Q. A solid sphere (mass $2\ M$) and a thin hollow spherical shell (mass $M$) both of the same size, roll down an inclined plane, then
A) Solid sphere will reach the bottom first
B) Hollow spherical shell will reach the bottom first
C) Both will reach at the same time
D) None of these
Explanation
(a) Time of descent $t = \frac{1}{\sin\theta}\sqrt{\frac{2h}{g}\left( 1 + \frac{K^{2}}{R^{2}} \right)}$
For solid sphere $\frac{K^{2}}{R^{2}} = \frac{2}{5}$
For hollow sphere $\frac{K^{2}}{R^{2}} = \frac{2}{3}$
As $\left( \frac{K^{2}}{R^{2}} \right)_{Hollow} > \left( \frac{K^{2}}{R^{2}} \right)_{Solid}$
$i.e.$ solid sphere will take less time so it will reach the bottom first
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.