Power — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Power MCQs with step-by-step solutions (8 questions). Part of Work, Power and Energy. Practise online on Prepizo — no login needed.
▶ Practise Power online (free)
Questions with solutions
Q1 — Power · easy · theory
Instantaneous power delivered by a force F⃗ to a body moving with velocity v⃗ is:
A. P = F⃗ × v⃗
B. P = Fv always (any angle)
C. P = F⃗·v⃗ ✓ Correct
D. P = F/v
Solution: P = F⃗·v⃗ = Fv cosθ. It is a scalar; 1 W = 1 J/s and 1 hp ≈ 746 W.
Q2 — Power · medium · theory
A machine delivers constant POWER to a body. The work done by it in time t is:
A. W = ½Pt²
B. Constant, independent of t
C. W = P/t
D. W = Pt (grows linearly with time) ✓ Correct
Solution: P = dW/dt constant ⇒ W = Pt. (The body's speed then grows as √t.)
Q3 — Power · easy · numerical
A motor lifts a 100 kg load through 3 m in 10 s at constant speed. The power delivered is:
A. 300 W ✓ Correct
B. 1000 W
C. 30 W
D. 3000 W
Solution: P = mgh/t = 100×10×3/10 = 300 W.
Q4 — Power · medium · numerical
A car moves at a constant 20 m/s against a total resistance of 500 N. The power developed by the engine is:
A. 100 kW
B. 2.5 kW
C. 25 kW
D. 10 kW ✓ Correct
Solution: Constant speed ⇒ drive force = resistance. P = Fv = 500×20 = 10⁴ W = 10 kW.
Q5 — Power · hard · numerical
A pump raises 200 kg of water per minute to a height of 10 m and ejects it at 2 m/s. The power required is:
A. ≈ 340 W ✓ Correct
B. ≈ 3400 W
C. ≈ 333 W
D. ≈ 400 W
Solution: Per minute: mgh + ½mv² = 20000 + ½×200×4 = 20400 J. P = 20400/60 = 340 W. Trap: don't forget the kinetic-energy term.
Q6 — Power · medium · numerical
A machine gun fires 60 bullets per minute, each of mass 10 g at 600 m/s. The power delivered to the bullets is:
A. 1800 W ✓ Correct
B. 3600 W
C. 600 W
D. 1080 W
Solution: One bullet: ½×0.01×360000 = 1800 J; 60 per minute = 1 per second ⇒ P = 1800 W.
Q7 — Power · hard · numerical
A body of mass 1 kg starts from rest under a machine delivering CONSTANT power 2 W. Its speed after 4 s is:
A. 2 m/s
B. 2√2 m/s
C. 4 m/s ✓ Correct
D. 8 m/s
Solution: Pt = ½mv² ⇒ v = √(2Pt/m) = √(2×2×4/1) = √16 = 4 m/s. Note v ∝ √t for constant power.
Q8 — Power · medium · numerical
A 1000 kg car climbs a slope of sinθ = 1/20 at a steady 72 km/h against 200 N of friction. The engine power is:
A. 14 kW ✓ Correct
B. 4 kW
C. 20 kW
D. 10 kW
Solution: v = 20 m/s; F = mg sinθ + f = 500 + 200 = 700 N; P = 700×20 = 14 kW.