In the figure shown, acceleration with which the mass $m$ falls down when released is (consider the string…
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)
$\frac{2 g}{3}$
$\frac{g}{2}$
$\frac{5 g}{6}$
g
Solution
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}
$
