A disc rotates freely with rotational kinetic energy E about a normal axis through centre. A ring having the same radius but double the mass of disc is now, gently placed on the disc. The new rotational kinetic energy of the system would be 

A ring is gently placed on a Rotating disc of energy E 

Let 'm' be the mass of the disc mass of the ring = 2m 

Since, the ring is gently placed, the Angular momentum of the system will remain conserved

$\overrightarrow{{\mathrm{L}}_{\mathrm{I}}}=\overrightarrow{{\mathrm{L}}_{\mathrm{F}}}$

${\mathrm{I}}_{\mathrm{D}}{\mathrm{W}}_{\mathrm{o}}=({\mathrm{I}}_{\mathrm{D}}+{\mathrm{I}}_{\mathrm{R}})\mathrm{\omega }\text{'}$

$\mathrm{Initial} \mathrm{kinetic} \mathrm{enrgy} =\mathrm{E}=\frac{1}{2}\frac{{\mathrm{L}}^2}{{\mathrm{I}}_{\mathrm{D}}+{\mathrm{I}}_{\mathrm{R}}}$

$\frac{\mathrm{E}\text{'}}{\mathrm{E}}=\frac{{\mathrm{I}}_{\mathrm{D}}}{{\mathrm{I}}_{\mathrm{D}}+{\mathrm{I}}_{\mathrm{R}}}$

${\mathrm{I}}_{\mathrm{D}}=\frac{{\mathrm{mR}}^2}{2}$, ${\mathrm{I}}_{\mathrm{R}}=2{\mathrm{mR}}^2$

$\frac{\mathrm{E}\text{'}}{\mathrm{E}}=\frac{\frac{{\mathrm{mR}}^2}{2}}{\frac{{\mathrm{mR}}^2}{2}+2{\mathrm{mR}}^2}=\frac{1}{5}$

$\Rightarrow \mathrm{E}\text{'}=\frac{\mathrm{E}}{5}$

Hence the correct answer is $\frac{E}{5}.$