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) (b)

Ans. (a) a A = g ; a

B = 0 ; x = mg/K ; (b) a

A = g ; x = 0

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



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Page # 10

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. t =g3

8

54. 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*)4

u3(B)

2

u3(C)

4

u(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) N

g = N

1 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. V AP

= 8 m/s ()


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Page # 11

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

mM

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

T

T

N1

Mg

N1 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).

a

mg

T

N2

N2

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 =g3

4

(b) S A

=

3

2

g3V4 , S

B =

3

5

g3V4 ,

64. 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 ]


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Page # 12

(A) (B) (C) (D)

65. In the figure is shown the top view of two horizontal forces pulling a box along the floor.A batch DPP 46_19.7.05

(a) How much work does each force do as the box is displaced 70 cm along the vertical line?

(b) What is the total work done by the two forces in pulling the box this distance?

[ Ans.: (a) w1 = 29.7 J, w

2 = 51.89 J (b) 81.2 J]

66. The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is

performed with the above spr ing of half length, the line OA will [3 marks] USS_Pg.296_22

O

F

A

x

(A*) Shift towards F-axis (B) Shift towards X-axis

(C) Remain as it is (D) Become double in length

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

DPP 47_ACJ_05-06

(A) 28 kg (B) 35 kg (C*) 24 kg (D) 42 kg

68. Two blocks of masses m1 = 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 m1 after it has fallen 40 cm ?

DPP 48_ACJ_05-06

[ Ans. :724 m/s ]


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Page # 13

69. A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass . Constant forces F1 = 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 _______ . DPP 48_ACJ_05-06

[ Ans. : 0.5 sec ]

70. A ladder that is 3.0 m long and weighs 200 N has its center of gravity 120 cm from the foot. At its topend is 50 N weight. Compute the work required to raise the ladder f rom a horizontal position on theground to a vertical position. DPP 49_ACJ_05-06

[ Ans : 390 J ]

71. Two blocks with masses m1 = 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 m2 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. I f 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 = ĵ110î50

(B) Interaction force on the walls due to the block = ĵ110î50

(C) If F

were reversed, now interactron force on the block due to wall = ĵ110î50

(D*) If F

were reversed, now the acceleration of the block = î6

50 m/sec2.


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Page # 14

Sol.

Ny = 110

Nx = 50

ĵ110î50Nblock

So A is correct

wallN

will be opposite of Nblock

= ĵ110î50

So B is incorrct.

If F

were reversed

it will loose contact with verticle wall and Ny = 10 ĵ

So option C is incorrect.

Net force = a6î50

î6

50a

So D is correct.

74. 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, N

2, N

3 and N

4

respectively and acceleration with which the block slides on the wedge in situations are b1, b

2, b

3 and

b4respectively then : [5 Marks] P&Q_DPP 44_9

(A*) N3 > N

1 > N

2 > N

4(B) N

4 > N

3 > N

1 > N

2(C*) b

2 > b

3 > b

4 > b

1(D) b

2 > b

3 > b

1 > b

4

Sol. (1)

Balancing forces perpendicular to incline N = mg cos37° + ma sin37°

N1 =

54 mg +

53 ma


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Page # 15

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

b1 =

5

3g –

5

4a

(2)

Similarly for this case get N2 =

5

4mg –

5

3ma and b

2 =

5

3g +

5

4a

N2 =

5

4mg –

5

3ma

(3)


5

4mg +

5

4ma and b

3 =

5

3 g +

5

3a

(d)


5

4mg –

5

4ma

and b4 =

5

3g –

5

3a

75. 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 (b) string (2) is cut (c) string (3) is cut

Ans. (a) T1 = 0 ; T

2 = mg ; T

3 = 2 mg ; a = g

(b) T1

= mg ; T2

= 0 ; a = 0

(c) T1 = mg ; T

2 = 0 ; T

3 = 0 ; a = 0


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Page # 16

76. A body of mass 10 kg is on a rough inclined plane of inclination = sin 1 (3/5) with the horizontal. Whena 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 (B) OB (C) OC (D) OD

77. 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)2

u3419 m/s (B)

2

u13m/s (C) u3 m/s (D*) u7 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 (B*) the acceleration of A is 3 × 0.98 m/s2

(C) the extension in the spring is 0.1 m (D) the extension in the spring is 0.2 m


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Page # 17

81. 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.

82. 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 thespring. JP DPP_52_2

Ans. [2 mg /k ]

83. In the figure shown neglecting friction and mass of pulleys, what is the acceleration of mass B?

fp=k esa n'kkZ, vuqlkj ?k"kZ.k ,oa iqyh ds nzO;eku dks ux.; ekusa] nzO;ekuBdk Roj.k D;k gksxk?[ 3.46_NLM ]

(A)g

3(B)

5

2

g(C)

2

2

g(D*)

2

5

g

84. 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

nks xqVds ArFkkBftuds nzO;eku leku gS] ,d gYdh vrU; jLlh }kjk ¼jLlh tks xqVdksa ls ca/ks nks fpduh iqyh esals xqtjrh gSA½ tqM+s gq, gSaA jLlh dk og Hkkx tks iqyh ls tqM+k ugha gS] {kSfrt gSA ,d {kSfrt cyFxqVds Aij fp=kkuqlkjyxk;k tkrk gS] 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]Bds Roj.k ds cjkcj gksxk(B*) Adk Roj.k]Bds Roj.k ls vf/kd gksxk(C) Adk Roj.k]Bds Roj.k ls de gksxk(D)buesa ls dksbZ ugha

85. A weightless inextensible rope on a stat ionary 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

,d gYdh vrU; jLlh fLFkj ost ij {kSfrt ls dks.k cukrs gq, j[kh gSA jLlh dk ,d fljk fcUnq

A

ij fLFkj gSA fcUnqB

ij ,d NksVk lk nzO;eku tksM+ j[kk gSA ost nk;ha vksj fu;r Roj.k ls xfr djuk 'kq: dj nsrk gSABij tqM+s nzO;eku dk


18/101

Page # 18

Roj.ka1 Kkr dhft,] tcfd ;g vHkh Hkh ost ij gks- 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;s x;s fp=k esa ArFkkCCykWdksa dks fu;r osxuls [khapk tkrk gSA CykWdBdk Roj.k gksxk :[ Made 2003] M.Bank_NLM_8.13

(A)u

b

2

tan2 sec (B*)u

b

2

tan3

(C)u

b

2

sec2 tan (D) zero‘'kwU;

87. In the figure shown the car moves down with a constant accelerat ion 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 (B*) m

g (C) 2 m

g (D) none of these

88. In the figure shown, find out the value of [ assume string to be tight ] [ Made 2004]fn[kk;s x;s fp=k esadk eku crkb;s[jLlh dks dlk gqvk ekus] [ Made 2004] M.Bank_NLM_8.25

(A*) tan1 4

3(B) tan1

3

4(C) tan1

8

3(D) none of thesebuesa ls dksbZ ugha

89. 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/s2. [M.Bank_NLM_4.3]

9.8NHkkj okyk ,d xqVdk est ij j[kk gqvk gSA est dh lrg xqVds ij Åij dh vksj10 Ndk cy yxkrh gSA ekfu,g = 9.8 m/s2

(A*) The block exerts a force of 10N on the table xqVdk est ij10Ndk cy yxkrk gSA

(B) The block exerts a force of 19.8N on the table xqVdk est ij19.8Ndk cy yxkrk gSA(C) The block exerts a force of 9.8N on the table xqVdk est ij9.8Ndk cy yxkrk gSA(D*) The block has an upward acceleration. xqVds dk Åij dh vksj Roj.k gksxkA


19/101

Page # 19

90. Two weights W1 & W

2 in equillibrium and at rest, are suspended as shown in figure. Then the ratio W

1/

W2 is about : [M.Bank_NLM_3.57]

nks HkkjW1 rFkkW

2 fp=kkuqlkj yVds gq, gS rFkk lkE;koLFkk vkSj fojke esa gS rksW

1/W

2 dk vuqikr gksxk:

(A*) 5/4 (B) 4/5 (C) 8/5 (D) none of thesebuesa ls dksbZ ugha

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

iwoZ ç'u esa nhokj ls tqM+h jLlh esa ruko gksxk & [M.Bank_NLM_3.58](A*) 0.6 W

1(B) 0.8 W

1(C) W

1(D) none of thesebuesa ls dksbZ ugha

92. 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.Bank(07-

08)_NLM_4.2]

8lseh0 f=kT;k ds fpdus fLFkj csyu ds vUnj nks fpdus xksys çR;sd f=kT;k5lseh0 rFkkWHkkj okys fLFkj j[ks gSAxksyksa dh m/oZ nhokjksa ds lkFk çfrfØ;k,sa Kkr djks\

