Two blocks of masses $m_1$ and $m_2$ are connected as shown in the figure. The acceleration of the block $m_2$ is:

The correct option is B $\frac{m_{2}g}{m_1+m_2}$
Step 1:Draw $FBD$ of both the blocks
Step 2:Find relation between acceleration of both blocks
Let $a$ be acceleration of $m_1$
Then acceleration of middle pulley
$=\frac{(a+0)}{2}=\frac{a}{2}$
Let acceleration of $m_2$ be $b$
Then by pulley constraint motion
$\frac{(0+b)}{2}=\frac{a}{2}$
$a=b$
Step 3:Write force equation and find acceleration
For $m_1$ mass
$T=m_{1}a\dots \left(i\right)$
For $m_2$ mass
$m_{2}g–T=m_{2}a\dots \left(ii\right)$
By adding above equation $\left(i\right)$ and $\left(ii\right)$
$a=\frac{m_{2}g}{m_1+m_2}$
FINAL ANSWER: (a)