Q. If voltage across a bulb rated 220 V–100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
A) 10%
B) 20%
C) 2.5%
D) 5%
Explanation
[Solution from source] For a fixed resistance, P = V²/R, so ΔP/P ≈ 2ΔV/V = 2 × 2.5% = 5%. (Exactly, (0.975)² = 0.951, a 4.9% drop.) 2.5% wrongly assumes power proportional to voltage. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.