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NEET • PHYSICS • LAWS OF MOTION

Laws of motion practice · Hard

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1 Hard
1 of 14
Application +4 | -1
A particle of mass mm is released from top of a smooth hemisphere of radius RR placed on a horizontal surface. Find the contact force acting between the particle to the sphere when the line joining the particle to the centre of sphere makes an angle 60∘60^{\circ} with upward vertical (Given g=10{g}=10 m/s2)\mathrm{m} / \mathrm{s}^2)
View answer and explanation
Correct option: D
None of the above
None of the above is correct because the contact force is calculated using the normal force equation, which results in a specific numerical value that is not present in the options.
2 Hard
2 of 14
Application +4 | -1
The apparent weight of a mass in a lift moving up is 80 kg when its weight in the stationary lift is 60 kg. If the same lift now moves up with same retardation the weight of the mass will be
View answer and explanation
Correct option: A
40 kg
Option A (40 kg) is correct because the change in apparent weight is proportional to the acceleration, and the retardation is the same magnitude as the acceleration.
3 Hard
3 of 14
Analyzing +4 | -1
Two persons of equal weight are hanging by their hands from the ends of a light rope hung over a frictionless pulley. They begin to climb the rope. One person can climb with twice the speed of the other (with respect to the rope). Who will get to the top first?
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Correct option: C
Both will get to the top simultaneously
Option C (Both will get to the top simultaneously) is correct. The center of mass of the system remains stationary, so both reach the top at the same time.
4 Hard
4 of 14
Application +4 | -1
An object flying in the air with velocity (20i^+25j^−12k^)(20 \hat{i}+25 \hat{j}-12 \hat{k}) suddenly breaks into two pieces whose masses are in the ratio of 1:5.1: 5. The smaller mass flies off with a velocity (100i^+35j^+8k^).(100 \hat{i}+35 \hat{j}+8 \hat{k}). The velocity of the larger piece will be:
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Correct option: A
4i^+23j^−16k^4 \hat{i}+23 \hat{j}-16 \hat{k}
Option A (4i^+23j^−16k^4 \hat{i}+23 \hat{j}-16 \hat{k}) is correct because it is the velocity of the larger piece calculated using the conservation of momentum.
5 Hard
5 of 14
Analyzing +4 | -1
A body of mass M is kept on a rough horizontal surface (friction coefficient μ\mu). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is F, where
View answer and explanation
Correct option: C
Mg≤F≤Mg1+μ2Mg \le F \le Mg\sqrt {1 + {\mu ^2}}
Option C (Mg≤F≤Mg1+μ2Mg \le F \le Mg\sqrt {1 + {\mu ^2}}) is correct because the force by the surface must balance both the weight and the applied horizontal force, up to the maximum static friction.
6 Hard
6 of 14
Application +4 | -1
A ball of mass mm moves with speed vv and it strikes normally with a wall and reflected back normally, if its time of contact with wall is tt then find force exerted by ball on wall
View answer and explanation
Correct option: A
2mvt\frac{{2mv}}{t}
Option A (2mvt\frac{{2mv}}{t}) is correct because the force exerted by the ball on the wall is equal to the change in momentum divided by the time of contact.
7 Hard
7 of 14
Application +4 | -1
A block of mass 30 kg is pushed with a horizontal force of 100 N if the coefficient of static and kinetic friction are 0.5 and 0.4. The frictional force acting on the block is (g=10 m/s2)\left( {g = 10\,m/{s^2}} \right)
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Correct option: A
100 N
The frictional force acting on the block is equal to the applied force because the applied force is less than the maximum static friction, so the block doesn't move. Thus, the correct answer is A (100 N).
8 Hard
8 of 14
Application +4 | -1
A particle moves in x−yx-y plane under the influence of a force F→\overrightarrow{\mathrm{F}} such that its linear momentum is P→(t)=i^cos⁡(kt)−j^sin⁡(kt).\overrightarrow{\mathrm{P}}(\mathrm{t})=\hat{\mathrm{i}} \cos (\mathrm{kt})-\hat{\mathrm{j}} \sin (\mathrm{kt}). If k\mathrm{k} is constant, the angle between F→\overrightarrow{\mathrm{F}} and P→\overrightarrow{\mathrm{P}} will be:
View answer and explanation
Correct option: A
π2\frac{\pi}{2}
Option A (π2\frac{\pi}{2}) is correct because the force is perpendicular to the momentum, indicating uniform circular motion.
9 Hard
9 of 14
Application +4 | -1
A car is negotiating a curved road of radius 120 m120 \mathrm{~m}. The road is banked at 7∘7^{\circ}. The coefficient of friction between the tyres and the road is μs=13\mu_{\mathrm{s}}=\frac{1}{\sqrt{3}}. The maximum safe velocity on the road in m/s\mathrm{m} / \mathrm{s} is (g=10 m/s2)\left({g}=10 \mathrm{~m} / \mathrm{s}^2\right) :-
View answer and explanation
Correct option: A
30
Option A (30) is correct because it is the maximum safe velocity calculated using the banking angle, radius of the curve, coefficient of friction, and gravitational acceleration.
10 Hard
10 of 14
Application +4 | -1
A Car takes a circular turn of radius 20 m at a speed of 54 km/h.54 \mathrm{~km} / \mathrm{h}. The least coefficient of friction between tyres and road that can prevent slide slipping is
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Correct option: A
1.15
Option A (1.15) is incorrect. The correct answer should be 1.125. There is a mistake in the provided correct answer. The question asks for the least coefficient of friction, which can be calculated using the formula for centripetal force and friction.
11 Hard
11 of 14
Understanding +4 | -1
Given below are two statements: In the light
of the given statements, choose the correct
answer from the options given.

Statement I: Mass is a measure of inertia of the body.

Statement II: Greater the mass, greater is the force required to change its
state of rest or of uniform motion.
View answer and explanation
Correct option: A
Both statement I and II are correct
Option A (Both statement I and II are correct) is correct because mass is indeed a measure of inertia, and a greater mass requires a greater force to change its state of motion due to its higher inertia.
12 Hard
12 of 14
Understanding +4 | -1
A metre scale is moving with uniform velocity. This implies
View answer and explanation
Correct option: B
the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero
Option B is correct because uniform velocity implies both zero net force and zero net torque acting on the scale.
13 Hard
13 of 14
Application +4 | -1
The angle of inclination of an inclined plane is 30∘30^\circ coefficient of friction between 12 kg body on it and its surface is 0.3. The acceleration of the body down the plane in (g=10 m/s2)\left( {g = 10\,m/{s^2}} \right)
View answer and explanation
Correct option: B
2.5 m/s22.5\,m/{s^2}
Option B (2.5 m/s22.5\,m/{s^2}) is correct because it is the acceleration calculated by considering the component of gravity along the inclined plane and the opposing frictional force.
14 Hard
14 of 14
Application +4 | -1
A bike of mass 300 kg taking turn on a banked road of radius 100 m and banking angle 37∘.37^\circ. If coefficient of static friction is 0.25. Then the maximum speed with which car can negotiate the turn safely. (g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)
View answer and explanation
Correct option: B
35 m/s
Option B (35 m/s) is correct because it is the maximum speed calculated using the formula for safe turn on a banked road with friction.
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