 # Moment of inertia `
Question:

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 :

1. $\mathrm{I}_{1}+\mathrm{I}_{3}<\mathrm{I}_{2}+\mathrm{I}_{4}$

2. $\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{3}+\frac{5}{2} \mathrm{I}_{4}$

3. $\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}_{3}>\mathrm{I}_{4}$

4. $\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}_{3}<\mathrm{I}_{4}$

Correct Option: , 3

Solution:

$\operatorname{Ring} \mathrm{I}_{1}=\frac{\mathrm{MR}^{2}}{2}$ about diameter

$\operatorname{Disc} \mathrm{I}_{2}=\frac{\mathrm{MR}^{2}}{2}$

Solid cylinder $\mathrm{I}_{3}=\frac{\mathrm{MR}^{2}}{2}$

Solid sphere $\mathrm{I}_{4}=\frac{2}{5} \mathrm{MR}^{2}$

$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}_{3}>\mathrm{I}_{4}$