In the figure shown, acceleration with which the mass $m$ falls down when released is (consider the string to be massless, $g$-acceleration due to gravity)

Let tension in string is $T$ and tension $T$ is rotating the hollow cylinder. Torque produced in hollow cylinder, $$ \tau=I \alpha $$ Moment of inerita of cylinder, $I=M R^2$ where, $M=$ mass of cylinder and $R=$ radius of cylinder.
Angular acceleration, $\alpha=\frac{a}{R}$, where $a=$ acceleration.
Equating Eqs. (i) and (ii), we get $T R=M R a$ $$ \Rightarrow \quad T=M a $$ Now, $\quad M g-T=M a$ $$ \begin{aligned} & M g-M a=M a \Rightarrow 2 M a=M g \\ \Rightarrow \quad a= & g / 2 \end{aligned} $$