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The moment of inertia of a wheel about the axis of rotation is 3 joules. Its kinetic energy will be 600 joule if the period of rotation is:

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0.05 s

0.314 s

3.14 s

20 s

answer is B.

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Detailed Solution

$\begin{array}{l} RKE = \frac{1}{2} {Iω}^{2} \Rightarrow ω = \sqrt{\frac{2 \times 600}{3}} = 20 rad / s \\ Period of rotation = \frac{2 π}{ω} = \frac{2 \times 3 .14}{20} = 0 .314 s \end{array}$

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