The moment of inertia of a uniform thin rod of mass m and length l about two axis PQ and RS passing through centre of rod C and in the plane of the rod are $I_{PQ}$ and $I_{RS}$ respectively. Then $I_{PQ} + I_{RS}$ is equal to

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$\frac{{ml}^{2}}{12}$

$\frac{{ml}^{2}}{3}$

$\frac{{ml}^{2}}{4}$

$\frac{{ml}^{2}}{2}$

answer is D.

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Detailed Solution

The MI of of rod about axis PQ figure (a) and MI of rod about axis $P^{'} Q^{'}$ figure (b) are same by symmetry.

$∴$ $I_{PQ} + I_{RS} = I_{PQ} + I_{RS} = \frac{{ml}^{2}}{12}$
by perpendicular axis theorem.

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