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 with acceleration a, the acceleration of mass $m_{1}$ will be 2a because it covers double distance in the same time in comparison to $m_{2}$ .
Let T is the tension in the string.
By drawing the free body diagram of A and B
$T = m_{1} 2 a - - - - - - - - - - (i)$
$M_{2} g - 2 T = m_{2} a - - - - - - - - - - - (i i)$
By solving (i) and (ii)
By solving (i) and (ii)
$a = \frac{m_{2} g}{(4 m_{1} + m_{2})}$

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