Question

Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ; $I_1$ = M.I. of thin circular ring 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 :

• A. $I_1 + I_2 = I_3 + \frac{5}{2} I_4$
• B. $I_1 + I_3<I_2 + I_4$
• C. $I_1 = I_2 = I_3<I_4$
• D. $I_1 = I_2 = I_3>I_4$

Hint:

Memorize the M.I. of basic shapes. A disc and a solid cylinder have the same formula for M.I. about their symmetry axis.

Answer 1

The Correct Option is D

Step 1: $I_1$ (Ring, diameter) $= \frac{1}{2}MR^2$.
Step 2: $I_2$ (Disc, central axis) $= \frac{1}{2}MR^2$.
Step 3: $I_3$ (Solid cylinder, axis) $= \frac{1}{2}MR^2$.
Step 4: $I_4$ (Solid sphere, diameter) $= \frac{2}{5}MR^2 = 0.4 MR^2$.
Step 5: Clearly $I_1 = I_2 = I_3 = 0.5 MR^2$, and $0.5>0.4$. So $I_1 = I_2 = I_3>I_4$.