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 t₁ and t₂, respectively, then

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$t_{1} < t_{2}$

$t_{1} = t_{2}$

$t_{1} = 2 t_{2}$

$t_{1} > t_{2}$

answer is A.

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Detailed Solution

$\begin{aligned} t = \sqrt{\frac{2 ℓ}{a_{cm}}} \\ a_{cm} = \frac{gsin θ}{1 + \frac{I_{cm}}{{MR}^{2}}} \end{aligned}$
$\begin{aligned} a_{cm} = \frac{gsin θ}{1 + \frac{I_{cm}}{{MR}^{2}}} \\ a_{1} = a_{{cm}_{1}} = \frac{5 gsin θ}{7} \dots . . Solid \\ a_{2} = a_{{cm}_{2}} = \frac{3 gsin θ}{5} \dots . . Hollow \\ a_{1} > a_{2} \\ t_{1} < t_{2} \end{aligned}$