Consider the shown system comprising of blocks A and B (each having mass m) connected by an ideal string as shown in the figure. The system is in equilibrium under influence of an external horizontal force F (Equal to $\frac{3mg}{4}$ ) acting on  block B. The force F is now withdrawn. If acceleration of block A just after withdrawing F is a. Then acceleration of B is equal to $\frac{xa}{3}.$ Then the value of $x$ is

 

Let tension in strings connecting A and B is T. From FBD of block A and pulley to which it is attached, for vertical equilibrium, we get 

$2T=mg\Rightarrow T=\frac{mg}{2}\mathrm{...............}\left(i\right)$

From FBD of B, for horizontal equilibrium, we get  $T+T\cos \theta =\frac{3}{4}mg$

Substituting value of $T$ from Eq. (i), We get $\Rightarrow \frac{mg}{2}+\frac{mg}{2}\cos \theta =\frac{3}{4}mg\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta ={60}^0$

Just after F is withdrawn, the block will accelerate as shown. Let’s assume tension in string connected to A be T. From FBD of pulley connected to block A, $T=2T\mathrm{...........}\left(ii\right)$

As sum of work by tension on the system is zero, therefore 

$Ta\cos {180}^0+Ta\text{'}+Ta\text{'}\cos {60}^0=0\Rightarrow -2Ta+Ta+\frac{Ta}{2}=0$ [Using eq. (ii)]

$\Rightarrow a=\frac{4a}{3}=1.33a$