Question 41 of 280

A uniform solid cylinder with radius $R$ and length $L$ has moment of inertia ${I}_{1}$, about the axis of cylinder. A concentric solid cylinder of radius ${R}^{'}=\dfrac{R}{2}$ and length ${L}^{'}=\dfrac{L}{2}$ is carved out of the original cylinder. If ${I}_{2}$ is the moment of inertia of the carved out portion of the cylinder then $\dfrac{{I}_{1}}{{I}_{2}}=$ __________.
(Both ${I}_{1}$ and ${I}_{2}$ are about the axis of the cylinder)

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Answer: 32

Moment of inertia of solid cylinder of radius $R$ and length $L$ about the axis is ${I}_{1}=\dfrac{{M}_{1}{R}^{2}}{2}$.
Moment of inertia of carved out portion of the cylinder of radius $\dfrac{R}{2}$ and length $\dfrac{L}{2}$ is ${I}_{2}=\dfrac{{M}_{2}{\left(\dfrac{R}{2}\right)}^{2}}{2}=\dfrac{{M}_{2}{R}^{2}}{8}$
Now, the ratio $\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{4{M}_{1}}{{M}_{2}}$
As, $mass=density\times volume$
So, $\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{4\times \rho \pi {R}^{2}L}{\dfrac{\rho \pi {R}^{2}L}{4\times 2}}=32$