JEE MAIN - Physics (2002 - No. 27)
Moment of inertia of a circular wire of mass $$M$$ and radius $$R$$ about its diameter is
$${{M{R^2}} \over 2}$$
$$M{R^2}$$
$$2M{R^2}$$
$${{M{R^2}} \over 4}$$
Explanation
Moment of Inertia of a circular wire about an axis $$nn'$$ passing through the centre of the circle and perpendicular to the plane of the circle $$ = M{R^2}$$
As shown in the figure, $$X$$-axis and $$Y$$-axis lies in the plane of the ring. Then by perpendicular axis theorem
$${I_X} + {I_Y} = {I_Z}$$
$$ \Rightarrow 2{I_X} = M{R^2}\,$$ $$\left[ \, \right.$$ as $${I_X} - {I_Y}$$ (by symmetry) and $${I_Z} = M{R^2}$$ $$\left. \, \right]$$
$$\therefore$$ $${I_X} = {1 \over 2}M{R^2}$$
As shown in the figure, $$X$$-axis and $$Y$$-axis lies in the plane of the ring. Then by perpendicular axis theorem
$${I_X} + {I_Y} = {I_Z}$$
$$ \Rightarrow 2{I_X} = M{R^2}\,$$ $$\left[ \, \right.$$ as $${I_X} - {I_Y}$$ (by symmetry) and $${I_Z} = M{R^2}$$ $$\left. \, \right]$$
$$\therefore$$ $${I_X} = {1 \over 2}M{R^2}$$
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