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 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: , 4
(4)
Given $\Rightarrow I_{1}=M . I$. of thin circular ring about its diameter $\mathrm{I}_{2}=$ M.I. circular disc about an axis perpendicular to disc and going through the centre. $I_{3}=M . I$. of solid cylinder about its axis
$\mathrm{I}_{4}=$ M. I, of solid sphere about its diameter we know that,
$I_{1}=\frac{M R^{2}}{2}, I_{2}=\frac{M R^{2}}{2}, I_{3}=\frac{M R^{2}}{2}$
$I_{4}=\frac{2}{5} M R^{2}$
So, $I_{1}=I_{2}=I_{3}>I_{4}$
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