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JEE • PHYSICS • WORK-ENERGY-POWER

Work-energy-power practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 481-496 of 496 questions
481 Medium
481 of 496
Understanding +4 | -1
Two observers are situated in different inertial reference frames. Then
View answer and explanation
Correct option: A
The momentum of a body by both observers may be same
Option A (The momentum of a body by both observers may be same) is correct because if both observers are at rest relative to each other, the momentum of a body they measure can be the same.
482 Medium
482 of 496
Understanding +4 | -1
A body revolving in a circle with uniform speed possesses
View answer and explanation
Correct option: A
normal acceleration.
Option A (normal acceleration) is correct because uniform circular motion involves acceleration directed towards the center.
483 Medium
483 of 496
Application +4 | -1
A particle of mass 100 g tied to a string is rotated along circle of radius 0.5 m. The breaking tension of string is 10 N. The maximum speed with which particle can rotated without breaking the string is
View answer and explanation
Correct option: D
7.07 ms−27.07\,\mathrm{m}{\mathrm{s}^{ - 2}}
Option D (7.07 ms−27.07\,\mathrm{m}{\mathrm{s}^{ - 2}}) is correct because the maximum speed is calculated using the centripetal force formula and the breaking tension of the string.
484 Medium
484 of 496
Understanding +4 | -1
Assertion: Work done by centripetal force is zero.
Reason: Centripetal force acts perpendicular to the displacement.
View answer and explanation
Correct option: A
If both Assertion & reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & reason are True & the Reason is a correct explanation of the Assertion) is correct because centripetal force is always perpendicular to the displacement, hence the work done is zero.
485 Medium
485 of 496
Understanding +4 | -1
A body in equilibrium may not have
View answer and explanation
Correct option: C
Acceleration
Option C (Acceleration) is correct because a body in equilibrium has a net force of zero, which implies zero acceleration according to Newton's first law.
486 Medium
486 of 496
Analyzing +4 | -1
Which of the following statements is wrong?
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Correct option: C
If two protons are brought towards each other the PE of the system decreases.
Option C (If two protons are brought towards each other the PE of the system decreases) is wrong because bringing two protons closer increases the potential energy due to electrostatic repulsion.
487 Medium
487 of 496
Analyzing +4 | -1
A body of mass m having an initial velocity v, makes head on collision with a stationary body of mass M. After the collision, the body of mass m comes to rest and only the body having mass M moves. This will happen only when
View answer and explanation
Correct option: C
m=Mm = M
Option C is correct because the velocity exchange only happens when the masses are equal, ensuring momentum and kinetic energy are conserved in the specific way described.
488 Medium
488 of 496
Analyzing +4 | -1
Statement-I: If both the speed of a body and radius of its circular path are doubled, then centripetal force also gets doubled.
Statement-II: Centripetal force is directly proportional to both speed of a body and radius of circular path.
View answer and explanation
Correct option: B
Statement-I is true, statement-II is false
Option B is correct because if both speed and radius are doubled, the centripetal force doubles, but the statement that centripetal force is directly proportional to speed is false.
489 Medium
489 of 496
Remembering +4 | -1
Work done by a conservative force on a system is equal to
View answer and explanation
Correct option: B
The change in potential energy of the system
Option D (None of the above) is technically more accurate, since the work done by a conservative force is the *negative* of the change in potential energy, or the *negative* change in potential energy of the system.
490 Medium
490 of 496
Analyzing +4 | -1
Assertion (A): Law of conservation momentum is a consequence of Newton's III law
Reason(R): Action, Reaction pairs are internal forces in collisions
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Option A is correct because the law of conservation of momentum is indeed a consequence of Newton's III law, and action-reaction pairs are internal forces in collisions, providing the necessary framework for momentum conservation.