(A) W/4 & 3W/4 (B) W/4 & W/4 (C*) 3W/4 & 3W/4 (D) W & W

93. 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:4fdxzk0 rFkk12fdxzk0 nzO;eku ds nks CykWd Øe'k% ArFkkBfpdus {kSfrt lrg ij j[ks gq, gSaA ,d16 Ndk cyF,

Aij fp=kkuqlkj yxk;k tkrk gSA ArFkkBds e/; yxus okys lEidZ cy dk eku gksxk &[M.Bank(07-08)_N.LM_4.7]

(A) 4 N (B) 8 N (C*) 12 N (D) 16 N

94. 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’xqVds dk nzO;eku100 kggS vkSj bls {kSfrt cyF = 2 mgtks fd fp=kkuqlkj yxrk gS]ls [khapk tkrk gSA(g = 10 m/s2ysa)A iqyh nzO;ekughu gS rFkk fLFkj est ds fdukjs ij c¡/kh gqbZ gSA est }kjk iqyh ij yxk;s x, cy dk eku(U;wVu esa)gksxkA

Made_MPS_2005 [M.Bank(07-08)_NLM_3.56]

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


20/101

Page # 20

Let 1T

, 2T

and SF

be the forces exerted by the horizontal string, vertical string and the support on the

massless pulley respectively. Then

1T

+ 2T

+ SF

= 0

or | SF

| = | 1T

+ 2T

| = 2 2 mg

( Tension in each string is | 1T

| = | 2T

| = 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 :

,dmnzO;eku dh oLrq ijFcy fp=kkuqlkj yxk;k tkrk gSA xqVds ij ifj.kkeh vfHkyEc çfrfØ;k cy dk ifjek.k gksxk:

[M.Bank(07-08)_NLM_4.9]

(A*) 10 2 N (B)10

2 N (C) 10 N (D) none of thesebuesa ls dksbZ ugha

96. 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/s2).

fp=k esa n'kkZ;s vuqlkj fLFkfr dks ekfu,A iqyh rFkk jLlh nzO;ekughu gS rFkk lHkh lrg ?k"kZ.kghu gSA jLlh esa ruko gksxkA(g = 10 m/s2). [M.Bank(07-08)_NLM_3.30]

(A) 0 N (B) 1 N (C) 2 N (D*) 5 N

97. 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/s2, then: [M.Bank(07-08)_NLM_3.45]

(fp=kkuqlkj ,d isUVj Lo;a cy yxkdj vius vki dks rFkk ck¡Dl dks5 m/s2ds Roj.k ls ,d nzO;ekughu jLlh o iqyh O;oLFkk}kjk fp=kkuqlkj Åij mBkrk gSA ;fn isUVj dk nzO;eku100 kgrFkk ck¡Dl dk50 kggS rFkk jLlh dk nzO;eku ux.; gS(;fn g = 10 m/s2)rks

(A*) tension in the rope is 1125 N (jLlh esa ruko1125 N)(B) tension in the rope is 2250 N (jLlh esa ruko2250 N)(C*) force of contact between the painter and the floor is 375 N

(isUVj rFkk Q'kZ ds chp lEidZ cy375 NgS)(D) none of these (buesa ls dksbZ ugha)

Sol. For painter ;

R + T – mg = ma

R + T = m(g + a) ............(1)

For the system ;


21/101

Page # 21

2T – (m + M)g = (m + M)a

2T = (m + M) (g + a) ..............(2)

where ; m = 100 kg

M = 50 kg

a = 5 m/sec2

T =2

15150 = 1125 N

and ; R = 375 N.

98. 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’ respect ively 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 cyFo2FØe'k% xqVds Ao‘B’ij fp=kkuqlkj vkjksfir gSA xqVdk A,xqVdsBij ugha fQlyrk gS rks nks xqVdksa ds e/; dk;Zjr vfHkyEc izfrfØ;k cygSA¼ ArFkk Bdh laifdZr lrg ?k"kZ.kghu gS½ [Q.33/RK_BM/Constrained Motion] [M.Bank(07-08)_NLM_8.29]

(A) F (B) F/2 (C)3

F(D*) 3F

Sol. Acceleration of two mass system is a =m2

F leftward

nks nzO;eku fudk; dk Roj.k gSa =m2

F ck;ha vksj

FBD of block A xqVds Adk eqDr oLrq js[kkfp=k 60°30°

F

N

N cos 60° – F = ma =m2

mF solving N = 3 F

99. 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=k esa n'kkZ;s f?kjuh nzO;eku fudk; esa f?kjuh rFkk jLlh nzO;ekughu gSA jLlh dk ,d fljk cyF = mg}kjk [khapktkrk gSA xqVds dk Roj.k gksxk &

(A) g/2 (B) 0 (C*) g (D) 3g

100. 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; fojke ls izkjEHk gksrk gS rFkk Ank;ha vksj5 mpyus ds i'pkr~5 m/sdk osx izkIr djrk gSA O;oLFkk dks lHkh txg?k"kZ.k jfgr o f?kjfu;ksa o Mksfj;ksa dks gYdh ekurs gq, Aij vkjksfir fu;r cyFdk eku gS-

(A) 50 N (B*) 75 N (C) 100 N (D) 96 N

Sol. (B)


22/101

Page # 22

a =10

25

s2

v2 = 2.5 m/s2

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;s x;s fp=k esadk eku crkb;s[jLlh dks dlk gqvk ekus] [ Made 2004] [M.Bank(07-08)_NLM_8.23]

(A*) tan1 4

3(B) tan1

3

4(C) tan1

8

3(D) none of thesebuesa ls dksbZ

ugha102. 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=k esa ,d fudk; iznf'kZr gSA lHkh pky /kjkry ds lkis{k gSaA rks /kjkry ds lkis{k CykWdBdh pky gksxhA

[Q.31/RK_BM/Constrained Motion]

(A) 5 m/s (B*) 10 m/s (C) 15 m/s (D) 7.5 m/s

Sol. (B)

1 + 2 2 = constant

dt

d 1 +

dt

d2 2 = 0

( 5 + 5) + 2 (5 + vB) = 0 or v

B = 10 m/s

103. Three blocks are connected by strings as shown in figure and pulled by a force F = 60 N. If m A

= 10 kg,

mB

= 20 kg and mC

= 30 kg, then :

rhu CykWdksa dks fp=kkuqlkj jLlh ls tksM+djF = 60 Ncy ls [khapk tkrk gSA ;fnm A = 10 kg, m

B = 20 kgrFkkm

C

= 30 kggks rks: [M.Bank_NLM_3.34]


23/101

Page # 23

(A) acceleration of the system is 2 m/s2 fudk; dk Roj.k2m/s2gksxkA(B*) T

1 = 10 N

(C*) T2 = 30 N (D) T

1 = 20 N & T

2 = 40 N

104. 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 ,d xkM+h]0.2fdxzk- nzO;eku ds fi.M ls jLlh }kjk tqM+h gS] izkjEHk esa xkM+h fpdus {kSfrt leryij ck;aha vksj7eh-/ls-. dh pky ls xfreku gS(g = 9.8eh-/ls2 ysa)[M.Bank_NLM_3.47]

0.2 kg

0.5 kg

[Modified AKS_2007]

(A*) The acceleration of the cart is7

g2 towards right.

(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*)xkM+h dk Roj.k7

g2 nk;ha rjQ gksxkA

(B*)xkM+h2.5lsd.M i'pkr~ {kf.kd :dsxhA(C*)igys5 lsd.M esa xkM+h }kjk r; dh xbZ nwjh17.5eh- gSA(D) 5lsd.M i'pkr~ xkM+h dk osx izkjfEHkd osx ds leku gksxkA

Sol. 0.2 g = 0.7 a a =7

g2 m/s2

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

ml fLFkfr ds fy, tc ;g fLFkj gksrk gSV = 0

0 = 7 +

7

g2t t =

g2

49 = 2.5 s

a

7 m/s

T = 0.5 a

T

T

0.2g

a

0.2 - T = 0.2 aDistance travelled till it comes to rest

fLFkj voLFkk vkus rd pyh xbZ nwjh

0 = 72 + 2

7g2

s

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 ;g nk;ha vksj8.75eh- nwjh r; djsxkAdqy nwjh = 2 x 8.75 = 17.5 m5ls- i'pkr~ bldh pky] izkjfEHkd pky(7eh-/ls-)ds cjkcj gksxh ijUrq fn'kk foijhr gksxhA

105. A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass. Constant forces F1 = 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 Breaches the pulley is _______. (Assume that friction is absent every where. Pulley and string are light.

