Work, Energy and Power NEET PYQReal NEET & AIPMT 2006–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Work, Energy and Power question from the NEET & AIPMT papers on this site (2006–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Work, Energy and Power previous year questions
29 questions. Tap an option to check your answer.
Q1NEET 2021
A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively :
Not quite — the answer is D.
At height h, PE = mgh and KE = mg(S − h). KE = 3PE gives S − h = 3h, so h = S/4. Then v² = 2g(S − h) = 2g(3S/4) = 3gS/2, so v = √(3gS/2); the forms √(3gS)/2 or 3gS/2 are dimensionally or numerically wrong.
Q2NEET 2021
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine ? (g = 10 m/s²)
Not quite — the answer is B.
Input power = (mass per second) × g × h = 15 × 10 × 60 = 9000 W. With 10% loss, output = 0.9 × 9000 = 8100 W = 8.1 kW.
Q3NEET 2019 · Q4
Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:
Not quite — the answer is B.
For a 1-D elastic collision v_A = (m₁ − m₂)u/(m₁ + m₂) = (4m − 2m)u/6m = u/3. A keeps (1/3)² = 1/9 of its kinetic energy, so it loses 8/9. 1/9 is the fraction retained, not lost.
Q4NEET 2019 · Q22
A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is:
Not quite — the answer is C.
W = ∫₀¹ (20 + 10y) dy = 20 + 5 = 25 J. Using the final force value 30 N times 1 m gives 30 J, ignoring that the force varies.
Q5NEET 2018 · Q32
A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
Not quite — the answer is B.
Momentum: mv = 4mv′, so v′ = v/4. e = (relative speed of separation)/(relative speed of approach) = (v/4 − 0)/v = 0.25. A value of 0.8 would not conserve momentum with the lighter block stopping.
Q6NEET 2016 Phase 2 · Q5
A bullet of mass 10 g moving horizontally with a velocity of 400 m s⁻¹ strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be
Not quite — the answer is C.
The block's speed just after impact is √(2gh) = √(2 × 10 × 0.1) ≈ 1.41 m/s. Momentum conservation: 0.01 × 400 = 0.01v + 2 × 1.41, so 0.01v = 4 − 2.83 = 1.17 and v ≈ 117 ≈ 120 m/s. Energy is not conserved in the bullet–block collision, so equating kinetic energies is the trap.
Q7NEET 2016 Phase 2 · Q6
Two identical balls A and B having velocities of 0.5 m/s and −0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be
Not quite — the answer is B.
In a one-dimensional elastic collision between equal masses the velocities are exchanged, so B moves with 0.5 m/s and A with −0.3 m/s. The question asks for B first, so the order is 0.5 m/s and −0.3 m/s; reversing the order is the common slip.
Q8NEET 2016 Phase 2 · Q7
A particle moves from a point (−2î + 5ĵ) to (4ĵ + 3k̂) when a force of (4î + 3ĵ) N is applied. How much work has been done by the force?
Not quite — the answer is C.
Displacement d = (4ĵ + 3k̂) − (−2î + 5ĵ) = 2î − ĵ + 3k̂. Work W = F·d = 4 × 2 + 3 × (−1) + 0 × 3 = 8 − 3 = 5 J. Forgetting the negative y-component gives 11 J.
Q9NEET 2016 Phase 1 · Q28
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?
Not quite — the answer is D.
At the top, the minimum speed is √(gR) (tension or normal force zero). Energy conservation from bottom to top: v² = gR + 2g(2R) = 5gR, so v = √(5gR). √(gR) is the minimum speed at the top, not at entry.
Q10NEET 2016 Phase 1 · Q39
A body of mass 1 kg begins to move under the action of a time dependent force F⃗ = (2t î + 3t² ĵ) N, where î and ĵ are unit vectors along x and y axis. What power will be developed by the force at the time t?
Not quite — the answer is D.
With m = 1 kg, a⃗ = F⃗, and starting from rest v⃗ = ∫a⃗ dt = t² î + t³ ĵ. Power P = F⃗·v⃗ = 2t·t² + 3t²·t³ = 2t³ + 3t⁵ W. Multiplying force by time instead of integrating gives the wrong powers of t.
Q11AIPMT 2015
Two particles of masses m₁, m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v₁ and v₂ then we must have:
Not quite — the answer is D.
Total energy is conserved: the initial kinetic energy equals the final kinetic energy plus the energy ε absorbed in exciting the particle, so ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² + ε. Writing −ε on the final side would mean the system gained energy, which is wrong.
Q12AIPMT 2015
Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs W_P and W_Q are related as, in case (a) and case (b), respectively:
Not quite — the answer is D.
Same extension: W = ½kx², so the stiffer spring P does more work (W_P > W_Q). Same force: x = F/k and W = F²/2k, so the softer spring Q does more work (W_Q > W_P).
Q13AIPMT 2015
A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is:
Not quite — the answer is B.
Constant power: ½mv² = kt, so v = √(2kt/m). Then F = m dv/dt = m × √(2k/m) × ½t^(−1/2) = √(mk/2) t^(−1/2). Forgetting the ½ from differentiation gives √(2mk) t^(−1/2).
Q14AIPMT 2015
A block of mass 10 kg, moving in x direction with a constant speed of 10 ms⁻¹, is subjected to a retarding force F = 0.1x J/m during its travel from x = 20 m to 30 m. Its final KE will be:
Not quite — the answer is B.
Initial KE = ½ × 10 × 10² = 500 J. Work by the retarding force = −∫₂₀³⁰ 0.1x dx = −0.05(900 − 400) = −25 J. Final KE = 500 − 25 = 475 J.
