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PHY 102 Test Compilation PHY 102 10 Objective Question
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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.
Options
A 20%
Correct Answer
B 80%
C 12.5%
D 60%
Correct Answer

Option A is the correct answer.

Detailed Explanation

Efficiency (η) is defined as the ratio of useful output energy to the total input energy.Given:Input Power (Pin) = 50 WTime (t) = 60 sTotal Input Energy (Ein) = Pin × t = 50 W × 60 s = 3000 JEnergy produced as heat (Eheat) = 2400 JFor a light bulb, the useful output energy is typically light, and heat is considered wasted energy.Useful Output Energy (Elight) = Total Input Energy - Energy produced as heatElight = 3000 J - 2400 J = 600 JEfficiency (η) = (Useful Output Energy / Total Input Energy) × 100%η = (600 J / 3000 J) × 100%η = (1/5) × 100% = 20%

Hint

Efficiency is the ratio of useful output energy to total input energy. For a light bulb, light is the useful output.

Question Info

Course: PHY 102
Topic: Test Compilation PHY 102 10
Type: Objective
Difficulty: Not specified
Question 892 of 1819
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