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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 31-60 of 496 questions
31 Medium
31 of 496
Application +4 | -1
A force (4i^+j^−2k^)N\left( {4\hat i + \hat j - 2\hat k} \right)N acting on a body maintains its velocity at (2i^+2j^+3k^)ms−1\left( {2\hat i + 2\hat j + 3\hat k} \right)m{s^{ - 1}}. The power exerted is:-
View answer and explanation
Correct option: A
4 W
Option A (4 W) is correct because power is the dot product of force and velocity, and the calculation yields 4 Watts.
32 Medium
32 of 496
Understanding +4 | -1
The angle between the linear momentum and angular momentum of particle moving along a circle is
View answer and explanation
Correct option: C
90∘90^{\circ}
Option C is correct because the angular momentum is perpendicular to the plane of the circle, while the linear momentum is tangent to the circle.
33 Medium
33 of 496
Understanding +4 | -1
A bullet is fired from a rifle. If the riffle recoils freely, then the kinetic energy of the rifle is
View answer and explanation
Correct option: A
Less than that of the bullet
Option A (Less than that of the bullet) is correct because the kinetic energy is inversely proportional to the mass when momentum is constant, and the rifle has a much larger mass than the bullet.
34 Medium
34 of 496
Understanding +4 | -1
One object of mass 20 kg20 \mathrm{~kg} is moving with speed 10 ms−110 \mathrm{~ms}^{-1} in west direction and another object of mass 10 kg10 \mathrm{~kg} is moving with 15 ms−115 \mathrm{~ms}^{-1} in north direction. Both collide and stick together. Choose the correct alternative.
View answer and explanation
Correct option: C
Their momentum is conserved as it is inelastic collision.
Option C (Their momentum is conserved as it is inelastic collision) is correct because in an inelastic collision, kinetic energy is not conserved, but momentum is always conserved.
35 Medium
35 of 496
Application +4 | -1
The maximum and minimum tensions in the string whirling in a circle of radius 2.5 m are in the ratio 5:3, then its velocity is
View answer and explanation
Correct option: A
98    ms−1\sqrt {98} \;\;\mathrm{m}{\mathrm{s}^{ - 1}}
Option A (98  ms−1\sqrt {98} \;\mathrm{m}{\mathrm{s}^{ - 1}}) is correct because it is the velocity calculated from the ratio of maximum to minimum tension in the string.
36 Medium
36 of 496
Analyzing +4 | -1
Assertion : In an elastic collision of two bodies, the momentum and energy of each body is conserved.
Reason : If two bodies stick to each other, after colliding, the collision is said to be perfectly elastic.
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option D (If both assertion and reason are false) is correct because in an elastic collision, momentum and kinetic energy of the *system* is conserved, not necessarily each individual body. Also, if bodies stick together after collision, it's perfectly *inelastic*.
37 Medium
37 of 496
Understanding +4 | -1
If ar and at{a_r}\,and\,{a_t} represent radial and tangential accelerations, the motion of a particle will be uniformly circular if
View answer and explanation
Correct option: C
ar≠0  but  at=0{a_r} \ne 0\mathrm{\;but\;}{a_t} = 0
Option C (ar≠0  but  at=0{a_r} \ne 0\mathrm{\;but\;}{a_t} = 0) is correct because uniform circular motion requires a non-zero radial acceleration and zero tangential acceleration.
38 Medium
38 of 496
Analyzing +4 | -1
A particle moves along a curve of unknown shape but magnitude of force F⃗\vec{F} is constant and always acts along the tangent to the curve. Then,
View answer and explanation
Correct option: C
F⃗\vec{F} must be non-conservative
Option C (F⃗\vec{{F}} must be non-conservative) is correct because if the force always acts along the tangent, the work done depends on the path taken, which is a characteristic of non-conservative forces.
39 Medium
39 of 496
Understanding +4 | -1
Statement-A: Energy is the cause for doing work and work is the effeet
