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

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Answer : $mg$

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)

From equations (1) and (2),we get

$T= m_2(g-a)=2m\times (g-\large\frac{g}{2})$$=mg$

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