A rigid massless beam is balanced by a body of mass $4m$ on the left hand side and a pulley-mass system on the right hand side. The value of $\frac{x}{y}$ is :
(Consider the rope and pulley to be massless).

The correct option is B $\frac{2}{3}$
Here the pulley and rope are massless, hence tension in the string on both sides of pulley will be same i.e $T$
The acceleration of masses $m$ and $2m$ connected through pulley will be in opposite direction and given by,
$a=\frac{(2m-m)g}{3m}=\frac{g}{3}$

From the F.B.D of components of system

Applying newton's second law on mass $m$,
$T-mg=ma$
$\Rightarrow T=mg+\frac{mg}{3}=\frac{4mg}{3}$
From the equilibrium condition on mass $4m$,
$T^{\prime }=4mg$
Applying balancing of moment on beam about hinge point $\left(o\right)$,
$\Rightarrow \text{Clockwise moment}=\text{Anti-clockwise moment}$
$\Rightarrow 2T\times y=T^{\prime }\times x$
$\Rightarrow \frac{x}{y}=\frac{2T}{T^{\prime }}$
$\Rightarrow \frac{x}{y}=\frac{\frac{8mg}{3}}{4mg}$
$\therefore \frac{x}{y}=\frac{2}{3}$