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A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is

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Solid sphere

Solid cylinder

Ring

Disc

answer is B.

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Detailed Solution

$\frac{1}{2} I ω^{2} = 40 % of \frac{1}{2} {mv}^{2}$

$\frac{1}{2} I ω^{2} = \frac{40}{100} (\frac{1}{2} {mr}^{2} ω^{2})$

$I = \frac{2}{5} {mr}^{2}$

So, the body is solid sphere

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