Question 49 of 280

Moment of inertia of a square plate of side $l$ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:

A$\dfrac{M{l}^{2}}{6}$B$\dfrac{2}{3}M{l}^{2}$C$M{l}^{2}$D$\dfrac{M{l}^{2}}{12}$

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Answer: (B) $\dfrac{2}{3}M{l}^{2}$

Use parallel axis theorem to get the magnitude of the moment of inertia of a square plate of side $l$ about the axis passing through one of the corners and perpendicular to the plane of a square plate,
${I}_{z}={I}_{cm}+M{\left(\dfrac{l}{\sqrt{2}}\right)}^{2}$
$\Rightarrow {I}_{z}=\dfrac{M{l}^{2}}{6}+\dfrac{M{l}^{2}}{2}=\dfrac{4M{l}^{2}}{6}=\dfrac{2M{l}^{2}}{3}$.

JEE Main Physics practice question on Rotational Motion (Moment of inertia of rigid bodies), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.