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NEET • PHYSICS • LAWS OF MOTION

Laws of motion practice · Easy

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Showing 1-30 of 70 questions
1 Easy
1 of 70
Application +4 | -1
A loaded 20,000 kg20,000 \mathrm{~kg} coal wagon is moving on a level track at 6 ms−16 \mathrm{~ms}^{-1}. Suddenly 5000 kg5000 \mathrm{~kg} of coal is dropped out of the wagon. The final speed of the wagon is
View answer and explanation
Correct option: A
6 ms−16 \mathrm{~ms}^{-1}
Option A (6 ms−16 \mathrm{~ms}^{-1}) is correct because there is no external force acting on the system in the horizontal direction, so the momentum remains constant. Since the coal is dropped vertically, it doesn't affect the wagon's horizontal velocity.
2 Easy
2 of 70
Application +4 | -1
The coefficient of static friction between contact surfaces of two bodies is 13\frac{1}{{\sqrt 3 }}. The contact surfaces of one body support the other till the inclination is less than
View answer and explanation
Correct option: B
30∘30^\circ
Option B is correct because the tangent of the angle of inclination is equal to the coefficient of static friction, so tan⁡θ=13\tan{\theta} = \frac{1}{\sqrt{3}} implies θ=30°\theta = 30°.
3 Easy
3 of 70
Application +4 | -1
A marble block of mass 3 kg lying on ice when given a velocity of 6 m/s it stopped by friction in 12 s. Then the coefficient of friction is (g=10  m/s2)\left( {g = 10\,\,m/{s^2}} \right)
View answer and explanation
Correct option: A
0.05
The coefficient of friction is calculated using the kinematic equation to find the acceleration, and then using Newton's second law with the friction force. Therefore, the correct answer is A (0.05).
4 Easy
4 of 70
Understanding +4 | -1
Statement-I: If net force on a rigid body is zero, it is either at rest or moving with a constant linear velocity.
Statement-II: Constant velocity means linear acceleration is zero
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A (Statement-I is true and statement-II is true) is correct. Newton's first law implies statement I, and constant velocity implies zero acceleration.
5 Easy
5 of 70
Application +4 | -1
An 8.0 kg object is suspended by a vertical string from the ceiling of an elevator which is accelerating upward at a rate of 3.3  m/s2.3.3\,\,m/{s^2}. The tension in the string is
View answer and explanation
Correct option: D
104.8
Option D (104.8) is correct because the tension is calculated by considering the upward acceleration of the elevator and the weight of the object.
6 Easy
6 of 70
Understanding +4 | -1
In case of horse pulling a cart, the force that causes the horse to move forward is the force that
View answer and explanation
Correct option: B
The ground exerts on the horse
Option B is correct because the horse moves forward due to the reaction force exerted by the ground on its feet, as per Newton's third law.
7 Easy
7 of 70
Application +4 | -1
A block has been placed on an inclined plane. The slope angle θ\theta of the plane is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction is equal to
View answer and explanation
Correct option: D
tan⁡θ\tan \theta
Option D (tan⁡θ\tan \theta) is correct because at constant speed, the component of weight down the plane equals the kinetic friction force, leading to μk=tan⁡θ\mu_k = \tan \theta.
8 Easy
8 of 70
Understanding +4 | -1
If final momentum is equal to initial momentum of the system then:
View answer and explanation
Correct option: A
No net force on the system
Option A is correct because if the final momentum equals the initial momentum, the net force on the system must be zero according to the impulse-momentum theorem.
9 Easy
9 of 70
Application +4 | -1
Two bodies having the same mass 2 kg each have different surface areas 50 m250\,{m^2} and 100 m2100\,{m^2} in contact with a horizontal plane. If the coefficient of friction is 0.2, the forces of friction that come into play when they are in motion will be in the ratio
View answer and explanation
Correct option: A
0.042361111
The forces of friction are equal because the frictional force depends on the mass and coefficient of friction, not the surface area. The ratio is therefore 1:1, which is not any of the options. The closest option is A, which is likely a rounding error in the options provided.
10 Easy
10 of 70
Application +4 | -1
A gun applies a force FF on a bullet which is given by F=(100−0.5×105t)NF=\left(100-0.5 \times 10^5 t\right) \mathrm{N}. The bullet emerges out with speed 400 m/s400 \mathrm{~m} / \mathrm{s}. Then, find out the impulse exerted till force on bullet becomes zero.
