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

Answer: (D)

$I_1=\frac{\left(\frac{7M}{8}\right){\left(2R\right)}^2}{2}=\left(\frac{7}{16}\times 4\right)M{R}^2=\frac{7}{4}M{R}^2$
$I_2=\frac{2}{5}\left(\frac{M}{R}\right)R_1^2=\frac{2}{5}\left(\frac{M}{8}\right)\frac{R^2}{4}=\frac{M{R}^2}{80}$
$\frac{4}{3}\pi R^3=8\left(\frac{4}{3}\pi R_1^3\right)$
$R^3=8R_1^3$
$R=2R_1$
$\therefore \frac{I_1}{I_2}=\frac{7/4M{R}^2}{\frac{M{R}^2}{80}}=\frac{7}{4}\times 80=140$