,d1fdxzk- nzO;eku dk CykWd ‘B’leku nzO;eku ds CykWd ‘A’ij fLFkj j[kk gS t = 0ij cyF1 = 20 NrFkk


24/101

Page # 24

F2 = 8 Nyxk;s tkrs gS bl le; CykWdBdh iqyh ls nwjh50 cmgS rks CykWdBfdrus le; esa iqyh rd igqapsxk _____(;g

ekfu, fd lHkh txg ?k"kZ.k vuqifLFkr gSA ¼iqyh rFkk jLlh gYdh gS) [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=k esa ;fn CykWd ArFkkBfdlh cká cy ds dkj.k ,d leku Roj.k ls xfr djsa rFkk ArFkkBds e/; dksbZ ?k"kZ.k cyugha gks rks nksuksa CykWdks ds e/; vfHkyEc çfrfØ;k cy dk eku D;k gksxkA

[M.Bank_NLM_4.10]

(A*) 2 mg/cos (B) 2 mg cos (C) mg cos (D) none of theseSol. N cos N = 2 mg/cos .

a

2mg

a

Nsin

N

Ncos

2m

A

107. A 2 kg toy car can move along an x axis. Graph shows force Fx, acting on the car which begins at rest

at time t = 0. The velocity of the particle at t = 10 s is :

,d 2 kgdh f[kykSuk dkjxv{k ds vuqfn'k xfr dj ldrh gSA xzkQ cyFx,dks çnf'kZr djrk gS] tks le; t = 0

ij fojke ij dkj ij yxuk çkjEk gksrk gSA t = 10 sij d.k dk osx gS % [Made A.K.S. sir][M.Bank_NLM_7.35]

F (N)x

4

0

-2

t(s)

4 8 9 10 11

(A) – i m/s (B) – 1.5 i m/s (C*) 6.5 i m/s (D) 13 i m/s

Sol. dp = pf – pi = dtF = Area under the curve.pi = 0

Net Area = 16 – 2 – 1 = 13 N-s

= Vf = 213 = 6.5 i m/s

[As momentum is positive, particle is moving along positive x axis.]


25/101

Page # 25

108. 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 inertialframe.The resultant force on the charged particle at A has the magnitude.

rhu lqǹ


26/101

Page # 26

fp=kkuqlkj50 kgdk ,d O;fDr gYdh Hkkj ekiu e'khu ij [kM+k gqvk gS tks30 kgds ckWDl es j[kh gqbZ gSA ckWDl dks jLlhds ,d fljs ls iqyh }kjk Vkaxk tkrk gS rFkk nwljk fljk O;fDr ds }kjk idM+k gqvk gSA ;fn O;fDr ckWDl dks fLFkj j[krk gSrks Hkkj ekiu dk ikB;kad _______U;wVu esa gksxkA [M.Bank(07-08)_NLM_2.1]

[ Ans: 100 N ]

Sol.

N

30g

T

T

N

T = N + 30 g ........(i)

T + N = 50 g ........(ii) N = 100 Nt.

112. Inside a horizontally moving box, an experimenter (who is stationary relative to box) 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. experi-

menter) will be. (Take g = 10 m/s2) [Made PKS 2005_GRSTU]

[M.Bank_NLM_5.11]

,d xfreku {kSfrt cDls esa] ,d iz;ksxdehZ ikrk gS fd tc oLrq dks {kSfrt fpduh est ij j[krs gS] rks ;g10eh-/ls-2 dsRoj.k ls xfreku gksrh gSA bl cDls esa tc1fdxzk- dh oLrq dks gYds /kkxs ls yVdkrs gSa] rks /kkxs esa lkE;oLFkk esa iz;ksxdehZds lkis{k ruko gksxk &(fn;k gSg = 10 m/s2)(A) 10 N (B*) 10 2 N (C) 20 N (D) zero

Sol. Acceleration of box = 10 m/s2

Inside the box forces acting on bob are shown in the figure

cDls dk Roj.k= 10eh-/ls-2

cDls ds vUnj oLrq ij dk;Zjr cy fp=k esa n'kkZ;sa x;s gSA

T = 22 )ma()mg( = 210 N

113. In the figure shown, A & B are free to move. All the surfaces are smooth. (0 < < 90º)fp=k esa n'kkZ, vuqlkj ArFkkBxfr djus ds fy, LorU=k gSA lHkh lrg fpduh gSA (0


27/101

Page # 27

Sol.

a

A

B

mg

a0

ma

(Pseudo)0

N

ma0 sin + N = mg cos mg cos N = mgcos – ma0sin

N < mg cosHence, (D) is true.

ma0 cos + mg sin = ma

a = g sin + a0 cos

A

a cos0

a0

a sin0 a

Hence acceleration of A = sing)sina()cosaa(2

0

2

0 .

114. Find the acceleration of the 6 Kg block in the figure. All the surfaces and pulleys are smooth. Also the strings

are inextensible and light. [Take g = 10 m/s2]

fp=k esa6 KgCykWd dk Roj.k D;k gksxk lHkh lrg rFkk f?kjuh ?k"kZ.kjfgr gS rFkk jLlh vfoLrkfjr gSA[Take g = 10 m/s2][Made_RA_2006_RSTG] [M.Bank_NLM_6.6]

[2]

Sol. All the blocks will be having the same

acceleration along the length of the string.

So, Applying Newtons law along the string on A,B & C.

6g – 2g sin300 – 2g = (6 + 2 + 2)a

3g = 10a a =10

g3

or a = 3 m/s2 Ans. a = 3 m/s2

115. System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder is - [M.Bank(07-08)_NLM_8.33]

fp=k esa ,d fudk; iznf'kZr gS ekfu;s fd csyu nks ostks ds e/; lEidZ esa jgrk gSA csyu dk osx gS-

csy u

2u m/su m/s30° 30°

[Made BKM, 2005]

(A)2

u3419 m/s (B)

2

u13m/s (C) u3 m/s (D*) u7 m/s

Sol. (D)

Method - I fof/k - As cylinder will remains in contact with wedge A D;ksafd csyu ost Ads laidZ esa gS


28/101

Page # 28

Vx = 2u

As it also remain in contact with wedge B

;g ostBds lkFk Hkh laidZ esa jgrk gSAu sin 30° = Vy cos30° – Vx sin30°

Vy = V

x

30cos

30sin +

30cos

30sinU

Vy = V

xtan30° + u tan 30°

Vy = 3u tan30° = 3 u

V = 2y

2x VV = 7 u Ans.

Method - II fof/k - IIIn the frame of A fof/k Ads ra=k esa

3u sin 30º = Vycos30º

Vy = 3u tan 30º = 3 u

and o Vx = 2u V = 2y

2x VV = u7 Ans.

116. A lift is falling with an acceleration 2 m/s2. A ball of mass 100 gm is attached at one end of the string and the

other end is fixed to the ceiling of the lift. The ball remains at rest relative to lift. The tension in the string is:

(g = 10 m/s2) [M.Bank_NLM_5.9]

,d fy¶V2 m/s2 ds Roj.k ls fxj jgh gSA jLlh ds ,d dksus ls100gmdh xsan tqM+h gS ,oa nwljk fljk Nr ls cU/kk gqvkgSA xsan fy¶V ds lkis{k fojke esa jgrh gSA jLlh esa ruko gS &(A) 1.2 N (B*) 0.8 N (C) 10 N (D) 0.2 N

Sol. T = ma (g – a)

= 0.1 (10 – 2) = 0.8 N

117. Three equal balls 1,2,3 are suspended on springs one below the other as shown in the figure. OA is a

weightless thread. The balls are in equilibrium [M.Bank(07-08)_NLM_1.2]

fp=k esa n'kkZ, vuqlkj rhu ,d tSlh xsan1,2,3fLizax dh lgk;rk ls fp=kkuqlkj ,d nwljs ls tqM+h gqbZ gSAOA,d

nzO;ekughu jLlh gSA xSansa lke;koLFkk esa gSA

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

instant

vxj jLlh dks dkV fn;k tkrk gS rks fudk; fxjuk 'kq: dj nsrk gSA izkjfEHkd {k.k ij lHkh xsanksa dk Roj.k Kkr dhft,A(b) Find the initial accelerations of all the balls if we cut the spring BC, which is supporting ball 3,


29/101

Page # 29

instead of cutting the thread.

vxj ge jLlh dh txg xsan3dks tksM+us okyh fLizax dks dkV nsrs gS rks lHkh xsanksa dk izkjfEHkd Roj.k Kkr dhft,A

[ Ans: (a) 3 g 0, 0, (b) 0, g g ]

Sol. (a) T AB

= 2mg, TBC

= mg

For A, 2mg + mg = ma A

a A

= 3g

For B, T AB

– mg – TBC

= maB

A m

T AB

mg

B m

T AB

mg

TBC

C m

mg

TBC

2mg – mg – mg = maB

maB = a

B = 0

TBC – mg = mac ac = 0.

(b) T AB

= 2mg

T AB

– mg = maB

2mg – mg = maB

B m

mg

T AB

m aB

aB = g ()

a A

= 0 & aC = g().

