Q. A ring starts to roll down the inclined plane of height $h$ without slipping. The velocity with 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) For a ring $K^{2} = r^{2}$ then
$v^{2} = \sqrt{\frac{2gh}{1 + \frac{K^{2}}{r^{2}}}}$
$\therefore$ $v^{2} = \frac{2gh}{2} = gh$
$v = \sqrt{gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.