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NEET • PHYSICS • WORK-ENERGY-POWER

Work-energy-power practice · Mixed difficulty

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Showing 331-347 of 347 questions
331 Easy
331 of 347
Application +4 | -1
A particle moves in a circle of radius 5 cm with constant speed and time period 0.2  π0.2\,\,\pi seconds. The acceleration of the particle is [in m/s2m/{s^2}]
View answer and explanation
Correct option: A
5
Option A (5) is correct because the centripetal acceleration is calculated using a=v2/ra = v^2/r or a=ω2ra = \omega^2 r, where ω=2π/T\omega = 2\pi/T.
332 Medium
332 of 347
Application +4 | -1
A particle of mass 2 kg starts moving in a straight line with an initial velocity of 2  ms−12\mathrm{\;m}{\mathrm{s}^{ - 1}} at a constant acceleration of 2  ms−22\mathrm{\;m}{\mathrm{s}^{ - 2}}. Then rate of change of kinetic energy
View answer and explanation
Correct option: A
Is four times the velocity at any moment
Option A (Is four times the velocity at any moment) is correct because the rate of change of kinetic energy is equal to the power, which is force times velocity, and the force is related to the acceleration.
333 Medium
333 of 347
Application +4 | -1
A machine gun mounted on a 2000 kg car on a horizontal frictionless surface fires 10 bullets per second. If 10 g be the mass of each bullet and 500 ms−1\mathrm{m}{\mathrm{s}^{ - 1}}, the velocity of each bullet, then the acceleration of the car will be
View answer and explanation
Correct option: C
140ms−2\frac{1}{{40}}\mathrm{m}{\mathrm{s}^{ - 2}}
Option C (140ms−2\frac{1}{{40}}\mathrm{m}{\mathrm{s}^{ - 2}}) is correct because the acceleration is calculated by dividing the force exerted by the bullets on the car by the mass of the car.
334 Easy
334 of 347
Application +4 | -1
A water pump lifts water at the rate of 5 litre per second through an average height of 8 m8 \mathrm{~m} from a well of depth 12 m12 \mathrm{~m}. Its power will be:
View answer and explanation
Correct option: B
980 watt
Option B (980 watt) is correct because the power is calculated by multiplying the mass flow rate, gravity, and the total height.
335 Medium
335 of 347
Analyzing +4 | -1
A certain body of mass MM moves under the action of a conservative force with potential energy UU given by U=Kxx2+a2(x→U=\frac{K x}{x^2+a^2}(x \rightarrow distance, a→a \rightarrow amplitude)) In this equation the units of K{K} will be equal to that of:
View answer and explanation
Correct option: C
Joule-metre
Option C (Joule-metre) is correct because the potential energy U has units of Joules, and the expression Kx/(x2+a2) must also have units of Joules. Therefore, K must have units of Joule-meter.
336 Easy
336 of 347
Application +4 | -1
A ball falls from a height of 16 m16 {~m} on a floor and rebounds to a height of 4 m4 {~m}. The coefficient of restitution for collision is :-
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (1/2) is correct because the coefficient of restitution is the square root of the ratio of the rebound height to the initial height.
337 Easy
337 of 347
Application +4 | -1
A uniform force of (3i^+j^)\left( {3\hat i + \hat j} \right) newton acts on a particle of mass 2 kg. The particle is displaced from position (2i^+k^)m\left( {2\hat i + \hat k} \right)m to position (4i^+3j^−k^)m\left( {4\hat i + 3\hat j - \hat k} \right)m. The work done by the force on the particle is
View answer and explanation
Correct option: D
9 J
Option D (9 J) is correct because the work done is the dot product of the force and the displacement vector.
338 Medium
338 of 347
Analyzing +4 | -1
Two blocks of masses m1 and m2{m_1}\,and\,{m_2} are connected by a massless spring and placed at smooth surface. The spring initially stretched and released. Then
View answer and explanation
Correct option: D
Both (2) and (3) are correct.
Option D (Both (2) and (3) are correct) is correct because the magnitude of momentum is conserved, and the mechanical energy of the system remains constant as there are no external forces.
339 Medium
339 of 347
Application +4 | -1
An iron nail is dropped from a height h from the level of a sand bed. If it penetrates through a distance x in the sand before coming to rest, then average force exerted by the sand on nail is
View answer and explanation
Correct option: A
mg(hx+1)m\mathrm{g}\left( {\frac{h}{x} + 1} \right)
Option A (mg(hx+1)m\mathrm{g}\left( {\frac{h}{x} + 1} \right)) is correct because it is derived from the work-energy theorem, considering both the gravitational potential energy and the work done by the sand.
340 Medium
340 of 347
Application +4 | -1
A force of 10i^−3j^+6k^10\hat i - 3\hat j + 6\hat k N acts on a body of mass 100 g and displaces it from 6i^+5j^−3k^6\hat i + 5\hat j - 3\hat k m to 10i^−2j^+7k^10\hat i - 2\hat j + 7\hat k m. The work done is
View answer and explanation
Correct option: B
121 J
Option B (121 J) is correct because the work done is the dot product of the force and displacement vectors.
341 Medium
341 of 347
Analyzing +4 | -1
Statement-1: A block of mass ' MM ' is placed on a moving railroad car moving with velocity ' vv '. The kinetic energy of the block is 12Mv2\frac{1}{2} M v^2 in all frames.

