A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is :

We know that

Translational kinetic energy

$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

Rotational kinetic energy

$\mathrm{KE}=\frac{1}{2}{\mathrm{I\omega }}^2$

According to the given condition,

$\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{2}\times \frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{v}=\mathrm{R\omega }$

$\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{4}{\mathrm{mR}}^2{\mathrm{\omega }}^2$

$\mathrm{I}=\frac{1}{2}{\mathrm{mR}}^2$

Hence the correct answer is solid cylinder.