118. A bob is hanging over a pulley inside a car through a string . The second end of the string is in the hand

of a person standing in the car . The car is moving with constant acceleration 'a' directed horizontally

as shown in figure . Other end of the string is pulled with constant acceleration ' a

' (relative to car)

vertically. The tension in the string is equal to [ Made 2003 ],d cksc dkj ds vUnj ,d Mksjh ds }kjk f?kjuh ls yVd jgk gSA ,d Mksjh ls ,d ckWc yVdk gSA Mksjh dk nwljk fljk dkjesa [kM+s ,d O;fDr ds gkFk esa gSA dkj {kSfrt fn'kk esa fu;r Roj.kafp=kkuqlkj xfr'khy gSA Mksjh dk nwljk fljk fu;r Roj.ka¼dkj ds lkis{k½ ls Å/okZ/kj uhps [khapk tkrk gSA Mksjh esa ruko cjkcj gS – [M.Bank(07-08)_NLM_5.10]

(A) m 22 ag (B) m 22 ag – ma (C*) m 22 ag + ma (D) m(g + a)

Sol. (C)

(Force diagram in the frame of the car) ¼dkj ds rU=k esa cy fp=k½

Applying Newton’s law perpendicular to string Mksjh ds yEcor~ U;wVu ds fu;e yxkus ij &mg sin = ma cos

tan =g

a

Applying Newton’s law along string T – m 22 ag = ma T = m + ma Ans.

Mksjh ds yEcor~ U;wVu ds fu;e yxkus ij T – m 22 ag = ma T = m + ma Ans.

119. Shown in the figure is a system of three particles of mass 1 kg, 2 kg and 4 kg connected by two springs. The

acceleration of A, B and C at an instant are 1 m/sec 2, 2 m/sec2 and 1/2 m/sec2 respectively directed as

shown in the figure external force acting on the system is :

fp=kkuqlkj nwljk fudk; es 1 kg, 2 kgrFkk4 kgdks nks fLçax ls tksM+k tkrk gSA fdlh {k.k A, BrFkkCdk Roj.k1 m/sec2, 2 m/sec2 rFkk1/2 m/sec2 fp=kkuqlkj fn'kk esa gSA fudk; ij yxus okyk ckº; cy gksxkA [M .Bank(07 -08)_NLM_1.10]


30/101

Page # 30

(A) 1 N (B) 7 N (C*) 3 N (D) 2N

120. Figure shows a 5 kg ladder hanging from a string that is connected with a ceiling and is having a spring

balance connected in between. A boy of mass 25 kg is climbing up the ladder at acceleration 1 m/s 2.

Assuming the spring balance and the string to be massless and the spring to show a constant reading, the

reading of the spring balance is : (Take g = 10 m/s2) [Made AA_2006_GRSTU]

,d5fdxzk0 dh lh


31/101

Page # 31

122. In the figure shown all contact surfaces are smooth. Acceleration of B block will be:

fp=k esa fn[kkbZ fLFkfr esa lHkh lEidZ lrg fpduh gSA xqVds Bdk Roj.k gksxk:

[M.Bank(07-08)_NLM_3.39]

(A*) 1 m/s2 (B) 2 m/s2 1 (C) 3 m/s2 (D) none of thesebuesa ls dksbZ ughaSol. Let a

B = a then a

A = 3a [by string constraint]

T = ma A

T = 3ma .........(i)mg – 3 T = m a

B mg – 3 T = ma .........(iI)

mg – 9 ma = ma a =10

g

a = 1 m/s2

123. In the figure a block ‘A’ of mass ‘m’ is attached at one end of a light spring and the other end of the spring is

connected to another block ‘B’ of mass 2m through a light string. ‘A’ is held and B is in static equilibrium.

Now A is released. The acceleration of A just after that instant is ‘a’. In the next case, B is held and A is in

static equilibrium. Now when B is released, its acceleration immediately after the release is 'b'. The value of

a/b is : (Pulley, string and the spring are massless) [M.Bank(07-08)_NLM_1.17]

fp=k esa nzO;eku'm'dk CykWd‘A’gYdh fLizax ds ,d fljs ls tqM+k gS rFkk fLizax dk nwljk fljk gYdh jLlh }kjk nzO;eku2mds nwljs CykWd‘B’ls tqM+k gSA‘A’dks fdlh us idM+k gS rFkk'B'fLFkj lkE;oLFkk esa gSA vc Adks NksM+rs gSaA ml {k.kds Bhd ckn CykWd'A'dk Roj.k'a'gSA nwljh fLFkfr esa‘B’dks idM+rs gS rFkk AfLFkj lkE;oLFkk esa gSA vc tcBdks NksM+rsgS] rks NksM+us djus ds rqjUr ckn bldk Roj.kbgSA

[Made RKV, AS 2005]

(A) 0 (B) undefined vLi"V (C*) 2 (D)2

1

Sol. For first case tension in spring will be

Ts = 2mg just after 'A' is released.

2mg – mg = ma a = g

In second case Ts = mg

2mg – mg = 2mb

b = g/2

a/b = 2

gy. izFke fLFkfr esa fLizax esa ruko gksxkT

s = 2mg 'A'dks eqDr djus ds rqjUr ckn

2mg – mg = ma a = gf}rh; fLFkfr esa T

s = mg

2mg – mg = 2mb

b = g/2

a/b = 2

124. In the figure shown all the surface are smooth. All the blocks A, B and C are movable, x-axis is horizontal and

y-axis vertical as shown. Just after the system is released from the position as shown. [M.Bank_N.LM_4.11]

n'kkZ;sa fp=k esa lHkh lrg fpduh gSA lHkh fi.M A, BrFkkCxfr dj ldrs gSAx-v{k {kSfrt rFkky-v{k n'kkZ;s vuqlkj m/okZ/kj gSA fudk; dks fn[kkbZ xbZ fLFkfr ls eqDr djrs gS rks blds rqjUr ckn &


32/101

Page # 32

B

A

C

y

x

Horizontal Surface

[Made RKV_2007]

(A*) Acceleration of 'A' relative to ground is in negative y-direction

(B*) Acceleration of 'A' relative to B is in positive x-direction

(C*) The horizontal acceleration of 'B' relative to ground is in negative x-direction.

(D*) The acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin .(A*)tehu ds lkis{k'A'dk Roj.k _.kkRedy-fn'kk esa gksxkA(B*) 'B'ds lkis{k'A'dk Roj.k /kukRedx-fn'kk esa gksxkA(C*)tehu ds lkis{k'B'dk {kSfrt Roj.k _.kkRedx-fn'kk esa gksxkA(D*)tehu ds lkis{k 'B'dk Roj.k urry 'C'ds vuqfn'kg sin ls T;knk gksxkA

Sol. (Tough) There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net

downward force (mg – N) is acting on it, making its acceleration along negative y-direction.

Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equaldue to constrain, thus w.r.t. B, A moves in positive x-direction.

fi.M ij dksbZ {kSfrt cy ugha gS]vr% ;gx-fn'kk esa ugha pyrk gS]tcfd bl ij uhps dh vksj ifj.kkeh cy, (mg –N)yxrkgS] tks fd blds Roj.k dks _.kkRedy-fn'kk esa nsrk gSAfi.MBuhps ds lkFk&lkFk _.kkRedx-fn'kk esa xfreku gksrk gSA ArFkkBdk uhps dh vksj Roj.k c)rk ls cjkcj gksrk gS]vr%Bds lkis{k] A/kukRedx-fn'kk esa xfreku gSA

B

Normal reaction due to C

B

C } kj k v fHky Ec i zfr fØ;

Due to the component of normal exterted by C on B, it moves in negative x-direction.CijB}kjk vfHkyEc ds ?kVd ds dkj.k ;g _.kkRed, x-fn'kk esa xfreku gSA

N A

NC

Mg

B

The force acting vertically downward on block B are mg and N A(normal reaction due to block A). Hence the

component of net force on block B along the inclined surface of B is greater than mg sin. Therefore theacceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin

CykWdBij Å/okZ/kj uhps dh vksj cymg + N A (CykWd Ads dkj.k vfHkyEc izfrfØ;k)gSA blfy;s CykWdBij dk;ZjrusV cy dkBdh ur lrg ds vuqfn'k ?kVdmg sin ls vf/kd gSA blfy;s Hkwfe ds lkis{kBdk Roj.k] ftldh fn'kkCdh ur lrg ds vuqfn'k gS]g sinls vf/kd gSA

125. Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown. When the displacementof 'B' w.r.t. 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that

moment is:

izkjEHk esa fLizax vladqfpr gSA vcBij n'kkZ;s vuqlkj cy'F'yxk;k tkrk gSA tc dqN le; ckn'B'dk'A'ds lkis{k nk;havksj foLFkkiu'x'gS rks ml {k.k'B'dk 'A'ds lkis{k Roj.k gksxk& [M.Bank_N.L.M._1.24]

(A)F

m2(B)

F kx

m

(C*)

F kx

m

2(D) none of these


33/101

Page # 33

Sol. F - Kx = mb and kx = maHence m (b – a) = F – 2kx

Ans. (C).

