JEE Main Physics question
A ball dropped from a height of 2 m rebounds to a height of 1.5 m after hitting the ground. Then the percentage of energy lost is
Answer: (A)
Solution
% energy lost = (U₁−U₂)/U₁ × 100 = (mgh₁ − mgh₂)/(mgh₁) × 100 = [(h₁−h₂)/h₁] × 100 = [(2−1.5)/2] × 100 = 25%.