Question

A solid sphere rolls down from the top of an inclined plane. Its linear velocity, on reaching the bottom of the plane, is $v$. When the same sphere slides down (without friction) from the top of the identical inclined plane, its linear velocity on reaching the bottom is $v^{\prime }$. The ratio $\frac{v^{\prime }}{v}$ is -

• A. $\sqrt{\frac{3}{5}}$
• B. $1$
• C. $\sqrt{\frac{7}{5}}$
• D. $\frac{3}{\sqrt{5}}$

Answer 1

The correct option is C $\sqrt{\frac{7}{5}}$

According to conservation of energy principle,

$\Delta PE=\Delta K{E}_{\text{translational}}+\Delta K{E}_{\text{rotational}}$

For rolling motion :

$Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$

$\Rightarrow Mgh=\frac{1}{2}M{v}^2+\frac{1}{2}\times \left(\frac{2}{5}M{R}^2\right)\times \frac{v^2}{R^2}$

$(\because I=\frac{2}{5}M{R}^2$ and $\omega =\frac{v}{R})$

$\Rightarrow Mgh=\frac{1}{2}M{v}^2+\frac{1}{5}M{v}^2=\frac{7}{10}M{v}^2...\left(1\right)$

For sliding motion without friction:

$Mgh=\frac{1}{2}M{v}^{\prime 2}...\left(2\right)$

Equating equations $1$ and $2$, we get,

$\frac{1}{2}M{v}^{\prime 2}=\frac{7}{10}M{v}^2$

$\Rightarrow \frac{v^{\prime }}{v}=\sqrt{\frac{7}{5}}$

Hence, option $\left(C\right)$ is the correct answer.