126. In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the casesif time required is same then in which case he has to exert more force. Assume pulleys and stringslight.

fp=k esa n'kkZ, vuqlkj] ,d vkneh xqVds dks tehu lshÅpk¡bZ rd mBkuk pkgrk gSA nksauks fLFkfr;ksa esa vxj le;

cjkcj yxrk gS rks dkSulh fLFkfr esa vko';d cy dk eku vf/kd gksxkA ;g ekfu, fd iqyh rFkk jLlh gYds gSA[M.Bank(07-08)_N.L.M._3.26]

[Made 2006, RKV, GRSTU] [3.35 NLM]

(A*) (i) (B) (ii) (C) same in bothnksuksa esa leku(D) Cannot be determinedfu/kkZj.k ugha dj ldrs

Sol. Since, h = 2

1at2 a should be same in both cases, because h and t are same in both cases as

given.

In (i) F1 – mg = ma.

F1 = mg + ma.

In (ii) 2F2 – mg = ma F

2 =

2

mamg F

1 > F

2.

127. A weight W is supported by two strings inclined at 60º and 30º to the vert ical . The tensions in the

strings are T1 & T2 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? (block is in equilibrium)

,d HkkjWnks jfLl;ksa tks Å/oZ ls60ºrFkk30ºij >qdh gS jLlh;ksa esa ruko fp=kkuqlkjT1 rFkkT2 gSA ;fn bu rukoksa dksWds :i esa cyksa dk f=kHkqt cukdj Kkr djus gS rks fuEu esa ls dkSulk f=kHkqt [khapsaxs? ¼Hkkj lkE;koLFkk esa gS½ [M.Bank_N.L.M._ 3.13]

(A) (B) (C) (D) (E*)


34/101

Page # 34

Sol.

= , = , =

+ + = 0

Ans. (E)

128. System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the

speed of the block will be: [Made 2004]

M.Bank_NLM_8.12

fp=k esa iznf'kZr fudk; esa ,d vkneh fu;r pky'u'ls jLlh dks nksuks fljksa ls [khap jgk gSA CykWd dh pky gksxhA

(A*)4u3 (B)

2u3 (C)

4u (D) none of thesebuesa ls dksbZ ugha

Sol. u =2

v0 1 ,

2

vv 21 = v ,

2

uv – 2 = v

Hence v = velocity of M =4

u3

NLM

129. System is shown in the figure. Velocity of sphere A is 9 m/s. Then speed of sphere B will be :

fp=k esa iznf'kZr fudk; esa ;fn xksys Adk osx9 m/sgks rks xksysBdh pky gksxhA [ Made 2004]

[M.Bank_NLM_8.14]

(A) 9 m/s (B*) 12 m/s (C) 9 4

5 m/s (D) none of thesebuesa ls dksbZ ugha

Sol. 9 cos = v sin (i)

12

R –19 = tan (ii)


35/101

Page # 35

(R + 5)2 = (12)2 + (19 – R)2

R = 10Hence from (i) and (ii)

v = 12 m/s2

130. 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 N4respectively and acceleration with which the block slides down on the wedge in situations are b1, b2, b3

and b4 respectively then : [Made M. Pathak] [M.Bank_NLM_5.18]m nzO;eku dk xqVdk ,d ost ij j[kk gSA n'kkZ;s vuqlkj ost pkj izdkj ls Rofjr gksrk gS(1), (2), (3)o(4)A ;fnfLFkfr;ksa(1), (2), (3)o(4)esa vfHkyEc izfrfØ;k Øe'k%N

1, N

2, N

3 oN

4 gSa ,oa Roj.k ftlls xqVdk ost ij bu fLFkfr;ksa

esa fQlyrk gS] osb1, b

2, b

3 vkSjb

4gSa rks:

(A*) N3 > N

1 > N

2 > N

4(B) N

4 > N

3 > N

1 > N

2(C*) b

2 > b

3 > b

4 > b

1(D) b

2 > b

3 > b

1 > b

4

Sol. (A), (C)

(1)

mg mgcos37°masin37°mgsin37°

macos37°

ma ( )t M+Roh; cy

N

a

Balancing forces perpendicular to incline N = mg cos37° + ma sin37°

ur ry ds yEcor~ cyksa dks larqfyr djus ijN = mg cos37° + ma sin37°

N1 =

5

4mg +

5

3ma

and along incline mg sin 37° – ma cos 37° = mb1

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

b1 =

5

3g –

5

4a

(2)

mg mgcos37°

masin37°

macos37°

mgsin37°

( ) mat M+Roh; cyN

a


5

4mg –

5

3ma

blh izdkj bl fLFkfr ds fy, ge ikrs gSa N2 =

5

4mg –

5

3ma

and vkSj b2 = 53 g +

54 a

N2 =

5

4mg –

5

3ma


36/101

Page # 36

(3) Similarly for this case get N3 =

5

4mg +

5

4ma

blh izdkj bl fLFkfr ds fy, ge ikrs gSa N3 =

5

4mg +

5

4ma

andvkSj b3 =

5

3g +

5

3a

(d) Similarly for this case get N4 =

5

4mg –

5

4ma

blh izdkj bl fLFkfr ds fy, ge ikrs gSaN4 =

5

4mg –

5

4ma

andvkSj b4 =

5

3g –

5

3a

131. 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]xv{k ds vuqfn'k xfr'khy ,d oLrq ij dk;Zjr cy d.k dh fLFkfr ds lkFk fp=kkuqlkj ifjofrZr gksrk gSA oLrq fdl fLFkfrij LFkk;h lkE;koLFkk esa gSA

(A) x = x1

(B*) x = x2

(C) both x1 and x

2(D) neither x

1 nor x

2

(A) x = x1

(B*) x = x2

(C) x1 rFkkx

2nksauks (D) x

1 rFkkx

2nksauks ugha

Sol. Body will be in equilibrium at both x1 & x

2 as at these points force will be zero.

At x2 on increasing x force becomes -ve & on decreasing x force becomes + ve. so force & displacementhave opposite signs. so it a pt. of stable eq.

132. S1 : Newton's third law depends on Newton's second law.

S2 : Newton's first law can be derived from Newton's second law.

S3 : All the three Newton's laws are independent of each other.

S4 : 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.

S1 : U;wVu dk rhljk fu;e] U;wVu ds f}rh; fu;e ij fuHkZj djrk gSA

S2 : U;wVu ds f}rh; fu;e ls] U;wVu ds izFke fu;e dks O;qRiUu fd;ktk ldrk gSA

S3 : U;wVu ds lHkh rhuksa fu;e ,d nwljs ls LorU=k gSA

S4 : ,d oLrq dks tehu ij fojke esa j[kk tkrk gS rc CykWd ij ì Foh }kjk yxk;k x;k xq:Rokd"kZ.k cy rFkk CykWd}kjk ì Foh ij yxk;k x;k vfHkyEc cy] U;wVu ds rhljs fu;e esa fØ;k o izfrfØ;k ;qXe dks crkrs gSA(A*) F T F F (B) F F T T (C) T T T F (D) T F F F

Sol. S1 : Newton’s third law is independent of f irst law. Hence S

1 is false.

S2 : Newton’s first law is a special case of second law when a = 0. S

2 is true.

S3 : Newton’s first law is derived from second law. S

3 is false.

S4 : The normal reaction by earth on block and by block on earth form action and reaction pair. S

4 is false.

S1 :U;wVu dk rhljk fu;e] igys fu;e ls Lora=k gSA vr%S

1 vlR; gSA

S2 :U;wVu dk izFke fu;e] U;wVu ds nwljs fu;e dk gh fo'ks"k fLFkfr gSA tca = 0. S

2 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%S

3 vlR; gSA

S4 :CykWd ij ì Foh }kjk yxk;k x;k vfHkyEc cy o CykWd }kjk ì Foh ij yxk;k x;k vfHkyEc cy fØ;k izfrfØ;k ;qXe

cukrs gSaA vr%S4 vlR; gSA


37/101

Page # 37

133. A part icle is projected with a veloci ty 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)

,d CykWd dks = 30°ds fpdus ur ry ds fuEure fcUnq ls20 m/sds osx ls iz{ksfir fd;k tkrk gSA ur ry dh Å¡pkbZ10 mgSA 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=kkuqlkj ijoy; dh lehdj.kx2 = 4aygSA bldh Å/okZ/kj v{k rFkk 'kh"kZ ewy fcUnq ij gS] bl ijoy; rkj ij mnzO;ekudh eudk fLFkr gSA rkj dk Ýse fLFkj gS rFkk eudk ¼eudk½ fcuk ?k"kZ.k ds ijoy; ij fQly ldrk gSA rkj Ýse ijy= 4afcUnq ls eudk fLFkjkoLFkk ls NksM+k tkrk gSA tc eudky = afLFkfr ij igq¡prk gS rks bldk Li'kZjs[kh; Roj.k gS:

B M _ C M _ 1 6 1[MB_Q. 1.8]

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

(A)2

g(B)

2

g3(C*)

2

g(D)

5

g

Sol. x2 = 4ay

Differentiating w.r.t. y, we get

yds lkis{k vodyu

dx

dy =

a2

x

At (2a, a),dx

dy = 1 hencevr% = 45°

the component of weight along tangential direction is mg sin .

