Q. A ring starts to roll down the inclined plane of height $h$ without slipping. The velocity with which it reaches the ground is
A) $\sqrt{\frac{10gh}{7}}$
B) $\sqrt{\frac{4gh}{7}}$
C) $\sqrt{\frac{4gh}{3}}$
D) $\sqrt{gh}$
Explanation
(d) $v = \sqrt{\frac{2gh}{1 + \frac{k^{2}}{R^{2}}}}$
Where $k$ is the radius of gyration
For ring, $\frac{k^{2}}{R^{2}} = 1$
$\therefore v = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.