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

• Correct answer
  A

  $\frac{2}{3} m a^2$
• B

  $\frac{4}{3} m a^2$
• C

  $3 m a^2$
• D

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

Solution

From theorem of perpendicular axes, we have $\begin{aligned} I & =I_C+m\left(\frac{a}{\sqrt{2}}\right)^2 \\ & =\left[\frac{m a^2}{12}+\frac{m a^2}{12}\right]+\frac{m a^2}{2} \\ & =\frac{2}{3} m a^2\end{aligned}$

Hence, the correct answer is option: 'A'.