### 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 on a

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 pebblevertically downwards

[/bg_collapse]**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 at a point in space (location of the particle) at a certain instant of time is related to at that point at that instant. Acceleration here and now is determined by the force here and now, not by any history or the motion of the particle.

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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 , how much force is being applied by rocket engine? (Take g = )

(a) (b)

(c) (d)

[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](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) (b)

(c) (d)

[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ](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 …. (ii)

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 , what would be the reading of the weighing scale? (Take )

(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

When the lift is ascending upwards with a acceleration of then

Or The reading of weight machine is 95 kg

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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

**(d)**

*a*=

*g*]

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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 in the upward direction in both the blocks. If *T* and be the tensions in the two parts of the string, then

(a) and

(b) and

(c) and

(d) and

[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ]**(a)**

From F.B.D. of mass 4 *kg * .….(i)

From F.B.D. of mass 2 *kg* .….(ii)

For total system upward force

= 70.8 N

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) 72*J *(c) 56* J *(d) 34 *J*

**(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^{-3}Ns (b) 3 × 10^{-3}Ns

(c) 5 × 10^{-3}Ns (d) 6 × 10^{-3}Ns

**(c)**

Impulse = = area under graph

∴ Total impulse from 4μs to 16 μs

= Area EBCD = 1/2 (200 + 800)^{2} × 10^{-6} + X 800 × 10 × 10^{-6} = 5 × 10^{-3} Ns

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**10.**** **Three blocks of masses m_{1}, m_{2} and m_{3} are placed on a horizontal frictionless surface. A force of 40 N pulls the system then calculate the value of T, if m_{1} = 10kg, m_{2} = 6 kg, m_{3} = 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 = = = 2m/s^{2}

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 F_{1} be the contact force between the block and the plane. Now the inclined plane stops and let F_{2} be the contact force between the two in this case. Then **F**_{1}**/F**** _{2}** is :

(a) 1 (b)

(c) 2 (d)

[bg_collapse view=”button-orange” color=”#4a4949″ expand_text=”Show Solution” collapse_text=”Solution” ]**(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 F_{1} sin 30º = ma

and F_{1} cos 30^{0} = mg

or F_{1} =

and in the second case

F_{2} = mg cos 30^{0}==

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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 = 0

⇒ = – ⇒ = mg

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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} = v^{2} – 2a (90)

⇒ 180a = ……(1)

0 = – 2ax

⇒ 2ax = v^{2}/4 ……..(2)

From (2)/(1), = ⇒ x = 30 cm

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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 = = finally a =

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