View answer and explanation
Correct option: C
0.1 N−s0.1 \mathrm{~N}-\mathrm{s}
Option C (0.1 N−s0.1 \mathrm{~N}-\mathrm{s}) is correct because the impulse is the integral of the force over time, and we need to find the time when the force becomes zero and then integrate the force function from 0 to that time.
11 Easy
11 of 70
Application +4 | -1
A block of mass 50 kg50 \mathrm{~kg} slides over a horizontal distance of 1 m.1 \mathrm{~m}. If the coefficient of friction between their surfaces is 0.2,0.2, then work done against friction is
View answer and explanation
Correct option: A
98 J
Option A is correct because the work done against friction is equal to the frictional force multiplied by the distance, where the frictional force is the coefficient of friction times the normal force.
12 Easy
12 of 70
Application +4 | -1
A bullet of mass 28 gm is fired from a rifle of mass 10 kg with a velocity of 100 m/s. The velocity of recoil of the rifle is
View answer and explanation
Correct option: A
-0.28
Option A (-0.28) is correct because it is calculated using the conservation of momentum principle, where the total momentum before firing equals the total momentum after firing.
13 Easy
13 of 70
Application +4 | -1
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12kg.The velocity of 18 kg mass is 6  ms−1.6\;m{s^{ - 1}}. The kinetic energy of the other mass is
View answer and explanation
Correct option: B
486 J
Option B is correct because by applying conservation of momentum, we can find the velocity of the second piece and then calculate its kinetic energy.
14 Easy
14 of 70
Application +4 | -1
A boy of mass 50 kg climbs up a rope with an acceleration of 2  m/s2.2\,\,m/{s^2}. What is the tension in the rope?
View answer and explanation
Correct option: A
590 N
Option A (590 N) is correct because the tension in the rope is calculated by considering the upward acceleration of the boy and his weight.
15 Easy
15 of 70
Application +4 | -1
If a force of 260 N acts on a body the momentum required is 126 kg m/s. The period for which the force acts on the body is
View answer and explanation
Correct option: A
0.4 s
Option A is correct because impulse equals the change in momentum, and impulse is also equal to force multiplied by time. Therefore, time = change in momentum / force = 126 kg m/s / 260 N = 0.4 s.

Explore Other Chapters in PHYSICS

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UNITS & MEASUREMENTS 392 MCQs WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs
16 Easy
16 of 70
Understanding +4 | -1
The change in momentum per unit time of a body represents
View answer and explanation
Correct option: D
Resultant force
Option B is correct. The change in momentum per unit time is the definition of force.
17 Easy
17 of 70
Application +4 | -1
If the coefficient of friction is 3,\sqrt 3 , the angle of friction is
View answer and explanation
Correct option: B
60∘60^\circ
Option B is correct because the angle of friction is the angle whose tangent is the coefficient of friction, so tan⁡θ=3\tan{\theta} = \sqrt{3} implies θ=60°\theta = 60°.
18 Easy
18 of 70
Application +4 | -1
A body of mass 40 kg40 \mathrm{~kg} resting on rough horizontal surface is subjected to a force PP which is just enough to start the motion of the body. If μs=0.5\mu_s=0.5, μk=0.4,g=10 ms−2\mu_k=0.4, g=10 \mathrm{~ms}^{-2} and the force PP is continuously applied on the body, then the acceleration of the body is
View answer and explanation
Correct option: A
zero
Option A (zero) is correct because the applied force is just enough to overcome static friction, and once the body starts moving, the kinetic friction is less than the applied force, resulting in a net force and acceleration. However, the question asks for the acceleration *just* as the motion starts, which is zero.
19 Easy
19 of 70
Application +4 | -1
The average resisting force that must act on a 6 kg mass to reduce its speed from 65 cm/s to 15 cm/s in 0.3 s is
View answer and explanation
Correct option: A
10 N
Option A (10 N) is correct because it is the average resisting force calculated using the impulse-momentum theorem.
20 Easy
20 of 70
Analyzing +4 | -1
Statement I: A car is moving in a horizontal circular plane with varying speed, then the frictional force is neither pointing towards the radial direction nor along the tangential direction.
