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Q.

The moment of inertia of a cube of mass m and side a about one of its edges is equal to

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a

$\frac{8}{3} m a^{2}$

b

$\frac{2}{3} m a^{2}$

c

$3 m a^{2}$

d

$\frac{4}{3} m a^{2}$

answer is A.

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Detailed Solution

From theorem of perpendicular axes, we have

$I = I_{C} + m {(\frac{a}{\sqrt{2}})}^{2} = (\frac{m a^{2}}{12} + \frac{m a^{2}}{12}) + \frac{m a^{2}}{2} = \frac{2}{3} m a^{2}$

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