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Momen of inertia (M.I) of four bodies, having same mass and radius, are reported as;
$I_{1} = M . I .$ of thin circular disc about its diameter,
$I_{2} = M . I$ of circular disc about an axis perpendicular to disc and going through the centre,
$I_{3} = M . I$ of solid cylinder about its axis and
$I_{4} = M . I$ of solid sphere about its diameter. Then

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$I_{1} + I_{3} < I_{2} + I_{4}$

$I_{1} = I_{2} = I_{3} > I_{4}$

$I_{1} + I_{2} = I_{3} + \frac{5}{2} I_{4}$

$I_{1} = I_{2} = I_{3} < I_{4}$

answer is A.

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Detailed Solution

$I_{1} = \frac{M R^{2}}{4}$

$I_{2} = \frac{M R^{2}}{2}$

$I_{3} = \frac{M R^{2}}{2}$

$I_{4} = \frac{2}{5} M R^{2}$

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