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(D) For an object rolling down an inclined plane of length $l$ and angle $	heta$, the acceleration is given by $a = \frac{g \sin 	heta}{1 + K^2/R^2}

Vedclass · Accepted Answer

$1.732$

Vedclass · Answer

(D) For an object rolling down an inclined plane of length $l$ and angle $	heta$, the acceleration is given by $a = \frac{g \sin 	heta}{1 + K^2/R^2}$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ with $u=0$ and $s=l$, we get $l = \frac{1}{2} \left( \frac{g \sin 	heta}{1 + K^2/R^2} ight) t^2$.Thus, $t = \sqrt{\frac{2l(1 + K^2/R^2)}{g \sin 	heta}}$, which implies $t \propto \sqrt{1 + K^2/R^2}$.For a hollow cylinder, the radius of gyration $K^2 = R^2$, so $K^2/R^2 = 1$. Thus, $t_1 \propto \sqrt{1 + 1} = \sqrt{2}$.For a solid cylinder, the radius of gyration $K^2 = R^2/2$, so $K^2/R^2 = 1/2$. Thus, $t_2 \propto \sqrt{1 + 1/2} = \sqrt{3/2}$.Taking the ratio, $\frac{t_2}{t_1} = \frac{\sqrt{3/2}}{\sqrt{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.Given $t_1 = 2 \ s$, we have $t_2 = \frac{\sqrt{3}}{2} 	imes 2 = \sqrt{3} \approx 1.732 \ s$.

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