Q. A body of mass $m$ slides down an incline and reaches the bottom with a velocity $v$. If the same mass were in the form of a ring which rolls down this incline, the velocity of the ring at bottom would have been
A) $v$
B) $\sqrt{2}v$
C) $\frac{1}{\sqrt{2}}v$
D) $\sqrt{\frac{2}{5}}v$
Explanation
(c) When body of mass $m$ slides down an inclined plane then $v = \sqrt{2gh}$
When it is in the form of ring then,
$v_{Ring} = \sqrt{\frac{2gh}{\left( 1 + \frac{K^{2}}{R^{2}} \right)}} = \sqrt{\frac{2gh}{1 + 1}} = \frac{\sqrt{2gh}}{\sqrt{2}} = \frac{v}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.