JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 23)
Moment of inertia of a disc of mass '$$M$$' and radius '$$R$$' about any of its diameter is $$\frac{M R^{2}}{4}$$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $$\frac{x}{2}$$ MR$$^{2}$$. The value of $$x$$ is ___________.
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توضیح
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$\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 \\\\ & =\frac{\mathrm{MR}^2}{2}+\mathrm{MR}^2 \\\\ & =\frac{3}{2} \mathrm{MR}^2 \\\\ & \mathrm{x}=3\end{aligned}$