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MHT CET 2019 (3 May, Shift 1) · Rolling Motion

Rolling Motion question from MHT CET 2019 (3 May, Shift 1)

Q66MHT CET 2019 · 3 MAY · SHIFT 1Easy

Rolling Motion

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is $\left(g = 10\text{ m/s}^{2}\right)$

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Answer: (d)

Solid sphere rolling: $v = \sqrt{\frac{10gh}{7}}$ .
$v = \sqrt{\frac{10\times 10\times 7}{7}} = \sqrt{100}$ m/s
So the answer is (d) $\sqrt{100}$ m/s $= 10$ m/s.

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