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BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Physics
Chapter: Rotational Motion

BITSAT 2024 · Physics · Rotational Motion

The moment of inertia of a cube of mass m and side a about one of...

The moment of inertia of a cube of mass

m

and side

a

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

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

Options

A $\frac{2}{3} m a^2$
B $\frac{4}{3} m a^2$
C $3 m a^2$
D $\frac{8}{3} m a^2$

Correct answer

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

Step-by-step solution

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

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}

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