A uniform solid sphere of mass m and radius R is suspended in vertical plane from a point on its periphery.The time period of its oscillation is:

2 π \sqrt{\frac{R}{g}}

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2 π \sqrt{\frac{3 R}{2 g}}

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2 π \sqrt{\frac{7 R}{5 g}}

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2 π \sqrt{\frac{2 R}{5 g}}

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Solution

The correct option is B $2 π \sqrt{\frac{3 R}{2 g}}$
$T = 2 π \sqrt{\frac{I}{m g L}}$
where $I = m R^{2} + \frac{m R^{2}}{2} = \frac{3 m R^{2}}{2}$
$∴ T = 2 π \sqrt{\frac{3 m R^{2}}{2 m g R}} = 2 π \sqrt{\frac{3 R}{2 g}}$

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