The following bodies are made to roll up (wit | vedclass

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(B) For a body rolling up an incline without slipping,the total mechanical energy is conserved. The initial kinetic energy at the bottom is $K = \frac

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$14 : 15 : 20$

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(B) For a body rolling up an incline without slipping,the total mechanical energy is conserved. The initial kinetic energy at the bottom is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.At the maximum height $h$,the kinetic energy is converted into potential energy: $mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$.Thus,$h = \frac{v^2}{2g}(1 + \frac{k^2}{R^2})$. Since $v$ and $g$ are constant,$h \propto (1 + \frac{k^2}{R^2})$.For $(i)$ a ring,$I = mR^2 \implies k^2 = R^2 \implies \frac{k^2}{R^2} = 1$. So,$h_1 \propto (1 + 1) = 2$.For $(ii)$ a solid cylinder,$I = \frac{1}{2}mR^2 \implies k^2 = \frac{1}{2}R^2 \implies \frac{k^2}{R^2} = \frac{1}{2}$. So,$h_2 \propto (1 + \frac{1}{2}) = \frac{3}{2} = 1.5$.For $(iii)$ a solid sphere,$I = \frac{2}{5}mR^2 \implies k^2 = \frac{2}{5}R^2 \implies \frac{k^2}{R^2} = \frac{2}{5}$. So,$h_3 \propto (1 + \frac{2}{5}) = \frac{7}{5} = 1.4$.The ratio $h_1 : h_2 : h_3 = 2 : 1.5 : 1.4 = 20 : 15 : 14$.Since the question asks for the ratio of heights for $(i), (ii), (iii)$,the ratio is $20 : 15 : 14$. Option $(B)$ is $14 : 15 : 20$,which is the inverse ratio. Given the options,$(B)$ is the intended answer.

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