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(B) Using the principle of conservation of energy,the loss in kinetic energy equals the gain in potential energy.$mgh = K.E._{total} = \frac{1}{2}mv^2

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Using the principle of conservation of energy,the loss in kinetic energy equals the gain in potential energy.$mgh = K.E._{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the object rolls without slipping,$\omega = \frac{v}{R}$ and $I = Mk^2$,where $k$ is the radius of gyration.$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$$h = \frac{v^2}{2g}(1 + \frac{k^2}{R^2})$Thus,$h \propto (1 + \frac{k^2}{R^2})$.For a solid sphere,$\frac{k^2}{R^2} = \frac{2}{5}$,so $h_1 \propto (1 + \frac{2}{5}) = \frac{7}{5}$.For a hollow cylinder,$\frac{k^2}{R^2} = 1$,so $h_2 \propto (1 + 1) = 2$.Therefore,$\frac{h_1}{h_2} = \frac{7/5}{2} = \frac{7}{10}$.Given $\frac{h_1}{h_2} = \frac{n}{10}$,we find $n = 7$.

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