491 Medium
491 of 496
Understanding +4 | -1
If v,P\mathrm{v}, \mathrm{P} and K\mathrm{K} denote the velocity, momentum and kinetic energy of a particle then
View answer and explanation
Correct option: B
P=dK/dt\mathrm{P}=\mathrm{dK} / \mathrm{dt}
Option A (P=dK/dv\mathrm{P}=\mathrm{dK} / \mathrm{dv}) is incorrect. Option B (P=dK/dt\mathrm{P}=\mathrm{dK} / \mathrm{dt}) is also incorrect. The correct answer is not among the options provided. However, if we consider the relation between Kinetic Energy (K) and momentum (P), we have K=P22mK = \frac{{P^2}}{{2m}}. Therefore, P=2mKP = \sqrt{{2mK}}. Differentiating K with respect to v, we get dKdv=mv=P\frac{{dK}}{{dv}} = mv = P. So, the closest correct answer would be A, but it should be dKdv=P\frac{{dK}}{{dv}} = P.
492 Medium
492 of 496
Analyzing +4 | -1
Assertion: A force is needed to move an object uniformly in a circle.

Reason: Acceleration is zero during uniform circular motion.
View answer and explanation
Correct option: C
Assertion is true, but Reason is false
Option C is correct because a force is indeed needed to maintain uniform circular motion due to the constant change in direction (centripetal acceleration), but the reason is false because acceleration is not zero in uniform circular motion.
493 Medium
493 of 496
Analyzing +4 | -1
A) When a ball is dropped on to the ground from certain height, it comes to rest after few bounces, loosing all of its energy. It is an example of the violation of conservation of energy.
B) In uniform circular motion, the acceleration of the body is constant.
View answer and explanation
Correct option: D
Both A and B are false
Option D (Both A and B are false) is correct because energy is conserved in the ball bouncing scenario (though converted to other forms), and acceleration in uniform circular motion is not constant due to changing direction.
494 Medium
494 of 496
Application +4 | -1
A 2 kg particle moves along x-axis such that its position (x)(x) varies with time (t)(t) as x=2t2+3.x = 2{t^2} + 3. During the initial 5 seconds, the work done by all the forces acting on the particle is
View answer and explanation
Correct option: A
400 J
Option A (400 J) is correct because the work done is equal to the change in kinetic energy, calculated using the velocities at the initial and final times.
495 Medium
495 of 496
Application +4 | -1
A particle of mass 1 g1 \mathrm{~g} moving with a velocity u⃗1=(3i^−2j^)ms−1\vec{u}_1=(3 \hat{i}-2 \hat{j}) \mathrm{ms}^{-1} experiences a perfectly inelastic collision with another particle of mass 2 g2 \mathrm{~g} moving with a velocity u⃗2=(4j^−6k^)ms−1\vec{u}_2=(4 \hat{j}-6 \hat{k}) \mathrm{ms}^{-1}. The velocity (in ms−1\mathrm{m} \mathrm{s}^{-1}) of the combined particle is
View answer and explanation
Correct option: A
i^+2j^−4k^\hat{i}+2 \hat{j}-4 \hat{k}
Option C (πa2μ0I2bωsin⁡ωt\frac{\pi a^2 \mu_0 I}{2 b} \omega \sin \omega t) is correct because it represents the induced emf in the smaller loop as a function of time, considering the changing magnetic flux due to rotation.
496 Medium
496 of 496
Application +4 | -1
Two perfectly elastic objects A and B of identical mass are moving with velocities 15ms−1\mathrm{m}{\mathrm{s}^{ - 1}} and 10ms−1\mathrm{m}{\mathrm{s}^{ - 1}} respectively, collide along the direction of line joining them. Their velocities after collision are respectively
View answer and explanation
Correct option: A
10ms−1, 15ms−1\mathrm{10m}{\mathrm{s}^{ - 1}},\,15m{s^{ - 1}}
Option A (10ms−1, 15ms−1\mathrm{10m}{\mathrm{s}^{ - 1}},\,15m{s^{ - 1}}) is correct because in a perfectly elastic collision with identical masses, the velocities are exchanged.
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