Q15AIPMT 2014
A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is:
Not quite — the answer is A.
The two m pieces have perpendicular momenta mv each, so the 2m piece carries √2 mv and moves at v/√2. Total KE = 2 × (1/2)mv² + (1/2)(2m)(v²/2) = (3/2)mv².
Q16NEET 2013
A uniform force of (3î + ĵ) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position (2î + k̂) metre to position (4î + 3ĵ − k̂) metre. The work done by the force on the particle is
Not quite — the answer is A.
Displacement d = (4î + 3ĵ − k̂) − (2î + k̂) = 2î + 3ĵ − 2k̂. W = F·d = 3 × 2 + 1 × 3 + 0 × (−2) = 9 J. The mass of the particle is not needed for the work done by a given force.
Q17AIPMT 2012 Mains
A stone is dropped from a height h. It hits the ground with a certain momentum P. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by
Not quite — the answer is B.
Momentum on hitting the ground is P = m√(2gh). Doubling the height multiplies P by √2 ≈ 1.414, an increase of about 41%. Doubling the momentum (100%) would need four times the height.
Q18AIPMT 2012 Mains
A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P₀. The instantaneous velocity of this car is proportional to
Not quite — the answer is B.
With constant power, the work done equals the kinetic energy gained: P₀t = ½mv², so v = √(2P₀t/m) and v ∝ t^(1/2). The velocity grows ever more slowly, not linearly or as t².
Q19AIPMT 2012 Prelims
The potential energy of a particle in a force field is U = A/r² − B/r, where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle is
Not quite — the answer is C.
Equilibrium needs dU/dr = −2A/r³ + B/r² = 0, giving r = 2A/B. At this r, d²U/dr² = 6A/r⁴ − 2B/r³ = B⁴/(8A³) > 0, so U is a minimum and the equilibrium is stable. A/B is where U itself is zero, not where the force is zero.
Q20AIPMT 2011 Mains
A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to a mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is
Not quite — the answer is C.
Momentum is conserved in this perfectly inelastic collision: m v î + 3m(2v) ĵ = 4m V⃗. So V⃗ = (v/4) î + (6v/4) ĵ = (1/4)v î + (3/2)v ĵ. Swapping the components, (3/2)v î + (1/4)v ĵ, is the common trap.
Q21AIPMT 2011 Prelims
The potential energy of a system increases if work is done
Not quite — the answer is C.
Potential energy is defined only for conservative forces: ΔU = −W(conservative). When the system does work against a conservative force (the conservative force does negative work), its potential energy increases. Work by non-conservative forces does not change potential energy.
Q22AIPMT 2010 Prelims · Q8
A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 then their velocities (in m/s) after collision will be
Not quite — the answer is B.
Momentum: m×2 = m v₁ + 2m v₂, so v₁ + 2v₂ = 2. Restitution: v₂ − v₁ = 0.5 × 2 = 1. Solving gives v₂ = 1 m/s and v₁ = 0, i.e. velocities 0 and 1 m/s. The pair 1 and 1 would need perfectly inelastic sticking, which conserves momentum only if v = 2/3.
Q23AIPMT 2009 Prelims · Q27
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?
Not quite — the answer is D.
Mass leaving per second = (mass per length) × speed = mv. KE delivered per second = ½(mv)v² = ½mv³. Taking ½mv² treats m as the mass per second instead of per unit length.
Q24AIPMT 2009 Prelims · Q41
A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s²)
Not quite — the answer is D.
Initial KE = ½ × 1 × 20² = 200 J. PE at 18 m = 1 × 10 × 18 = 180 J. Energy lost to air friction = 200 − 180 = 20 J.
Q25AIPMT 2009 Prelims · Q50
A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:
Not quite — the answer is A.
Released from rest with the spring unstretched, the block moves until it is momentarily at rest again. Energy conservation: Mgx = ½kx², so the maximum extension is x = 2Mg/k. Mg/k is only the equilibrium extension (where the block moves fastest), not the maximum.
Q26AIPMT 2008 Prelims · Q9
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine? (g = 10 m/s²)
Not quite — the answer is B.
Input power = (dm/dt)gh = 15 × 10 × 60 = 9000 W. With 10% lost, the output is 0.9 × 9000 = 8100 W = 8.1 kW. 9 kW without the loss is not offered; adding the loss instead of subtracting it would give about 9.9 kW.
Q27AIPMT 2006 Prelims · Q13
The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is :
Not quite — the answer is C.
Spring energy is U = ½kx², so U ∝ x². Increasing the stretch from 2 cm to 8 cm (4 times) multiplies the energy by 4² = 16, giving 16U. Assuming U ∝ x would give only 4U.
Q28AIPMT 2006 Prelims · Q15
A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation s = (1/3)t², where t is in s. Work done by the force in 2 s is :
Not quite — the answer is C.
From s = t²/3, v = ds/dt = 2t/3 and a = 2/3 m s⁻², so the force is F = 3 × 2/3 = 2 N. At t = 2 s, s = 4/3 m, so W = 2 × 4/3 = 8/3 J; this equals the gain in kinetic energy ½ × 3 × (4/3)² = 8/3 J. The reciprocal 3/8 J is a slip in the algebra.
Q29AIPMT 2006 Prelims · Q19
300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s², work done against friction is :
Not quite — the answer is B.
Work against gravity is mgh = 2 × 10 × 10 = 200 J. The remaining 300 − 200 = 100 J of the work done is spent against friction (assuming no gain in kinetic energy). Zero would be true only on a frictionless incline.
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