A solid sphere of mass and radius is divided into two unequal parts. The smaller part having mass is converted into a sphere of radius

Question

A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The smaller part having mass $\frac{M}{8}$ is converted into a sphere of radius $r$ and the larger part is converted into a circular disc of thickness $t$ and radius $2 R$ . If $I_{1}$ is moment of inertia of a sphere having radius $r$ about an axis through its centre and $I_{2}$ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia $I_{2} / I_{1} =$ ______. [2026]

Smart Achievers · Accepted Answer

(2)

$I_{1} = \frac{2}{5} (\frac{M}{8}) r^{2} and I_{2} = \frac{\frac{7 M}{8} \times {(2 R)}^{2}}{4}$

$M = ρ \times \frac{4 π}{3} \times R^{3}$ $. . . (i)$

$\frac{M}{8} = ρ \times \frac{4 π}{3} \times$ $| r |^{3}$ $. . . (i i)$

$From (i) and (ii), r = \frac{R}{2}$

$\frac{I_{1}}{I_{2}} = \frac{\frac{2}{5} (\frac{M}{8}) {(\frac{R}{2})}^{2}}{\frac{7 M}{8 \times 4} \times {(2 R)}^{2}} = \frac{2}{5} \times \frac{1}{4 \times 7} = \frac{1}{70}$

$\Rightarrow \frac{I_{2}}{I_{1}} = \frac{70}{1}$

Solution

1 35

2 70

3 140

4 210

Want Us to Email you About Special Offers & Updates?