Question 246 of 280

A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highest point of an inclined plane of length ' $L$ ' and makes an angle $30^{\circ}$ with the horizontal. The speed of the particle at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant. Then the new speed of the sphere at the bottom of the plane is $v_2$. The ratio $v_1^2: v_2^2$ is

A$1: \sqrt{2}$B$1: \sqrt{3}$C$1: 3$D$1: 2$

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Answer: (A) $1: \sqrt{2}$

using WET
$\begin{aligned}
& \mathrm{W}_{\mathrm{g}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \\
& \mathrm{Mg} \mathrm{~L} \sin \theta=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}}
\end{aligned}$
K.E. in pure rolling $\frac{1}{2} \mathrm{mV}_{\mathrm{cm}}^2+\frac{1}{2} \mathrm{I}_{\mathrm{cm}} \omega^2$
$\begin{aligned}
& =\frac{1}{2} \mathrm{mV}^2+\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^2 \frac{\mathrm{~V}^2}{\mathrm{R}^2} \\
& \frac{7}{10} \mathrm{mV}^2
\end{aligned}$
$\mathrm{mgL} \sin \theta=\frac{7}{10} \mathrm{mV}_{\mathrm{f}}^2-0$
$\begin{aligned}
& \mathrm{V}_{\mathrm{f}}^2 \propto \sin \theta \\
& \left(\frac{\mathrm{~V}_1}{\mathrm{~V}_2}\right)^2=\frac{\sin \theta_1}{\sin \theta_2}=\frac{\sin 30^{\circ}}{\sin 45^{\circ}}=\frac{1}{\sqrt{2}}
\end{aligned}$