$\begin{array}{1 1}\large\frac{g}{4}\\\large\frac{g}{3}\\\large\frac{g}{2}\\g\end{array} $

Answer : $\large\frac{g}{2}$

Since the inclined plane is smooth and $m_2 > m_1$,block $m_1$ will up the plane and block $m_2$ will move vertically with a common acceleration $a$.If $T$ is the tension in the string,the free-body diagrams of masses $m_1$ and $m_2$ are as shown in figure.

The equations of motion of the blocks are

$T-m_1g\sin \theta=m_1a$-------(1)

$m_2g-T=m_2a$--------(2)

Equations (1) and (2) give

$a= \large\frac{(m_2-m_1\sin \theta)g}{(m_1+m_2)}$

$\;\;\;=\large\frac{(2m-m\times \sin 30^{\large\circ})g}{(m+2m)}$

$\;\;\;=\large\frac{g}{2}$

Hence the correct choice is $\large\frac{g}{2}$

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