Power: NEET MCQs with Answers

15 free questions on Power for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
A body of mass 1 kg begins to move under the action of a time-dependent force F = (2tî + 3t²ĵ) N, where î and ĵ are unit vectors along the x and y axes. The power developed by the force at time t is [NEET 2016]
  1. (2t² + 3t³) W
  2. (2t² + 4t⁴) W
  3. (2t³ + 3t⁴) W
  4. (2t³ + 3t⁵) W
Show answer
Answer: (D) (2t³ + 3t⁵) W
Explanation: With m = 1 kg, v = ∫F dt = t²î + t³ĵ. Hence P = F·v = 2t(t²) + 3t²(t³) = 2t³ + 3t⁵ W.

Question 2

Previous-year question
A particle of mass m is driven by a machine that delivers a constant power of k watts. If the particle starts from rest, the force on it at time t is [AIPMT 2015]
  1. √(mk)·t^(–1/2)
  2. √(2mk)·t^(–1/2)
  3. ½√(mk)·t^(–1/2)
  4. √(mk/2)·t^(–1/2)
Show answer
Answer: (D) √(mk/2)·t^(–1/2)
Explanation: From kt = ½mv², v = √(2kt/m). Hence F = P/v = k/√(2kt/m) = √(mk/2)·t^(–1/2).

Question 3

Previous-year question
The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. Taking the density of mercury as 13.6 × 10³ kg/m³ and g = 10 m/s², the power of the heart in watt is [AIPMT 2015]
  1. 2.35
  2. 3.0
  3. 1.50
  4. 1.70
Show answer
Answer: (D) 1.70
Explanation: Pressure = hρg = 0.15 × 13.6 × 10³ × 10 ≈ 2.04 × 10⁴ Pa and the volume flow rate is 5 × 10⁻³/60 m³/s. Hence P = pressure × volume rate ≈ 1.70 W.

Question 4

Previous-year question
How much water can a pump of 2 kW raise in one minute to a height of 10 m? (Take g = 10 m/s²) [AIPMT 1990]
  1. 1000 L
  2. 1200 L
  3. 100 L
  4. 2000 L
Show answer
Answer: (B) 1200 L
Explanation: Work done in 1 minute, W = Pt = 2000 × 60 = 1.2 × 10⁵ J. W = Mgh ⟹ M = W/(gh) = 1.2 × 10⁵ / (10 × 10) = 1200 kg. Since 1 kg of water occupies 1 L, the volume raised is 1200 L.

Question 5

Previous-year question
If F = (60î + 15ĵ − 3k̂) N and v = (2î − 4ĵ + 5k̂) m/s, then the instantaneous power is [AIPMT 2000]
  1. 195 W
  2. 45 W
  3. 75 W
  4. 100 W
Show answer
Answer: (B) 45 W
Explanation: P = F · v = (60)(2) + (15)(−4) + (−3)(5) = 120 − 60 − 15 = 45 W.

Question 6

Previous-year question
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 the energy. Taking g = 10 m/s², how much power is generated by the turbine? [AIPMT 2008]
  1. 8.1 kW
  2. 10.2 kW
  3. 12.3 kW
  4. 7.0 kW
Show answer
Answer: (A) 8.1 kW
Explanation: Input power = (m/t)gh = 15 × 10 × 60 = 9000 W. After 10% loss, P = 9000 × 90/100 = 8100 W = 8.1 kW.

Question 7

Previous-year question
An engine pumps water continuously through a hose. The water leaves the hose with velocity v, and m denotes the mass per unit length of the water jet. The rate at which kinetic energy is imparted to the water is [AIPMT 2009]
  1. ½mv³
  2. mv³
  3. ½mv²
  4. ½m²v²
Show answer
Answer: (A) ½mv³
Explanation: Mass delivered per second, dM/dt = m × v (mass per unit length × length delivered per second). Rate of KE = ½ (dM/dt) v² = ½ (mv) v² = ½mv³.
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Question 8

Previous-year question
A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time t is v, the power delivered to the particle is [AIPMT 2010]
  1. Mv²/t
  2. ½Mv²/t²
  3. Mv²/t²
  4. ½Mv²/t
Show answer
Answer: (D) ½Mv²/t
Explanation: The work done in time t equals the gain in kinetic energy, ½Mv². The (average) power delivered over this interval is therefore P = W/t = ½Mv²/t.

