Question

A square frame ABCD is formed by four identical rods each of mass 'm' and length 'T'. This frame is in X - Y plane such that side AB coincides with X - axis and side AD along Y - axis. The moment of inertia of the frame about X - axis is

• A. $\frac{5m{l}^2}{3}$
• B. $\frac{2m{l}^2}{3}$
• C. $\frac{4m{l}^2}{3}$
• D. $\frac{m{l}^2}{12}$

Answer 1

Answer: (A)

$\frac{5m{l}^2}{3}$

Moment of inertia of rod AD and BC are each $\frac{1}{3}m{T}^2$

For rod AB, moment of inertia is zero since it is on X-axis and hence r=0.

For, rod CD, every element is at distance $T$ from X-axis and hence its moment of inertia is $m{T}^2$.

Hence, $I=2\times \frac{1}{3}m{T}^2+m{T}^2$

$⟹I=\frac{5}{3}m{T}^2$