A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined pla

Question

A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_{1}$ and $t_{2}$, respectively, then\

Professor PC Thomas Classes · Answer

(c) : The acceleration of a body of mass $M$ rolling down an inclined plane is, $$ a=\frac{g \sin 	heta}{\left(1+\frac{k^{2}}{r^{2}}ight)} $$ where $k$ is the radius of gyration.\ For solid sphere, $$ I=\frac{2}{5} M R^{2} \quad 	herefore a=\frac{5}{7} g \sin 	heta $$ For hollow sphere, $$ I=\frac{2}{3} M R^{2} \quad 	herefore a=\frac{3}{5} g \sin 	heta $$ As, acceleration is maximum for solid sphere hence, the solid sphere will reach the bottom first. Hence, $t_{1}<t_{2}$.\