A pendulum is constructed as a thin walled sphere of radius R filled up with water and suspended from a point O, $l$ apart . How many times will the time period of small oscillation change when the water freezes-

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$\frac{2}{5} {\frac{R}{l}}^{2}$

$\sqrt{1 + \frac{2}{5} {(\frac{R}{l})}^{2}}$

$\sqrt{\frac{1}{1 + \frac{2}{5} {(\frac{R}{l})}^{2}}}$

$1 + \frac{2}{5} {(\frac{R}{l})}^{2}$

answer is C.

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Detailed Solution

$C a s e - I \to W h e n w a t e r i s p r e s e n t T = 2 π \sqrt{\frac{l}{g}}$

$C a s e - I I \to W h e n w a t e r f r e e z e s T' = 2 π \sqrt{\frac{I}{m g l}} = 2 π \sqrt{\frac{m l^{2} + \frac{2}{5} m R^{2}}{m g l}} = 2 π \sqrt{\frac{l}{g}} {1 + \frac{2}{5} {(\frac{R}{l})}^{2}}^{1 / 2}$

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