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The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is:

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$\sqrt{2} : 1$

4 : 1

$1 : \sqrt{2}$

2 : 1

answer is A.

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Detailed Solution

$\frac{{MR}^{2}}{2} = I$ (about its axis through the centre)
$K_{1} = \frac{R}{\sqrt{2}}$
$\frac{{MR}^{2}}{4} = I^{'}$ (about its diameter)
$\begin{array}{l} K_{2} = \frac{R}{2} \\ \frac{K_{1}}{K_{2}} = \sqrt{\frac{2}{1}} \end{array}$

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