Question 26 of 280

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

A$t_1\gt t_2$B$t_1=t_2$C$t_1 \lt t_2$D$t_1=2 t_2$

Show answer and solution▾

Answer: (C) $t_1 \lt t_2$

$\begin{aligned} & \mathrm{t}=\sqrt{\frac{2 \ell}{\mathrm{a}_{\mathrm{cm}}}} \\ & \mathrm{a}_{\mathrm{cm}}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}_{\mathrm{cm}}}{\mathrm{MR}^2}} \\ & \mathrm{a}_1=\mathrm{a}_{\mathrm{cm}_1}=\frac{5 \mathrm{~g} \sin \theta}{7} \ldots . . \text { Solid } \\ & \mathrm{a}_2=\mathrm{a}_{\mathrm{cm}_2}=\frac{3 \mathrm{~g} \sin \theta}{5} \ldots . . \text { Hollow } \\ & \mathrm{a}_1\gt \mathrm{a}_2 \\ & \mathrm{t}_1 \lt \mathrm{t}_2\end{aligned}$