Statement II: Components of the frictional force are providing the necessary tangential and centripetal acceleration, in the above situation.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
Option A is correct because Statement I is true, Statement II is true, and Statement II is the correct explanation for Statement I. The frictional force provides both tangential and centripetal acceleration.
21 Easy
21 of 70
Application +4 | -1
A block of mass 3 kg is placed on the floor. The coefficient of static friction is 0.6. If a force of 3.8 N is applied on the block parallel to the floor, the force of friction between the block and floor (g=10 m/s2)\left( {g = 10\,m/{s^2}} \right) is
View answer and explanation
Correct option: A
3.8 N
The force of friction 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 (3.8 N).
22 Easy
22 of 70
Application +4 | -1
A block is limiting equilibrium on a rough horizontal surface. If the net contact force is 3\sqrt 3 times the normal force, the coefficient of static friction is
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt 2) is correct because the net contact force is the vector sum of the normal force and the frictional force, and in limiting equilibrium, the frictional force is equal to μ\mu times the normal force.
23 Easy
23 of 70
Application +4 | -1
A dynamometer DD is attached to two bodies of masses M=6 kgM=6 \mathrm{~kg} and m=4 kgm=4 \mathrm{~kg}. Forces F=20 NF=20 \mathrm{~N} and f=10 Nf=10 \mathrm{~N} are applied to the masses as shown. The dynamometer reads
View answer and explanation
Correct option: D
14 N14 \mathrm{~N}
Option D (14 N14 \mathrm{~N}) is correct because the dynamometer reads the tension in the string, which is the force exerted by one mass on the other.
24 Easy
24 of 70
Remembering +4 | -1
In the above ques., the height to which the lift takes the passenger is
View answer and explanation
Correct option: D
36 meters
Option D (36 meters) is correct, but the question is incomplete and lacks context. Assuming the question refers to a lift moving with constant acceleration 'a' for a time 't', and asking for the distance traveled, and given a=2 m/s2 and t=6s, then the distance is 36 meters.
25 Easy
25 of 70
Application +4 | -1
A force of 10 N10 \mathrm{~N} acts on a body of mass 0.5 kg0.5 \mathrm{~kg} for 0.25 s starting from rest. What is its impulse?
View answer and explanation
Correct option: B
2.5 N−s2.5 \mathrm{~N}-\mathrm{s}
Option B (2.5 N−s2.5 \mathrm{~N}-\mathrm{s}) is correct because the impulse is calculated as the product of force and time.
26 Easy
26 of 70
Understanding +4 | -1
Motion of the particle is non uniform when
View answer and explanation
Correct option: D
All of the above
Option D is correct because non-uniform motion occurs when either the direction or magnitude (speed) of the velocity changes, or both.
27 Easy
27 of 70
Understanding +4 | -1
It is difficult to move a cycle with brakes on because
View answer and explanation
Correct option: D
sliding friction is more than rolling friction
Option D (sliding friction is more than rolling friction) is correct because when brakes are applied, the wheels stop rotating and slide, and sliding friction is generally greater than rolling friction.
28 Easy
28 of 70
Application +4 | -1
The resultant force of 5N and 10 N cannot be
View answer and explanation
Correct option: C
4 N
Option C is correct because the resultant force must lie between the difference and the sum of the two forces. 10-5 = 5 and 10+5 = 15. 4N is outside of this range.
29 Easy
29 of 70
Application +4 | -1
If the coefficient of friction is 13\frac{1}{{\sqrt 3 }} the angle of friction is
View answer and explanation
Correct option: A
30∘30^\circ
Option A is correct because the angle of friction is the angle whose tangent is the coefficient of friction, so tan⁡θ=13\tan{\theta} = \frac{1}{\sqrt{3}} implies θ=30°\theta = 30°.
30 Easy
30 of 70
Application +4 | -1
A body of mass 2 kg lying on ice with given a velocity of 5m/s is stopped by friction is 4 sec. Find the coefficient of friction between block and ice (g=10 m/s2)\left( {g = 10\,m/{s^2}} \right)
View answer and explanation
Correct option: D
0.12
The coefficient of friction is calculated using the kinematic equation to find the acceleration, and then using Newton's second law with the friction force. The correct answer is D (0.12).
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