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

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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62.5 k g m^{2}

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Solution

The correct option is A $0.0625 k g m^{2}$
Moment of inertia of a thin ring about an axis passing through its center and perpendicular to its plane is
$I = M R^{2} .$

$I = 0.25 \times {0.5}^{2}$

$= 0.0625 k g m^{2}$

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A thin ring has mass 0.25 kg and radius 0.5 m. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is

The correct option is A 0.0625kgm2Moment of inertia of a thin ring about an axis passing through its center and perpendicular to its plane isI=MR2.I=0.25×0.52=0.0625kgm2

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

The correct option is A $0.0625 k g m^{2}$
Moment of inertia of a thin ring about an axis passing through its center and perpendicular to its plane is
$I = M R^{2} .$

$I = 0.25 \times {0.5}^{2}$

$= 0.0625 k g m^{2}$