Hkkj dk Li'kZ js[kh; fn'kk esa ?kVdmg sin.

hence tangential acceleration is g sin =2

g

vr% Li'kZ js[kh; Roj.k g sin =2

g

135. 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]

,d NM+ ABfp=k esa iznf'kZr gSA NM+ dk Afljk Hkwfe ij fLFkj ¼c¡/kk½ gS rFkk CykWd nk¡;h vkSj 3 m/sosx ls xfr dj jgk

gSA tc NM+ /kjkry ls60º dk dks.k cuk;s rks NM+ ds fljsBdk osx gksxk&


38/101

Page # 38

[Made 2004]

(A) 3 m/s (B*) 2 m/s (C) 2 3 m/s (D) 3 m/s

Sol. Let AB = , B = (x , y)

Bv

= vx î + vy ĵ

Bv

= î3 + ĵv y (i)

x2 + y2 = 2

2x vx = 2y v

y = 0 3 +

x

y v

y = 0

3 + (tan600) vy = 0

vy = – 1

Hence from (i)

Bv

= î3 – ĵ

Hence

vB = 2 m/s

136. 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;eku dk fi.M ,d cy tksfd fLFkfrt ÅtkZ&foLFkkiu oØ }kjk çnf'kZr gS fd mifLFkfr esaxfn'kk esa xfrdj jgk gSA ;fn bls fcUnqx=2m,ls NksM+k tk;s rksx = 5mfcUnq dks ikj djrs le; bldh pky gS&

[M_Bank(07-08)_WPE_7.7]

(A) zero (B) 1 ms1 (C*) 2 ms1 (D) 3 ms1

Sol. Loss in potential energy = gain in kinetic energy

6 – 2 =2

1 . 2 . (v2)

v = 2 m/s.

Alternate :

Sol. Slope of U – X graph isdx

du = – F

Since, Slope is constant and negative from x = 1 to x = 3.5 m, the force is accelerating and constant

(constant acceleration case)

Acceleration (x = 1 to 3.5) = F/m = tan/m =2

)5.2/10( = 2m/s2

Using ; v2 = u2 + 2aS from x = 2 to x = 3.5 m

v2 = 2(2)

2

3= 6

From x = 3.5 to x = 4.5 :

a2 =

2

)1/2( = – 1m/s2

& v2

2 = v12 – 2aS

2


39/101


40/101

Page # 40

u =3

v2

hence work done by man = change in K.E. of system

=2

1 mu2 +

2

1 2m (v – u)2 =

2

1m

2

3

v2

+

2

12m

2

3

v

=

3

mv2

Ans.

3

1 mv2

140. In the figure shown the blocks A & C are pulled down with constant velocities u . Acceleration of block

B is :

fn[kk;s x;s fp=k esa ArFkkCCykWdksa dks fu;r osxuls [khapk tkrk gSA CykWdBdk Roj.k gksxk :[ Made 2003] [M.Bank_NLM_8.11]

(A)u

b

2

tan2 sec (B*)u

b

2

tan3 (C)u

b

2

sec2 tan (D) zero‘'kwU;

Sol. Let vB = v()

u = vcos

dt

du=

dt

dvcos – v sin

dt

d ...........(1)

dt

d(b cot ) = – v

dt

d =

b

vsin2

Using ...... (1)

0 = aB cos – (v sin )

b

sinv 2

aB =

cosb

sinv 32

=

2

cos

u

cosb

sin3

aB = b

tanu 32 Ans. (B)

141. STATEMENT-1 : A man standing in a lift which is moving upward, will feel his weight to be greater than when

the lift was at rest. (F_only)

STATEMENT-2 : If the acceleration of the lift is ‘a’ upward, then the man of mass m shall feel his weight to be

equal to normal reaction (N) exerted by the lift given by N = m(g+a) (where g is acceleration due to gravity)

[Test_PQF combined_(CT-4_19-08-07)_Paper-1_Q.12]

oDrO;-1 : ,d vkneh fy¶V esa [kM+k gS rFkk fy¶V Åij dh rjQ py jgh gS rks og O;fDr bl ckj fy¶V ds fojke esagksus dh rqyuk esa ges'kk vf/kd Hkkj eglwl djsxkAoDrO;-2 : ;fn fy¶V dk Åij dh rjQ Roj.k 'a'gS rksmnzO;eku dk O;fDr }kjk eglwl fd;k x;k Hkkj fy¶V }kjk vknehij yxk;s x;s vfHkyEc cyN ds cjkcj gksxk tgkaN = m(g+a) (;gkagxq:Rokd"kZ.k ds dkj.k Roj.k gS)(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False(D*) Statement-1 is False, Statement-2 is True

(A)oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1dk lgh Li"Vhdj.k gSA(B)oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1dk lgh Li"Vhdj.k ugha gSA


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(C)oDrO;-1lR; gS] oDrO;-2vlR; gS ;(D*)oDrO;-1vlR; gS] oDrO;-2lR; gS

Sol. [Easy]

If the lift is retarding while it moves upward, the man shall feel lesser weight as compared to when lift was at

rest. Hence statement1 is false and statement 2 is true.

;fn Åij dh rjQ tkrh fy¶V eafnr gksrh gS rks vkneh }kjk eglwl fd;k x;k Hkkj bl ckj okLrfod Hkkj ls de gksxkAvr% oDrO;-1vlR; gS] oDrO;-2lR; gS

142. Two blocks, of masses M and 2M, are connected to a light spring of spring constant K that has one end

fixed, as shown in figure. The horizontal surface and the pulley are frictionless. The blocks are released

from rest when the spring is non deformed. The string is light. [Made A.K.S. sir] [M.Bank(07-08)_NLM_1.26]

Mo2MnzO;eku ds nks xqVds fLizax fu;rkadKds ,d gYds fLizax ls tqM+s gSa ftldk ,d fljk fp=kkuqlkj fLFkj gSA{kSfrt lrg ,oa iqyh ?k"kZ.k jfgr gSaA xqVds fojke ls NksM+s tkrs gSa tc fLizax vfod̀r ¼fcuk f[kapk½ gSA Mksjh gYdh gSA

2M

MK

(A*) Maximum extension in the spring isK

Mg4.

fLizax esa vf/kdre foLrkjK

Mg4 gSA

(B*) Maximum k inetic energy of the system isK

gM2 22

fudk; dh vf/kdre xfrt ÅtkZ KgM222

gS(C*) Maximum energy stored in the spring is four times that of maximum kinetic energy of the system.

fLizax esa laxzfgr vf/kdre ÅtkZ fudk; dh vf/kdre xfrt ÅtkZ dh pkj xquh gSA

(D) When kinetic energy of the system is maximum, energy stored in the spring isK

gM4 22

tc fudk; dh xfrt ÅtkZ vf/kdre gS] rks fLizax esa lxzfgr ÅtkZK

gM4 22 gSA

Sol. Maximum extension will be at the moment when both masses stop momentarily after going down.

Applying W-E theorem from starting to that instant.

kf – ki = Wgr. + Wsp + W ten.

0 – 0 = 2 M.g.x + 2Kx

21 + 0

x =K

Mg4

System will have maximum KE when net force on the system becomes zero. Therefore

2 Mg = T and T = kx x =K

Mg2

Hence KE will be maximum when 2M mass has gone down byK

Mg2.