Question 9

Previous-year question
An engine pumps water through a hose pipe. The water passes through the pipe and leaves it with a velocity of 2 m/s. If the mass per unit length of water in the pipe is 100 kg/m, the power of the engine is [AIPMT 2010]
  1. 400 W
  2. 200 W
  3. 100 W
  4. 800 W
Show answer
Answer: (D) 800 W
Explanation: Mass leaving per second = λv = 100 × 2 = 200 kg/s. Thrust developed, F = (dm/dt)v = λv² = 100 × (2)² = 400 N. Power of the engine, P = F · v = λv³ = 100 × (2)³ = 800 W.

Question 10

Previous-year question
A body projected vertically upward from the earth rises to a height equal to the earth's radius before returning. The power exerted by the gravitational force is greatest [AIPMT 2011]
  1. at the highest position of the body
  2. at the instant just before the body hits the earth
  3. it remains constant all through
  4. at the instant just after the body is projected
Show answer
Answer: (B) at the instant just before the body hits the earth
Explanation: P = F · v. At the highest point v = 0, so the power is zero there. The speed keeps increasing during the fall and the gravitational force also grows as the body approaches the surface, so their product is largest at the instant just before impact.

Question 11

Previous-year question
A car of mass m starts from rest and accelerates in such a way that the instantaneous power delivered to it has a constant magnitude P₀. The instantaneous velocity of the car is proportional to [AIPMT 2012]
  1. t² P₀
  2. t^(1/2)
  3. t^(−1/2)
  4. t/√m
Show answer
Answer: (B) t^(1/2)
Explanation: P₀ = F·v = mv (dv/dt) ⟹ ∫mv dv = ∫P₀ dt ⟹ mv²/2 = P₀t ⟹ v = √(2P₀t/m) ⟹ v ∝ √t = t^(1/2).

Question 12

Previous-year question
One coolie takes 1 minute to raise a suitcase through a height of 2 m, while a second coolie takes 30 s to raise the same suitcase through the same height. The ratio of the powers of the two coolies is [NEET 2013]
  1. 1 : 3
  2. 2 : 1
  3. 3 : 1
  4. 1 : 2
Show answer
Answer: (D) 1 : 2
Explanation: The work done is identical in both cases, so P = W/t ⟹ P ∝ 1/t. P₁/P₂ = t₂/t₁ = 30/60 = 1/2 Hence the ratio is 1 : 2.

Question 13

Previous-year question
A man's heart pumps 5 L of blood per minute through the arteries at a pressure of 150 mm of mercury. If the density of mercury is 13.6 × 10³ kg/m³ and g = 10 m/s², the power of the heart in watt is [AIPMT 2015]
  1. 1.70
  2. 2.35
  3. 3.0
  4. 1.50
Show answer
Answer: (A) 1.70
Explanation: Pressure, p = ρgh = 13.6 × 10³ × 10 × 0.15 = 2.04 × 10⁴ Pa. Volume flow rate, dV/dt = 5 × 10⁻³ / 60 = 8.33 × 10⁻⁵ m³/s. P = p × (dV/dt) = 2.04 × 10⁴ × 8.33 × 10⁻⁵ ≈ 1.70 W.

Question 14

Previous-year question
A particle of mass m is driven by a machine that delivers a constant power of k watt. If the particle starts from rest, the force acting on it at time t is [AIPMT 2015]
  1. √(mk/2) · t^(−1/2)
  2. √(mk) · t^(−1/2)
  3. √(2mk) · t^(−1/2)
  4. ½√(mk) · t^(−1/2)
Show answer
Answer: (A) √(mk/2) · t^(−1/2)
Explanation: Constant power: F·v = k, and F = m dv/dt, so mv (dv/dt) = k ⟹ ∫v dv = (k/m)∫dt ⟹ v²/2 = kt/m ⟹ v = √(2kt/m). Hence F = k/v = k √(m/2kt) = √(mk/2) · t^(−1/2).

Question 15

Previous-year question
A body of mass 1 kg starts moving under the action of a time-dependent force F = (2t î + 3t² ĵ) N, where î and ĵ are unit vectors along the X and Y axes. The power developed by this force at time t will be [NEET 2016]
  1. (2t² + 4t⁴) W
  2. (2t³ + 3t⁴) W
  3. (2t³ + 3t⁵) W
  4. (2t + 3t³) W
Show answer
Answer: (C) (2t³ + 3t⁵) W
Explanation: Since m = 1 kg, a = F = (2t î + 3t² ĵ) m/s². Integrating from rest: v = ∫a dt = (t² î + t³ ĵ) m/s. P = F · v = (2t î + 3t² ĵ) · (t² î + t³ ĵ) = 2t·t² + 3t²·t³ = (2t³ + 3t⁵) W.

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