Statement-B: Energy is a scalar but power is a vector quantity.
View answer and explanation
Correct option: C
A is correct and B is wrong
Option C is correct because energy is the cause for doing work, making work the effect, and both energy and power are scalar quantities.
40 Medium
40 of 496
Application +4 | -1
A block of mass m=0.5 kg{m}=0.5 \mathrm{~kg} gently placed on a conveyor belt moving with velocity of 4 m/s4 \mathrm{~m} / \mathrm{s}. The coefficient of friction between the conveyor belt and bag is 0.2.0.2. Initially the block slips on the conveyor belt before it stops due to friction. The work done by kinetic friction on the block is [g=10 m/s2]\left[{g}=10 \mathrm{~m} / \mathrm{s}^2\right]
View answer and explanation
Correct option: A
4 J4 \mathrm{~J}
Option A (4 J4 \mathrm{~J}) is correct because it is the work done by the kinetic friction on the block as it comes to rest relative to the belt.
41 Medium
41 of 496
Application +4 | -1
The turbine pits at the Niagra falls are 50 m50 \mathrm{~m} deep. The average horse power developed is 500. If the efficiency of the generator is 85%85 \%, how much water passes through the turbines per minute? Take g=10 ms−2g=10 \mathrm{~ms}^{-2}.
View answer and explanation
Correct option: A
5.39×105 kg5.39 \times 10^5 \mathrm{~kg}
Option A (5.39×105 kg5.39 \times 10^5 \text{ kg}) is correct because it correctly calculates the mass of water using the power developed and the efficiency of the generator.
42 Medium
42 of 496
Application +4 | -1
Two substances have masses 90 kg90 \mathrm{~kg} and 160 kg160 \mathrm{~kg} respectively. They possess equal kinetic energy. The ratio of their momenta will be
View answer and explanation
Correct option: A
3/43 / 4
Option A (3/43 / 4) is correct because the ratio of momenta is the square root of the ratio of masses when kinetic energies are equal.
43 Medium
43 of 496
Application +4 | -1
A force of constant magnitude F=10 NF=10 \mathrm{~N} acts on a particle moving in a plane such that it is perpendicular to the velocity v⃗(∣v⃗∣=v=5 m/s)\vec{v}(|\vec{v}|=v=5 \mathrm{~m} / \mathrm{s}) of the body, and the force is always directed towards a fixed point. Then the angle (in radian) turned by the velocity vector of the particle as it covers a distance S=10 mS=10 \mathrm{~m} is (take: mass of the particle as m=2 kg)m=2 \mathrm{~kg})
View answer and explanation
Correct option: A
2
Option A (2) is correct because the angle turned by the velocity vector is calculated using the formula derived from circular motion and the given conditions.
44 Medium
44 of 496
Understanding +4 | -1
A bullet fired from a gun can pierce a target due to its:
View answer and explanation
Correct option: C
Kinetic energy
Option C (Kinetic energy) is correct because the bullet's ability to pierce a target is directly related to its kinetic energy, which is the energy of motion.
45 Medium
45 of 496
Application +4 | -1
A groove is in the form of a broken line ABCA B C and the position vectors of the three points are respectively 2i^−3j^+2k^,3i^+2j^−k^2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} and i^+j^+k^\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}. A force of magnitude 24324 \sqrt{3} acts on a particle of unit mass kept at the point AA and moves it along the groove to the point CC. If the line of action of the force is parallel to the vector i^+2j^+k^\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} all along, the number of units of work done by the force is
View answer and explanation
Correct option: C
72272 \sqrt{2}
The correct answer is C (72272 \sqrt{2}) because it is the result of calculating the work done by the force using the dot product of the force and displacement vectors.
46 Medium
46 of 496
Understanding +4 | -1
Two moving particles collide and stick together on a smooth horizontal surface. The following statements are given below

