Q65 / 90 · KVPY 2016 (SB/SX) · Physics
Uniform ring of radius $\mathrm{R}$ is moving on a horizontal surface with speed $\mathrm{v}$ and then climbs up a ramp of inclination $30^{\circ}$ to a hcight $h$. Thcre is no slipping in the cntirc motion. Then h is
- A$\mathrm{v}^{2} / 2 \mathrm{~g}$
- Correct answerB$\mathrm{v}^{2} / \mathrm{g}$
- C$3 \mathrm{v}^{2} / 2 \mathrm{~g}$
- D$2 \mathrm{v}^{2} / \mathrm{g}$
Solution
\{Using conservation of energy)
$$
\begin{array}{l}
\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2} \mathrm{mR}^{2}\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=\mathrm{mgh} \\
\mathrm{m}\left(\frac{\mathrm{V}^{2}}{2}+\frac{\mathrm{V}^{2}}{2}\right)=\mathrm{mgh} \\
\mathrm{h}=\frac{\mathrm{V}^{2}}{\mathrm{~g}}
\end{array}
$$
Hence, the correct answer is option: 'B'.