Applying W/E theorem

kf – 0 = 2Mg. K

Mg2 –

2

22

K

gM4.K

2

1

kf = 2

22

K

gM2


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F2 =

22

22

Mm

m1

gm

; F =2

Mm

m1

mg

145. System shown in figure is in equilibrium. The magnitude of change in tension in the string just before

and just after, when one of the spring is cut. Mass of both the blocks is same and equal to m and spring

constant of both springs is k. (Neglect any effect of rotation)

fn, x, fp=k esa fudk; lkE;koLFkk esa gSA tc ,d fLçax dks dkV fn;k tkrk gS rks jLlh esa ruko esa dqy ifjorZufLçax ds dkVus ds Bhd igys vkSj Bhd ckn esa D;k gksxk nksuksa xqVdksa dk nzO;eku ,d tSlk gS vkSjmds cjkcj gSvkSj nksuksa fLçax dk fLçax fu;rkadKgSA ¼?kw.kZu ds izHkko dks ux.; ekusa½ M.Bank_NLM_1.14

[Made 2004]

(A*)2

gm(B)

4

gm(C)

4

gm3(D)

2

gm3

146. A boy and a block, both of same mass, are suspended at the same horizontal level, from each end of a light

string that moves over a frictionless pulley. The boy starts moving upwards with an acceleration 2.5 m/s 2

relative to the rope. If the block is to travel a total distance 10 m before reaching at the pulley, the time taken

by the block in doing so is equal to : [Made AKS_2007_GRSTU] [M.Bank_NLM_3.46]

/k"kZ.k jfgr f?kjuh ls ikfjr ,d gYdh jLlh ds izR;sd fljs ij] leku {kSfrt Lrj ij] ,d yM+dk ,oa ,d fi.M nksuksa lekunzO;eku ds yVds gSA yM+dk Åij dh vksj jLlh ds lkis{k2.5eh-/ls-2 ds Roj.k ls pyuk izkjEHk djrk gSA ;fn fi.M] f?kjuhrd igqapus esa dqy nwjh10eh-r; djrk gS] rks fi.M dks ,slk djus esa yxk gqvk le; cjkcj gSA

10 m

mm

(A) s8 (B*) 4s (C)2

10 s (D) 8s

Ans. (B)

Sol. Acceleration of boy and pulley will be same equal to 1.25 m/s2 w.r.t. ground. Hence

yM+ds ,oa f?kjuh nksuksa dk tehu ds lkis{k Roj.k cjkcj ,oa1.25eh-/ls-2 ds cjkcj gksxkA vr%

10 =2

1(1.25) t2 t = 4 sec.

147. In the position shown collar B moves to the left with a velocity of 150 mm/s. Determine:

fn[kk;s x;s fp=k esa dkWyjBck¡;h vkSj150 mm/sds osx ls xfr djrk gS rks crkb;sA [M.Bank_NLM_8.25][Made 2004]


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(a) the velocity of collar A dkWyj Adk osx(b) the velocity of portion C of the cable rkj dsCHkkx dk osx(c) the relative velocity of portion C of the cable with respect to collar B.

dkWyjBds lkis{k rkj dsCHkkx dk vkisf{kd osx[ Ans.: V

A = 300 right, V

C = 600 left, V

CB = 450 left ]

148. A rigid rod leans against a vertical wall at a point A. The other end of the rod is on the horizontal floor.

Somebody is pushing point A of the rod downwards with constant velocity. Will B move with constant velocity

? What is the path of the centre of the rod. M.Bank_NLM_8.35

,d ǹqdh gS] NM+ dk nwljk fljk {kSfrt Q'kZ ij gS] dksbZ O;fDr NM+ dks fcUnq Aij fu;r osx ls uhps dh vksj /kdsy jgk gS rks D;k fljsBdk osx fu;r gksxk \ NM+ ds dsUnz dk fcUnqiFk D;k gksxk \

[ Ans. No, Circular]

149. A wedge of height h is released from rest with a light particle P placed on it as shown . The wedge slides down

an incline which makes an angle with the horizontal . All the surfaces are smooth , P will reach the surface of the incline in time :hÅWPkkbZ dk ost ur ftl ij ,d gYdk d.kP j[kk gS dks fojke ls fp=kkuqlkj NksMk tkrk gSA ost dks.k ds ur ry ijfQlyrk gSA lHkh lrg fpduh gSA d.kP ur ry dh lrg ij fdrus le; ds ckn igqpsxkA

(A*) 2sing

h2

(B) cossingh2

(C) tangh2

(D) 2cosg

h2

150. A lift of total mass M kg is raised by cables from rest to rest through a height h . The greatest tension

which the cables can safely bear is nM kg wt . Show that the shortest interval of time in which the

ascent can be made is ,

2/1

g)1n(

hn2

.

MnzO;eku dh ,d fy¶V] rkjksa(cables)ds }kjk fojke ls fojke rdhÅpkbZ rd mBkbZ tkrh gSA rkjksa dk vf/kdrelqjf{kr ruko ftruk rkj lgu dj ldrk gS] nMfdxzkµHkkj gSA çnf'kZr djks fd U;wure le;kUrjky ftles fy¶V

Rofjr dj ldrs gSa]

2/1

g)1n(

hn2

ds cjkcj gSA

Sol.2

1v0 T= h T =

0v

h2

v0 = 0 + (h – 1) gt10 = v0 – g (T – t1)

v0 = gT – gt1 .

151. Two blocks A and B of mass 1 kg & 2 kg respectively are connected by a string, passing over a light

frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that

string remains just taut. At moment t = 0, a force F = 20 t newton starts acting on the pulley alongvertically upward direction, as in Figure. Calculate : [M.Bank(07-08)_NLM_4.13]

1 kgrFkk2 kgnzO;eku ds nks CykWdksa dks jLlh }kjk f?kjuh ls tksM+k tkrk gSA f?kjuh ?k"kZ.k jfgr rFkk gYdh gSA nksuksa CykWd{kSfrt lrg ij fojke esa gS rFkk f?kjuh bl rjg gS fd jLlh ruh gqbZ gSAt = 0ijF = 20 tU;wVu dk cy f?kjuh ij Åijdh vksj fp=kkuqlkj yxk;k tkrk gSA x.kuk djks:


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(a) velocity of A when B loses contact with the floor

(b) height raised by the pulley upto that instant(c) work done by the force F upto that instant.

(a) CykWd Adk osx tc CykWdB/kjkry ls laidZ NwV tkrk gSA(b) bl {k.k rd f?kjuh }kjk r; Å¡pkbZ(c) bl {k.k rd cyF}kjk fd;k x;k dk;Z[ Ans: 5 m/s, 5/6 m, 175/6 J ]

Sol. A looser contact at t = 1 sec and B at t = 2 sec. Between t = 1 to 2 sec. , a A = T – 1 g = 10 t – 10

dt

dv = 10 t – 10 10dv

t

1

)1 –t( dt

V = 10 2

)1 –t( 2

= 5 (t – 1)2

(1)

At t = 2 , V = 5 (2 – 1)2 = 5 mgs Ans. of (a)

(b) height raised by A = S ?

sol. (1) dt

ds = 5 (t – 1)2 ds =

2t

1t

3

3

)1 –t(5

s =3

5

Height raised by pulley between t = 1 to t = 2 sec. =

2

5 =

6

5 m Ans.

(c) WF + W

G = K W

F – 1 g (5/3) =

2

1 (1) (5)2

WF =

3

50 +

2

25 =

6

175 J

152. A block of mass m1 is placed on a wedge of an angle , as shown. The block is moving over the inclined

surface of the wedge. Friction coefficient between the block and the wedge is µ1, where as it is µ

2

between the wedge and the horizontal surface. If µ1 =

2

1, = 45°, m

1 = 4 kg, m

2 = 5kg and g = 10 m/s2,

find minimum value of µ2 so that the wedge remains stationary on the surface. Express your answer in

multiple of 10 –3.nzO;ekum

1 dk ,d fi.M dks.k okys ost ij fp=kkuqlkj j[kk gSa fi.M] ost dh ur&lrg ij Åij xfreku gSA ?k"kZ.k xq.kkad

fi.M rFkk ost ds chp µ1 gS tcfd ;g ost rFkk {kSfrt lrg ds chpµ

2 gSA ;fn µ

1 =

2

1, = 45°, m

1 = 4 kg, m

2 =

5kgrFkkg = 10eh-/ls-2 gks rksµ2 dk U;wure eku Kkr djks] ftlls ost] lrg ij fLFkj jgsA vius mRrj dks10 –3 ds xq.kd

esa O;Dr djksA [M_Bank_Friction_Q. 2.72]

m1µ1

m2

µ2

[Made RKV_2007]

Sol. Taking block + wedge as system and applying NLM in horizontal direction

fi.M $ ost dks fudk; ysdj rFkk {kSfrt fn'kk esa U;wVu dk fu;e yxkus ijf 2

= m1a cos


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Page # 46

= m1 [g(sin – µ

1 cos )] cos ........... (1)

Again applying NLM in vertical direction

iqu% U;wVu dk fu;e m/okZ/kj fn'kk esa yxkus ij(m

1 + m

2)g – N

2 = m

1 a sin

N2

= (m1 + m

2)g – m

1 sin (g sin – µ

1g cos )

For limiting conditionlhekUr fLFkfr ds fy, f 2 = µ

2N

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

m1µ1

m2

µ2

f 2

N2

a

From (1) and (2)

(1)

rFkk(2)

ls

µ2 =

)cosggsn(sinmg)mm(

)cosgsing(cosm

1122

11

Using valuesekuksa dk iz;ksx djus ij

µ2 =

8

1 = 125 x 10 –3

Ans. 125

Alt.

