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The moment of inertia of a cylinder about its own axis is equal to its moment of inertia about an axis passing through its centre and normal to its length. The ratio of length to radius is

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3 : 1

\sqrt{3} : 1

\sqrt{2} : 1

2 : 1

answer is B.

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Detailed Solution

$I_{A} = \frac{{MR}^{2}}{2}$ $∵ I_{A} = I_{B}$

$I_{C} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4}$

$\frac{{MR}^{2}}{2} = \frac{{ML}^{2}}{12} + \frac{{MR}^{2}}{4}$

$\frac{{MR}^{2}}{2} - \frac{{MR}^{2}}{4} = \frac{{ML}^{2}}{12}$

$\frac{{MR}^{2}}{4} = \frac{{ML}^{2}}{12}$

$R^{2} = \frac{L^{2}}{3}$

$\frac{L^{2}}{R^{2}} = 3$

$\frac{L}{R} = \frac{\sqrt{3}}{1}$ $(or) L : R = \sqrt{3} : 1$

$L = \sqrt{3} \times R$

$r = \sqrt{3} R$

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