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under influence of an external horizontal force F (Equal to $\frac{3mg}{4}$) acting on block B. The force F is now withdrawn.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

 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

$2T=mg\Rightarrow T=\frac{mg}{2}\mathrm{.........}\text{(i)}$
From FBD of B, for horizontal equilibrium, we get
$\text{T+T cos}\theta =\frac{3}{4}mg$
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Just after F is withdrawn, the block will accelerate as shown.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 $T^{\text{'}}$.
From FBD of pulley connected to block A,  $T^{\text{'}}=2T\mathrm{...........}\text{(ii)}$
 
As sum of work by tension on the system is zero, therefore 
$$\begin{gathered} T^{\text{'}}a\cos {180}^0+T{a}^{\text{'}}+T{a}^{\text{'}}\cos {60}^0=0 \\ \Rightarrow -2Ta+T{a}^{\text{'}}+\frac{T{a}^{\text{'}}}{2}=0\left[Using\text{ Eq. (ii)}\right] \\ \Rightarrow a^{\text{'}}=\frac{4a}{3}=1.33a \end{gathered}$$