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Moment of Inertia of a Disc

A thin ring h...

Question

A thin ring has mass $0.25 k g$ and radius $0.5 m$ . Its M.I. about an axis passing through its centre and perpendicular to its plane is ....

0.0265 g - c m^{2}

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0.0625 k g - m^{2}

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0.625 k g - m^{2}

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6.25 k g - m^{2}

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Solution

The correct option is D $0.0625 k g - m^{2}$
M.I of an ring passing through its cenre and perpendicular to its plane is given by $I =$ $M R^{2}$
$= 0.25 \times 0.5 = 0.0625 k g m^{2}$

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