A solid sphere rolls without slipping on an inclined plane at an angle \(\theta\). The ratio of total kinetic energy to its rotational kinetic energy is

Moment of Inertia of a solid sphere, \(I=\frac{2}{5} \mathrm{MR}^2\)
Since there is no slipping,
\(\mathrm{v}=\mathrm{R} \omega\)
\(\therefore \quad\) Rotational kinetic energy \(\mathrm{E}_{\mathrm{Tot}}=\frac{1}{2} \mathrm{I} \omega^2\)
\(\begin{aligned}
& =\frac{1}{2} \times \frac{2}{5} \times \mathrm{M} \times \mathrm{R}^2 \times \omega^2 \\
& =\frac{\mathrm{MR}^2 \omega^2}{5} \\
& =\frac{\mathrm{MV}^2}{5}... (i)
\end{aligned}\)
Total kinetic energy
\(\begin{aligned}
\mathrm{E}_{\mathrm{K}} & =\frac{1}{2} \mathrm{I}^2+\frac{1}{2} \mathrm{MV}^2 \\
& =\frac{\mathrm{MV}^2}{5}+\frac{\mathrm{MV}^2}{2} \\
& =\frac{7 \mathrm{MV}^2}{10}... (ii)
\end{aligned}\)
Dividing (ii) by (i), we get,
\(\frac{\mathrm{E}_{\mathrm{K}}}{\mathrm{E}_{\mathrm{me}}}=\frac{\left(\frac{7 \mathrm{MV}^2}{10}\right)}{\left(\frac{\mathrm{MV}^2}{5}\right)}=\frac{7}{2}\)