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(D) Let $T$ be the tension in the string and $a$ be the downward acceleration of the mass $m$. The equation of motion for the falling mass $m$ is: $mg

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$\frac{2g}{3}$

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(D) Let $T$ be the tension in the string and $a$ be the downward acceleration of the mass $m$. The equation of motion for the falling mass $m$ is: $mg - T = ma$ $(1)$For the rotating cylinder of mass $M$ and radius $R$,the torque $	au$ is given by $	au = T \cdot R$. The moment of inertia of the cylinder is $I = \frac{1}{2}MR^2$. Using $	au = I\alpha$,where $\alpha = \frac{a}{R}$ is the angular acceleration,we get: $T \cdot R = (\frac{1}{2}MR^2) \cdot (\frac{a}{R}) \implies T = \frac{1}{2}Ma$ $(2)$Substituting the value of $T$ from $(2)$ into $(1)$: $mg - \frac{1}{2}Ma = ma$$mg = ma + \frac{1}{2}Ma = a(m + \frac{M}{2})$$a = \frac{mg}{m + M/2} = \frac{2mg}{2m + M}$.If we assume the masses are equal $(M = m)$,then $a = \frac{2mg}{3m} = \frac{2g}{3}$.

$A$ mass $m$ is supported by a massless string wound around a uniform cylinder of mass $M$ and radius $R$. On releasing the mass from rest,it will fall with an acceleration of:

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