of cross-section of the cylinder is A and length is L then volume is,

$$ V=\pi R^2 L $$

If the density of the cylinder is $\rho$ and volume is $V$, then the mass of the cylinder is, $m=\rho V$

Putting in the expression for the moment of inertia,

$$ I=\frac{1}{2}\left[\rho \pi R^2 L\right] \times R^2=\frac{\pi \rho R^4 L}{2} $$

For the larger cylinder, $\rho_1=\rho$, the radius is $\mathrm{R}_1=\mathrm{R}$ and the length is $\mathrm{L}_1=\mathrm{L}$

So, the moment of inertia of the larger cylinder is,

$$ I_1=\frac{\pi \rho_1 R_1^4 L_1}{2}=\frac{\pi \rho R^4 L}{2} $$

For the smaller cylinder, as it is carved from the same material, so it has the same density ( $\rho_2=\rho$ ). The radius is $\mathrm{R}_2=\mathrm{R} / 3$ and the length is $\mathrm{L}_2=\mathrm{L} / 2$

So, the moment of inertia of the smaller cylinder is,

$$ \mathrm{I}_2=\frac{\pi \rho_2 \mathrm{R}_2^4 \mathrm{~L}_2}{2}=\frac{\pi \rho\left(\frac{\mathrm{R}}{3}\right)^4\left(\frac{\mathrm{~L}}{2}\right)}{2}=\frac{\pi \rho \mathrm{R}^4 \mathrm{~L}}{324} $$

Now, the ratio of the moment of inertia of larger cylinder to the smaller cylinder is,

$$ \frac{I_1}{I_2}=\frac{\left(\frac{\pi \rho R^4 L}{2}\right)}{\left(\frac{\pi \rho R^4 L}{324}\right)}=\frac{324}{2}=162 $$

Therefore, the value of ratio of moment of inertia is $\frac{I_1}{I_2}=162$.

Hence, the correct answer is 162 .

19:T669,

The moment of inertia of a uniform solid cylinder of mass M and radius R about its central axis is given by :

$$ \mathrm{I}=\frac{1}{2} \mathrm{MR}^2 $$

If area of cross-section of the cylinder is A and length is L then volume is,

$$ V=\pi R^2 L $$

If the density of the cylinder is $\rho$ and volume is $V$, then the mass of the cylinder is, $m=\rho V$

Putting in the expression for the moment of inertia,

$$ I=\frac{1}{2}\left[\rho \pi R^2 L\right] \times R^2=\frac{\pi \rho R^4 L}{2} $$

For the larger cylinder, $\rho_1=\

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