(a) Mechanical energy is conserved.

(b) Total energy is conserved.

(c) Work done by the system is positive

(d) Work done by the system is negative.

Choose the correct option.
View answer and explanation
Correct option: D
(b) and (d) are correct.
Option D ((b) and (d) are correct) is correct because total energy is always conserved, but mechanical energy is not conserved in an inelastic collision, and the work done by the system is negative.
47 Medium
47 of 496
Application +4 | -1
A 2kg mass is rotating on a circular path of radius 0.8 m with angular velocity of 44 rad / sec. If radius of path becomes 1 M, then the value of angular velocity will be [rad / sec]
View answer and explanation
Correct option: A
28.1628.16
Option A (28.16) is correct because the angular velocity is calculated using the conservation of angular momentum as the radius changes.
48 Medium
48 of 496
Analyzing +4 | -1
The displacement x in metre of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in second by the equation t=x+3t = \sqrt x + 3, the work done by the force (in joule) in first six seconds is
View answer and explanation
Correct option: B
Zero
Option B (Zero) is correct because the initial and final velocities are equal in magnitude, resulting in no net change in kinetic energy and thus zero work done.
49 Medium
49 of 496
Application +4 | -1
A mass of 1 kg1 \mathrm{~kg} is acted upon by a single force F=(4i^+4j^)N{F}=(4 \hat{{i}}+4 \hat{j}) \mathrm{N}. Under this force, it is displaced from (0,0)(0,0) to (1 m,1 m)(1 \mathrm{~m}, 1 \mathrm{~m}). If initially, the speed of the particle was 2 ms−12 \mathrm{~ms}^{-1}, its final speed should be
View answer and explanation
Correct option: B
4.5 ms−14.5 \mathrm{~ms}^{-1}
Option B (4.5 ms−14.5 \mathrm{~ms}^{-1}) is correct because it is derived from the work-energy theorem, considering the force and displacement.
50 Medium
50 of 496
Application +4 | -1
If nn bullets each of mass mm are fired with a velocity vv per second from a machine gun, the force required to hold the gun in position is
View answer and explanation
Correct option: D
mnvm n v
Option D (mnvm n v) is correct because the force is the rate of change of momentum, which is the number of bullets per second times the momentum of each bullet.
51 Medium
51 of 496
Application +4 | -1
A body of mass 6 kg is acted upon by a force which causes a displacement in it given by xt24      x\frac{{{t^2}}}{{4\;}}\;\; metre where t is the time in second. The work done by the force in 2s is
View answer and explanation
Correct option: D
3 J
Option D (3 J) is correct because the work done is equal to the change in kinetic energy, calculated using the given displacement function.
52 Medium
52 of 496
Understanding +4 | -1
A ball of mass mm collides with a wall with speed vv
and rebounds on the same line with the same
speed. If mass of the wall is taken infinite at rest,
then work done by the ball on the wall is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the wall is infinitely massive and at rest, so it does not move during the collision, and therefore no work is done on the wall.
53 Medium
53 of 496
Application +4 | -1
A ball of mass 0.1 kg is suspended by a string. It is displaced through an angle of 600{60^0} and left. When the ball passes through the mean position, the tension in the string is
View answer and explanation
Correct option: B
1.96 N
Option B (1.96 N) is correct because the tension at the mean position is calculated using the conservation of energy and centripetal force.
54 Medium
54 of 496
Application +4 | -1
An engine accelerates a car of mass 800 kg to a speed of 72  kmh−172\mathrm{\;km}{\mathrm{h}^{ - 1}}. If the frictional force is 10 M per ton, the power developed by the engine is
View answer and explanation
Correct option: D
5 kW
Option D (5 kW) is correct because the power developed by the engine is calculated using the kinetic energy and the time taken to reach the final speed, considering the frictional force.
55 Medium
55 of 496
Application +4 | -1
The potential energy of a certain spring when stretched through a distance ‘S’ is 10 joule. The amount of work (in joule) that must be done on this spring to stretch it through an additional distance ‘S’ will be
View answer and explanation
Correct option: A
30
Option A (30) is correct because the work required to stretch the spring an additional distance S is three times the initial potential energy, resulting in 30 J.
56 Medium
56 of 496
Analyzing +4 | -1
Assertion: The gain in potential energy of an object of mass mm raised to height equal to the radius of earth is 12 mgR\frac{1}{2} \mathrm{~mgR}.
Reason: Kinetic energy at surface 12mv2=\frac{1}{2} m v^2= P.E at the top and at the top, v=gRv=\sqrt{g R}.

∴PE=12 mgR\therefore \mathrm{PE}=\frac{1}{2} \mathrm{~mgR}
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option C is correct because the assertion is true, but the reason is false; the kinetic energy at the surface is not directly related to the potential energy at the top in this way.
57 Medium
57 of 496
Analyzing +4 | -1
Assertion (A): When a body moving with certain velocity collides head on with another body of same mass at rest, the collision being perfectly elastic head on collision, 100%100 \% of its K.E is transferred to the later
Reason (R): Both momentum and KE are conserved in the case of perfectly elastic collision
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is correct explanation of Assertion.
Option A (If both Assertion and Reason are true and Reason is correct explanation of Assertion) is correct because both momentum and kinetic energy are conserved in a perfectly elastic collision, leading to maximum energy transfer when masses are equal.
58 Medium
58 of 496
Application +4 | -1
In a nuclear reactor, the fuel is consumed at the rate of 1 mg/s. The power generated in kilowatt is
View answer and explanation
Correct option: B
9×1079 \times {10^7}
The correct answer is B (9×1079 \times {10^7}) because it is the power generated calculated using Einstein's mass-energy equivalence.
59 Medium
59 of 496
Application +4 | -1
A gun fires bullet each of mass 1 g with velocity of 10  ms−110\;\mathrm{m}{\mathrm{s}^{ - 1}} by exerting a constant force of 5 g weight. Then the number of bullets fired per second is (Take g=10  ms−2)g = 10\;\mathrm{m}{\mathrm{s}^{ - 2}})
View answer and explanation
Correct option: B
5
Option B (5) is correct because the number of bullets fired per second is calculated using the formula derived from the rate of change of linear momentum.
60 Medium
60 of 496
Application +4 | -1
A force of (7i^+6k^)\left( {7\hat i + 6\hat k} \right) N makes a body move on a rough plane with a velocity of (3j^+4k^)ms−1\left( {3\hat j + 4\hat k} \right)m{s^{ - 1}}. Calculate the power in watt
View answer and explanation
Correct option: A
24
Option A (24) is correct because the power is calculated as the dot product of the force and velocity vectors, resulting in 24 watts.
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