The acceleration of block $B$ in the figure will be :

\frac{m_{2} g}{(4 m_{1} + m_{2})}

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\frac{{2 m}_{2} g}{(4 m_{1} + m_{2})}

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\frac{{2 m}_{1} g}{(m_{1} + 4 m_{2})}

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\frac{{2 m}_{1} g}{(m_{1} + m_{2})}

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Solution

The correct option is A $\frac{m_{2} g}{(4 m_{1} + m_{2})}$
When the block $m_{2}$ moves downward acceleration $a$ , the acceleration of mass $m$ be $2 a$ , because it covers double distance at same time in comparison to $m_{2}$
Let $T$ is the tension in the string
From the free body diagram of $A$ and $B$ we get
$T = m_{1} 2 a . . . . . . . . . . . . . . . (1)$
$m_{2} g - 2 T = m_{2} a . . . . . . . . . . . . . . . (2)$
From Eqs.(i) and (ii) we get
$a = \frac{m_{2} g}{(4 m_{1} + m_{2})}$

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