Q113 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
A solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is '$V_1$'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is '$V_2$'. The ratio $V_1 : V_2$ is (neglect friction)
- A$\sqrt{7} : \sqrt{5}$
- B$\sqrt{7} : \sqrt{3}$
- C$\sqrt{3} : \sqrt{5}$
- Correct answerD$\sqrt{5} : \sqrt{7}$
Solution
When a solid sphere rolls down an inclined plane of height $h$, by conservation of mechanical energy:
$mgh = \dfrac{1}{2}mV_1^2 + \dfrac{1}{2}I\omega^2$
For a solid sphere, the moment of inertia is $I = \dfrac{2}{5}mr^2$ and for pure rolling, $\omega = \dfrac{V_1}{r}$.
$mgh = \dfrac{1}{2}mV_1^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\left(\dfrac{V_1}{r}\right)^2$
$mgh = \dfrac{1}{2}mV_1^2 + \dfrac{1}{5}mV_1^2 = \dfrac{7}{10}mV_1^2$
$V_1 = \sqrt{\dfrac{10gh}{7}}$
When the sphere slides down the same plane without friction, it only acquires translational kinetic energy. By conservation of mechanical energy:
$mgh = \dfrac{1}{2}mV_2^2$
$V_2 = \sqrt{2gh}$
The ratio of the velocities is:
$\dfrac{V_1}{V_2} = \dfrac{\sqrt{\dfrac{10gh}{7}}}{\sqrt{2gh}} = \sqrt{\dfrac{10}{7 \times 2}} = \sqrt{\dfrac{5}{7}}$
Thus, $V_1 : V_2 = \sqrt{5} : \sqrt{7}$.
$mgh = \dfrac{1}{2}mV_1^2 + \dfrac{1}{2}I\omega^2$
For a solid sphere, the moment of inertia is $I = \dfrac{2}{5}mr^2$ and for pure rolling, $\omega = \dfrac{V_1}{r}$.
$mgh = \dfrac{1}{2}mV_1^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\left(\dfrac{V_1}{r}\right)^2$
$mgh = \dfrac{1}{2}mV_1^2 + \dfrac{1}{5}mV_1^2 = \dfrac{7}{10}mV_1^2$
$V_1 = \sqrt{\dfrac{10gh}{7}}$
When the sphere slides down the same plane without friction, it only acquires translational kinetic energy. By conservation of mechanical energy:
$mgh = \dfrac{1}{2}mV_2^2$
$V_2 = \sqrt{2gh}$
The ratio of the velocities is:
$\dfrac{V_1}{V_2} = \dfrac{\sqrt{\dfrac{10gh}{7}}}{\sqrt{2gh}} = \sqrt{\dfrac{10}{7 \times 2}} = \sqrt{\dfrac{5}{7}}$
Thus, $V_1 : V_2 = \sqrt{5} : \sqrt{7}$.
Hence, the correct answer is option: 'D'.