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# The moment of inertia of a cube of mass m and side (a) about one of its edges is equal to:A. $\dfrac{2}{3}m{a^2}$B. $\dfrac{4}{3}m{a^2}$C. $3m{a^2}$D. $\dfrac{8}{3}m{a^2}$

Last updated date: 13th Jun 2024
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Hint: The perpendicular axis theorem is defined as that moment of inertia of a planar lamina about an axis perpendicular to the lamina which is equal to the sum of the moment inertia of the Lamina about the two axes at the right angle to each other in its own plane of intersection to each other.

According to perpendicular axis:
$I = {I_C} + m{(\dfrac{a}{{\sqrt 2 }})^2}$
$\Rightarrow (\dfrac{{m{a^2}}}{{12}} - \dfrac{{m{a^2}}}{{12}}) + \dfrac{{m{a^2}}}{2} = \dfrac{2}{3}m{a^2}$
$I = \dfrac{2}{3}m{a^2}$

The term moment of inertia is also known as mass moment of inertia. It is defined as the ratio of net angular momentum of a system to its angular velocity around the principal axis. Moment of inertia plays a very important role in physics which means that in physics problems that involve the mass in rotation motion and that are calculated by angular momentum.

So, the correct answer is “Option A”.