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QJEEM23Ph3874: JEE Main 2023 Physics - Class 11 JEEM23Ph3874 Solution: Moment of inertia of a disc of mass ' ' and radius ' ' about…

Moment of inertia of a disc of mass ' MM ' and radius ' RR ' about any of its diameter is MR24\frac{M R^{2}}{4} . The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, x2\frac{x}{2} MR 2^{2} . The value of xx is ___________.

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JEE Main 2023 (Online) 1st February Evening Shift Physics - Rotational Motion Question 58 English Explanation

amp;I=Icm+Md2amp;=MR22+MR2amp;=32MR2amp;x=3\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 \\\\ & =\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2 \\\\ & =\frac{3}{2} \mathrm{MR}^2 \\\\ & \mathrm{x}=3\end{aligned}

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