A square lamina of side ' $b$ ' has same mass as a disc of radius ' $R$ ' the moment of inertia of the two…

A square lamina of side ' $b$ ' has same mass as a disc of radius ' $R$ ' the moment of inertia of the two objects about an axis perpendicular to the plane and passing through the centre is equal. The ratio $\frac{\mathrm{b}}{\mathrm{R}}$ is
  1. $1:1$
  2. $\sqrt{3}:1$
  3. $\sqrt{6}:1$
  4. $1:\sqrt{3}$

Solution

$\begin{aligned} & \mathrm{I}_{\text {lamina }}=\frac{\mathrm{Mb}^2}{6} \\ & \mathrm{I}_{\text {disc }}=\frac{\mathrm{MR}^2}{2} \\ & \text { Given } \frac{\mathrm{Mb}^2}{6}=\frac{\mathrm{MR}^2}{2} \\ & \quad \frac{\mathrm{b}^2}{\mathrm{R}^2}=3 \\ & \therefore \quad \frac{\mathrm{b}}{\mathrm{R}}=\frac{\sqrt{3}}{1}\end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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