Question

A uniform solid sphere of mass $M$ , radius $R$ and having moment of inertia about an axis passing through center of mass is $I$ , is recast into a uniform circular disc of thickness $t$ , whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$ . Then radius of the disc will be

• A. $R\sqrt{\frac{2}{15}}$
• B. $\frac{2 R}{\sqrt{15}}$
• C. $\frac{4 R}{\sqrt{15}}$
• D. $\frac{R}{4}$

Answer 1

Answer: (B)

$I=\frac{2}{5}MR^{2}$ and $\rho =\frac{3 M}{4 \pi R^{3}}$
For circular disc $\pi r^{2}\times t\rho =M\Rightarrow \frac{\pi r^{2} \cdot t \cdot 3 M}{4 \pi R^{3}}=M$
$\Rightarrow \, \, t=\frac{4 \pi R^{3}}{3 \pi r^{2}}=\frac{4 R^{3}}{3 r^{2}} \, \Rightarrow \, r^{2}=\frac{4 R^{3}}{3 t}$
$\frac{2}{5}MR^{2}=\frac{3}{2}Mr^{2}$
$\Rightarrow \, \frac{2}{5}R^{2}=\frac{3}{2}\cdot \frac{4 R^{3}}{3 t}\Rightarrow \, t=5R$
$\Rightarrow \, r^{2}\cdot 3\cdot 5R \, = \, 4R^{3} \, \Rightarrow \, r^{2}=\frac{4 R^{2}}{15}$
$\Rightarrow \, r \, =\frac{2 R}{\sqrt{15}}$