f 2

N2

µ

m g c o s

1

1

N = m g cos 1

Applying NLM in vertical direction

N2 = m

2g + m

1g cos2 + µ

1m

1g sin .......... (1)

Applying NLM in horizontal direction

f 2 = µ

2N

2 = m

1g cos sin – µ

1m

1g cos2

On Solving

µ2 =

8

1 Ans.

153. In the figure shown 'P' is a plate on which a wedge B is placed and on B a block A is placed. The plate is

suddenly removed and system of B and A is allowed to fall under gravity. Neglecting any force due to air on

A and B, prove mathematically that the normal force on A due to B is zero.

fp=kkuqlkj IysV'P'ij CykWd ArFkkBj[ks tkrs gSaA IysVPvpkud gVk yh tkrh gS rFkk fudk;(BrFkk A)dks xq:Ro dsizHkko esa fxjus fn;k tkrk gSA gok ds }kjk ArFkkBij cy ux.; gSA xf.krh; :i ls fl) dhft, fd A ijB ds }kjkvfHkyEc izfrfØ;k cy 'kwU; gSA [ M_Bank_NLM_Q. 4.16(a)]

Sol. F.B.D. of block A

Applying newtons second law for block A in vertical direction

m Ag – N cos = m

Ag ; where is the angle of the wedge

N cos = 0as < 90° N = 0

154. In the figure PQRS is a frictionless horizontal plane on which a particle A of mass m moves in a circle

of radius r with an angular velocity such that 2 r = g/3. Another particle of mass m is tied to A through

an inextensible massless string. O is the hole through which string passes down to B. B can moveonly vertically. The tension in the string at this instant will be:


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Page # 47

fp=k esaPQRS,d ?k"kZ.kjfgr {kSfrt ry gS ftl ijmnzO;eku dk ,d d.k A,rf=kT;k ds ò Ùk esa dks.kh; osx lsbl çdkj xfr djrk gS fd2r = g/3AmnzO;eku dk ,d nwljk d.k ,d nzO;ekughu vforkU; Mksjh }kjk Als ca/kk gqvk gSAOog fNnz gS ftlesa gksdj MksjhBrd tkrh gSABdsoy m/okZ/kj xfr dj ldrk gSA bl {k.k Mksjh esaa rukogksxk:

(A) mg/3 (B*) 2 mg/3 (C) mg/6 (D) none

Sol.T

T

m

m

a

a

mg

m w r (C.F.F)2

T – m w2r = ma

T – m

3

g = ma (1)

mg – T = ma (2)

T –3

mg = mg – T 2T = 4 mg/ 3

T = 2 mg /3 Ans. (B)

155. A bus is moving with a constant acceleration a = 3g/4 towards right. In the bus, a ball is tied with a rope

and is rotated in vertical circle as shown. The tension in the rope will be minimum, when the ropemakes an angle = _____ . [Made M. Pathak Sir] M.Bank_Circu_M_3.57,d cla = 3g/4fu;r Roj.k ls nka;h rjQ xfr dj jgh gSA cl esa ,d xsan tks jLlh ls tqM+h gSA m/okZ/kj ry esa fp=kkuqlkjòÙkh; xfr dj jgh gSA jLlh esa ruko U;wure gksxk tc jLlh dks.k cukrh gSA rks = _____gksxk

a=3g/4

(A*) 53° (B) 37° (C) 180 – 53° (D) 180 + 37°

Sol.

Tmax A

Fnet mg

530

B Tmin

inertial force

M(3g/4)53

0

Fnet is shown in the figure. So, tension will be max. at point A and will be min. at point B.

156. In the figure shown C is a fixed wedge on horizontal surface. Blocks A and B are of masses m and 2m

respectively are kept as shown in figure. They can slide along the inclined plane smoothly. The pulley and

string are massless. Take = 30° and g = 10m/s2. The inclined planes are very long. A and B are released

from rest. 2 seconds after the release, 'B' is caught for a moment and released again. Find out the speed of 'A' just before the instant when the string becomes tight again.


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Page # 48

fp=k esa fLFkj ostC{kSfrt lrg ij j[kk gSAmo2mnzO;eku ds CykWd ArFkkBfp=kkuqlkj j[ks tkrs gSA CykWd ur ryds vuqfn'k fQly ldrs gSA f?kjuh rFkk jLlh nzO;eku jfgr gSA = 30°rFkkg = 10m/s2gSA ur ry cgqr yEck gSA ArFkkBdks fojke ls NksM+k tkrk gSA NksM+us ds 2lsd.M i'pkr~'B'dks idM+k tkrk gS rFkk fQj NksM+k tkrk gS rks CykWd'A'dh bl {k.k ls igys pky D;k gksxh tc jLlh nwljh ckj ruh gksxhA

[4]

[M.Bank(07-08)_NLM_4.16] [Made_RKV_2006_RSTG]

Sol. Initially acceleration of B and A is same along the string which is given by

a =m3

30sinmg30sinmg2 =

6

g

After 2 second, speed of both A and B is

VB = V

A =

6

g × 2 =

3

10 m/s

Now,

When B is caught for a moment and released again, speed of B becomes zero,

while A is still having a speed3

10 m/s up the inclined due to which string will become slack. But A will

deccelerate and B will accelerate. Because of this the string will become tight again after A and B travel the

same distance

Let this time interval be t

3

10 × t –

2

1 g sin 30° × t2 =

2

1g sin 30° × t2

3

10 =

2

g t ; as t 0

t =3

2 sec

At this time speed of A is given by

V =3

10 –

2

g × t =

3

10 –

2

10 ×

3

2 = 0 Ans. (b) speed of A = 0

157. A system is shown in the figure. A man standing on the block is pulling the rope. Velocity of the point of string

in contact with the hand of the man is 2 m/s downwards. The velocity of the block will be: [ assume that the

block does not rotate ] [ Made 2004] [M.Bank(07-08)_NLM_8.13]

fp=k esa ,d fudk; iznf'kZr gSA CykWd ij [kM+k O;fDr jLlh dks [khap jgk gSaA O;fDr ds gkFk ls lEidZ okyh jLlh dk osx2 m/suhps dh vksj gSA CykWd dk osx gksxk[ekfu;s fd CykWd ?kw.kZu ugha djrk gSA] [ Made 2004]

(A) 3 m/s (B*) 2 m/s (C) 1/2 m/s (D) 1 m/s

Sol. (B)

12

34

2 m/s

v

1 + 2 + 3 + 4 = C


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dt

d 1 +

dt

d 2 +

dt

d 3 +

dt

d 4 = 0

–v – v + 0 + v + 2 = 0 v = 2m/s

158. A ring of radius R lies in vertical plane. A bead of mass ‘m’ can move along the ring without friction. Initially the

bead is at rest at the bottom most point on ring. The minimum constant horizontal speed v with which the ring

must be pulled such that the bead completes the vertical circle [M.Bank(07-08)_Circular Motion_3.9]

R

f=kT;k dk oy; m/okZ/kj ry esa fLFkr gSA‘m’

nzO;eku dk ,d eudk oy; ds vuqfn'k fcuk ?k"kZ.k ds xfr dj ldrk gSAizkjEHk esa eudk oy; ds U;wure fcUnq ij fojke esa gSA oy; dks fdl U;wurefu;r {kSfrtpky ls [khapuk pkfg, ftllseudk ò Ùkh; xfr dj ldsA

[Made MPS-2005]

(A) gR3 (B*) gR4 (C) gR5 (D) gR5.5

Sol. In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is v at the lowest position.

The condition for bead to complete the vertical circle is, its speed at top position

vtop

0From conservation of energy

2

1 m 2topv + mg (2R) = 2

1mv2

or v = gR4

159. A person wants to slide down a rope whose breaking strength is4

3th of his weight. He can come down

safely. [NLM]

,d O;fDr jLlh ds }kjk uhps vkus dh dksf'k'k dj jgk gSA jLlh esa vf/kdre lguh; ruko(breaking strength)

mlds Hkkj dk4

3Hkkx gSA D;k og lqjf{kr uhps vk ik;sxkA

Sol. True

160. System is shown in figure and wedge is moving towards left with speed 2 m/s. Then velocity of the block B

will be: [ Made 2004]

fp=k esa iznf'kZr fudk; esa xqVdk osx2 m/spky ls ck¡;h vksj xfr dj jgk gS] rks CykWdBdk osx gksxkAM.Bank_NLM_8.21

(A) 3 m/s (B) 1 m/s (C*) 2 m/s (D) 0 m/s

Sol. Assume that angle of indination = = 60º


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Page # 50

VB =

2

2120º

= º120cos)2)(2(222 22 = 2 m/s Ans. (C)

161. Column I describes the motion of the object and

Newton's Laws of Motion Resonance

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