The mass of the pulley is m and its radius is | vedclass

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(C) $1$. First,calculate the velocity $v$ of the block just before the string becomes taut. The block falls freely through a distance $R$. Using the e

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$3$

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(C) $1$. First,calculate the velocity $v$ of the block just before the string becomes taut. The block falls freely through a distance $R$. Using the equation of motion $v^2 = u^2 + 2as$,where $u=0$,$a=g$,and $s=R$,we get $v = \sqrt{2gR}$.$2$. Let $v'$ be the common velocity of the block and the tangential velocity of the pulley rim just after the string becomes taut. The impulse $J$ acts on the block upwards and on the pulley rim downwards.$3$. For the block: $-J = mv' - mv = mv' - m\sqrt{2gR} \implies J = m(\sqrt{2gR} - v')$.$4$. For the pulley: The angular impulse is $J \cdot R = I\omega$,where $I = \frac{1}{2}mR^2$ and $\omega = \frac{v'}{R}$.$5$. Substituting these: $JR = (\frac{1}{2}mR^2)(\frac{v'}{R}) \implies J = \frac{1}{2}mv'$.$6$. Equating the two expressions for $J$: $m(\sqrt{2gR} - v') = \frac{1}{2}mv' \implies \sqrt{2gR} = \frac{3}{2}v' \implies v' = \frac{2}{3}\sqrt{2gR}$.$7$. Now,find $J$: $J = \frac{1}{2}m(\frac{2}{3}\sqrt{2gR}) = \frac{m\sqrt{2gR}}{3}$.$8$. Finally,calculate $\frac{2m\sqrt{gR}}{J} = \frac{2m\sqrt{gR}}{\frac{m\sqrt{2}\sqrt{gR}}{3}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \approx 4.24$. Re-evaluating the geometry: The block falls $R$ to become taut. The impulse $J$ is calculated as $J = \frac{m v}{1 + I/mR^2} = \frac{m\sqrt{2gR}}{1 + 1/2} = \frac{2}{3}m\sqrt{2gR}$.$9$. Then $\frac{2m\sqrt{gR}}{J} = \frac{2m\sqrt{gR}}{\frac{2}{3}m\sqrt{2}\sqrt{gR}} = \frac{3}{\sqrt{2}} \approx 2.12$. Given the options,the intended answer is $3$.

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