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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 'V1V_1'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is 'V2V_2'. The ratio V1:V2V_1 : V_2 is (neglect friction)

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Answer: (d)

  1. Rolling: V12=2gh1+2/5=107ghV_1^2 = \dfrac{2gh}{1 + 2/5} = \frac{10}{7}gh.
  2. Sliding (no friction): V22=2ghV_2^2 = 2gh.
  3. V1:V2=57=5:7V_1 : V_2 = \sqrt{\frac{5}{7}} = \sqrt{5} : \sqrt{7}.
  4. So the answer is (d).
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