JEE Main 2025 — 23 January, Morning Shift — Question 61

$\operatorname{Lsin} \theta \rightarrow$ height of plane

using WET

$\mathrm{W}_{\mathrm{g}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}}$

$m g L \sin \theta=k_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}}$

K.E. in pure rolling $\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}+\frac{1}{2} \mathrm{I}_{\mathrm{cm}} \omega^{2}$

$=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \frac{\mathrm{~v}^{2}}{\mathrm{R}^{2}}=\frac{7}{10} \mathrm{mv}^{2}$

$\implies\mathrm{mgL} \sin \theta=\frac{7}{10} \mathrm{mv}_{\mathrm{f}}^{2}-0$

$\mathrm{v}_{\mathrm{f}}^{2} \propto \sin \theta$

$\left(\frac{\mathrm{v}_{1}}{\mathrm{~v}_{2}}\right)^{2}=\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{\sin 30^{\circ}}{\sin 45^{\circ}}=\frac{1}{\sqrt{2}}$