JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 23)
A uniform solid cylinder with radius R and length L has moment of inertia I$$_1$$, about the axis of the cylinder. A concentric solid cylinder of radius $$R'=\frac{R}{2}$$ and length $$L'=\frac{L}{2}$$ is carved out of the original cylinder. If I$$_2$$ is the moment of inertia of the carved out portion of the cylinder then $$\frac{I_1}{I_2}=$$ __________.
(Both I$$_1$$ and I$$_2$$ are about the axis of the cylinder)
Answer
32
Explanation
$I_{1}=\frac{\left(\rho \pi R^{2} L\right) R^{2}}{2} \quad$ ( $\rho$ : density of cylinder)
$$ \begin{aligned} & I_{2}=\frac{\left[\rho \pi\left(\frac{R}{2}\right)^{2} \frac{L}{2}\right]\left(\frac{R}{2}\right)^{2}}{2} \\\\ & \therefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1} \end{aligned} $$
$$ \begin{aligned} & I_{2}=\frac{\left[\rho \pi\left(\frac{R}{2}\right)^{2} \frac{L}{2}\right]\left(\frac{R}{2}\right)^{2}}{2} \\\\ & \therefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1} \end{aligned} $$
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