Question

Moment of Inertia $\left(MI\right)$ of four bodies, having same mass and radius, are reported as -

$I_1=MI$ of thin circular ring about its diameter.

$I_2=MI$ of circular disc about an axis perpendicular to the disc and passing through the centre.

$I_3=MI$ of solid cylinder about its axis.

$I_4=MI$ of solid sphere about its diameter.

Then :

• A. $I_1=I_2=I_3<I_4$
• B. $I_1+I_2=I_3+\frac{5}{2}{I}_4$
• C. $I_1+I_3<I_2+I_4$
• D. $I_1=I_2=I_3>I_4$

Answer 1

The correct option is D $I_1=I_2=I_3>I_4$

$I_1=MI$ of thin circular ring about its diameter.
$\Rightarrow I_1=\frac{M{R}^2}{2}=0.5M{R}^2$

$I_2=MI$ of circular disc about an axis perpendicular to the disc and passing through the centre.
$\Rightarrow I_2=\frac{M{R}^2}{2}=0.5M{R}^2$

$I_3=MI$ of solid cylinder about its axis.
$\Rightarrow I_3=\frac{M{R}^2}{2}=0.5M{R}^2$

$I_4=MI$ of solid sphere about its diameter.
$\Rightarrow I_4=\frac{2M{R}^2}{5}=0.4M{R}^2$

$\therefore I_1=I_2=I_3>I_4$