Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ;
$\mathrm{I}_{1}=$ M.I. of thin circular ring about its diameter.
$\mathrm{I}_{2}=$ M.I. of circular disc about an axis
perpendicular to the disc and going through the centre,
$\mathrm{I}_{3}=$ M.I. of solid cylinder about its axis and
$\mathrm{I}_{4}=$ M.I. of solid sphere about its diameter.
Then :
Correct Option: , 3
$\operatorname{Ring} \mathrm{I}_{1}=\frac{\mathrm{MR}^{2}}{2}$ about diameter
$\operatorname{Disc} \mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}$
Solid cylinder $\mathrm{I}_{3}=\frac{\mathrm{MR}^{2}}{2}$
Solid sphere $\mathrm{I}_{4}=\frac{2}{5} \mathrm{MR}^{2}$
$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}_{3}>\mathrm{I}_{4}$
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