Statement-2: The kinetic energy of a system or body is different in different frames.
View answer and explanation
Correct option: D
Statement-1 is false, Statement- 2 is true.
Option D (Statement-1 is false, Statement- 2 is true) is correct. Kinetic energy depends on the frame of reference, so Statement-1 is false, and Statement-2 correctly states that kinetic energy varies with different frames of reference.
342 Medium
342 of 347
Application +4 | -1
Two springs A and B are identical but A is harder than B(kA>kB)B({k_A} > {k_B}). Let WA and WB{W_A}\,and\,{W_B} represent the work done when the springs are stretched through the same distance and WA′ and WB′{W'_A}\,and\,{W'_B} are the work done when these are stretched by equal forces, then which of the following is true
View answer and explanation
Correct option: B
WA>WB and WA′ < WB′{W_A} > {W_B}\,and\,{W'_A}\, < \,{W'_B}
Option B is correct because with the same displacement, a stiffer spring (A) requires more work, and with the same force, a less stiff spring (B) stretches more, requiring more work.
343 Medium
343 of 347
Application +4 | -1
The string of a pendulum of length l is displaced through 900{90^0} from the vertical and released. Then the minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is
View answer and explanation
Correct option: C
3 mg
Option C (3 mg) is correct because at the mean position, the tension in the string is the sum of the weight and the centripetal force.
344 Medium
344 of 347
Remembering +4 | -1
One kilowatt hour is a unit of:
View answer and explanation
Correct option: A
Energy
The correct answer is A. One kilowatt-hour (kWh) is a unit of energy, specifically the energy consumed by a 1 kW device operating for 1 hour.
345 Medium
345 of 347
Understanding +4 | -1
One man takes 1 minute to raise a box to a height of 1 m and another man takes 12\frac{1}{2} minute to do so. The energy of the two is
View answer and explanation
Correct option: B
Same
The correct answer is B. The energy required to lift the box to the same height is the same, regardless of the time taken, as energy depends on force and displacement, not time.
346 Medium
346 of 347
Analyzing +4 | -1
A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break
View answer and explanation
Correct option: B
When the mass is the lowest point of the circle
Option B (When the mass is the lowest point of the circle) is correct because the tension in the wire is greatest at the bottom due to the combined effect of the weight of the mass and the centripetal force required for circular motion.
347 Medium
347 of 347
Application +4 | -1
A rope ladder with a length ll carrying a man with a mass m at its end is attached to the basket of balloon with a mass M. The entire system is in equilibrium in the air. As the man climbs up the ladder into the balloon, the balloon descends by a height h. Then the potential energy of the man
View answer and explanation
Correct option: A
Increase by mg(l−h)m\mathrm{g}\left( {l - h} \right)
Option A (mg(l−h)m\mathrm{g}\left( {l - h} \right)) is correct because the man's potential energy increases by the difference between the ladder length and the balloon's descent.
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