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A square ABCD is fabricated by four indentical thin rods each of mass m and length l. Then moment of inertia of the square about an axis passing through its centre O and perpendicular to its plane is

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

$\frac{1}{6} m l^{2}$

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

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

answer is D.

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Detailed Solution

$I_{0} = 4 [\frac{1}{12} m l^{2} + m {(\frac{l}{2})}^{2}] = 4 \times \frac{1}{3} m l^{2} = \frac{4}{3} m l^{2}$

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