Question Detail
Question
Light bulb has input power consumption of 50 watts. The light bulb was activated for 60 seconds and produced heat of 2400 joules. Find the efficiency of the light bulb.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Total energy = 50 × 60 = 3000J. Useful (light) energy = 3000 - 2400 = 600J. Efficiency = (600/3000) × 100 = 20%
Hint
Efficiency is the ratio of useful energy output to total input energy.