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A uniform solid cylinder with radius \mathrm{R} and length \mathrm{L} has moment of inertia \mathrm{I_{1}}, about the axis of the cylinder. A concentric solid cylinder of radius \mathrm{R'=\frac{R}{2}} and length \mathrm{L'=\frac{L}{2}} is carved out of the original cylinder. If \mathrm{I_{2}} is the moment of inertia of the carved out portion of the cylinder then \mathrm{\frac{I_{1}}{I_{2}}=} _______
(Both \mathrm{I_{1}} and \mathrm{I_{2}} are about the axis of the cylinder)

Option: 1

32


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{I}_1=\frac{\mathrm{MR}^2}{2} \\ \end{aligned}

\begin{aligned} & \text { mass }=\rho \pi \frac{\mathrm{R}^2}{4} \cdot \frac{\mathrm{L}}{2} \\ \end{aligned}

\begin{aligned} & \mathrm{~m}_2=\frac{\mathrm{M}}{8} \\ \end{aligned}

\begin{aligned} & \mathrm{I}_2=\frac{\mathrm{m}_2 \mathrm{R}_2^2}{2}=\frac{\mathrm{MR}^2}{8 \times 4 \times 2} \\ \end{aligned}

\begin{aligned} & \frac{\mathrm{I}_1}{\mathrm{I}_2}=32 \end{aligned}

Posted by

rishi.raj

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