Q143 / 150 · MHT CET 2023 (09 May Shift 1) · Physics
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
- Correct answerA$\frac{7}{2}$
- B$\frac{5}{2}$
- C$\frac{7}{3}$
- D$\frac{5}{4}$
Solution
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,
$\frac{\mathrm{E}_{\mathrm{K}}}{\mathrm{E}_{\mathrm{me}}}=\frac{\left(\frac{7 \mathrm{MV}^2}{10}\right)}{\left(\frac{\mathrm{MV}^2}{5}\right)}=\frac{7}{2}$
Hence, the correct answer is option: 'A'.