Now get the best preparation questions and solutions for jee mains and Advanced by Jeemain.guru, Law of motion physics
1. Which of the following cases the net force acting on the body is not zero?
(a) A drop of rain falling down with a constant speed
(b) A cork of mass 10 g floating on the surface of water
(c) A car moving with a constant speed of 
Rough road
(d) A pebble of mass 0.05 kg is thrown vertically upwards
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](d) : As the rain drop is falling down with a constant speed, its acceleration, a=0 Hence, net force on the drop = ma=0. As the cork is floating on the surface of water, its weight is balanced by the upthrust. Hence net force on the cork is zero.
The force exerted by the engine is balanced by the friction due to rough road. As the car is moving with constant velocity, it’s acceleration a=0 Hence. Net force on the car, F=ma=0
Whenever a body is thrown vertically upwards gravitational pull of earth gives it a uniform acceleration a=g in downward direction.
Hence, net force on the pebble
2. Which one of the following statements is not true about Newton’s second law of motion
(a) The second law of motion is consistent with the first law
(b) The second law of motion is a vector law
(c) The second law of motion is applicable to a single point
particle
(d) The second law of motion is not a local law
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](d) : The second law of motion is a local law which means that force 
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3. A rocket is going upwards with accelerated motion. A man sitting in it feels his weight increased 5 times his own weight. It the mass of the rocket including that of the the man is 
(a) 
(c) 
(b) : As the weight of man increased by 5 times , so acceleration of the rocket,
Force applied by rocket engine is
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4. A trolley of mass 20 kg is attached to a block of mass 4 kg by a massless string passing over a frictionless pulley as shown in the figure. If the coefficient of kinetic friction between trolley and the surface is 0.02, then the acceleration of the trolley and block system is (Take 
(a) 
(c) 
(c):
As the string is inextensible, and the pulley is smooth, therefore both the 4kg block and the 20 kg trolley have the same acceleration a
Let T be tension in the string
The equation of motion of the block is 40-t = 4a…… (i)
The equation of motion of the trolley is
Here
Or 
Adding (i) and (ii) we get
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5. A person of mass 50 kg stands on a weighing scale on a lift. If the lift is ascending upwards with a uniform acceleration of 
(a) 50 kg (b) 60 kg (c) 95 kg (d) 100 kg
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](c): The reading on the scale is a measure of the force on the floor by the person. By the Newton’s third law this is equal and opposite to the normal force N on the person by the floor
Or
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6. A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity ‘g’, the reading on the spring balance will be
(a) 2 kg (b) (4 ×g) kg (c) (2 ×g) kg (d) Zero
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](d)
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7. Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force F applied on the upper string produces an acceleration of 
(a) 
(b) 
(c)
(d)
(a)
From F.B.D. of mass 4 kg 
From F.B.D. of mass 2 kg 
For total system upward force
by substituting the value of T in equation (i) and (ii) and solving we get
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8. A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction between their surfaces is 0.2, then work done against friction is
(a) 98 J (b) 72J (c) 56 J (d) 34 J
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](a)
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9. The magnitude of force (in N) acting on a body varies with time t (in N) as shown. AB, BC and CD are straight line segments. The magnitude of total impulse of force on the body from t = 4 μs to t = 16 μs is
(a) 6 × 10-3Ns (b) 3 × 10-3Ns
(c) 5 × 10-3Ns (d) 6 × 10-3Ns
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](c)
Impulse = 
∴ Total impulse from 4μs to 16 μs
= Area EBCD = 1/2 (200 + 800)2 × 10-6 + 
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10. Three blocks of masses m1, m2 and m3 are placed on a horizontal frictionless surface. A force of 40 N pulls the system then calculate the value of T, if m1 = 10kg, m2 = 6 kg, m3 = 4 kg –
(a) 40 N (b) 20 N (c) 10 N (d) 5 N
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](b)
a = 
40 – T = 10 × 2
T = 20 N
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11. A block is kept on a smooth inclined plane of angle of inclination 30º that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let F1 be the contact force between the block and the plane. Now the inclined plane stops and let F2 be the contact force between the two in this case. Then F1/F2 is :
(a) 1 (b)
(c) 2 (d)
(b)
Surface is smooth. Therefore, force of friction between the block and the plane is zero. So, contact force is really the normal reaction between the two.
In the first case F1 sin 30º = ma
and F1 cos 300 = mg
or F1 =
and in the second case
F2 = mg cos 300
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12. A block of mass ‘m’ is slipping down a rough inclined plane with constant speed. The force on block by plane is –
(a) mg
(b)
(c) Depends upon coefficient of friction
(d) Depends upon angle of inclination
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ]12. (d)
Net force on block 
As block is slipping with constant speed
⇒ 
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13. A bullet fired into a wooden block, loses half of its velocity after penetrating 90 cm. Before coming to rest it penetrates a further distance of –
(a) 10 cm (b) 20 cm (c) 30 cm (d) 60 cm
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](c)
(v/2)2 = v2 – 2a (90)
⇒ 180a = 
0 = 
⇒ 2ax = v2/4 ……..(2)
From (2)/(1), 
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14. A trolley T (mass 5 kg) on a horizontal smooth surface is pulled by a load L (2 kg) through a uniform rope ABC of length 2 m and mass 1 kg. As the load falls from BC = 0 to BC = 2m, its acceleration (in m s-2) changes from –
(Take g = 10 m s-2)
(a) 20/6 to 20/5 (b) 20/8 to 30/ 8
(c) 20/5 to 30/ 6 (d) none of these
[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](b)
Initially a = 
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