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MHT CET 2026 (20 Apr, Shift 1) · Rolling Motion

Rolling Motion question from MHT CET 2026 (20 Apr, Shift 1)

Q4MHT CET 2026 · 20 APR · SHIFT 1Medium

Rolling Motion

A solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is ' $V_1$ '. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is ' $V_2$ '. The ratio $V_1 : V_2$ is (neglect friction)

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Answer: (d)

Rolling: $V_1^2 = \dfrac{2gh}{1 + 2/5} = \frac{10}{7}gh$ .
Sliding (no friction): $V_2^2 = 2gh$ .
$V_1 : V_2 = \sqrt{\frac{5}{7}} = \sqrt{5} : \sqrt{7}$ .
So the answer is (d).

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