Question

The moment of inertia of a uniform thin rod of mass $m$ and length $l$ about the axis $PQ$ and $RS$ passing through center 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

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

Answer 1

The correct option is D $\frac{m{l}^2}{12}$



Due to symmetry,
$I_{PQ}=I_{P^{\prime }{Q}^{\prime }}$
By perpendicular axes theorem,
$I_{PQ}+I_{RS}=I_{P^{\prime }{Q}^{\prime }}+I_{RS}=\frac{m{l}^2}{12}$