Skip to main content
Back to WORK-ENERGY-POWER

JEE PHYSICS MCQ: WORK-ENERGY-POWER

1 Medium
1 of 1
Application +4 | -1
A body dropped from a height HH reaches the ground with a speed of 1.2gH1.2 \sqrt{g H}. Calculate the work done by air-friction.
View answer and explanation
Correct option: D
−0.28 mgH-0.28 \mathrm{~mgH}
Option D (−0.28 mgH-0.28 \mathrm{~mgH}) is correct because the work done by air friction is the difference between the total work done and the work done by gravity.

More Questions from WORK-ENERGY-POWER

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.