Question

A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio $I_1/I_2$ is given by :

• A. 185
• B. 65
• C. 285
• D. 140

Answer 1

The Correct Option is D

$I_{1}=\frac{\left(\frac{7M}{8}\right)\left(2R\right)^{2}}{2} = \left(\frac{7}{16}\times4\right)MR^{2} = \frac{7}{4} MR^{2} $
$ I_{2} = \frac{2}{5} \left(\frac{M}{R}\right)R^{2}_{1} = \frac{2}{5} \left(\frac{M}{8}\right) \frac{R^{2}}{4} = \frac{MR^{2}}{80} $
$\frac{4}{3} \pi R^{3} = 8 \left(\frac{4}{3 } \pi R_{1}^{3}\right) $
$ R^{3} = 8 R_{1}^{3} $
$ R = 2 R_{1} $
$\therefore \frac{I_{1}}{I_{2}} = \frac{7/4MR^{2}}{\frac{MR^{2}}{80}} = \frac{7}{4} \times80 = 140 $