Find the acceleration of the block of mass $M$ in the situation shown in the figure. All the surfaces are frictionless and the pulley and the string are light.

The correct option is B $\frac{g}{3}$
Assuming the accelerations as shown,
$\text{T - Mg sin}\theta =\text{M (2a)}$
$\Rightarrow T=2\text{Ma + Mg sin}\theta$ ......(1)
$\text{2Mg-2T}=\text{2Ma}$......(2)
On multiplying by 2 on both sides in equation (1)
$2T=\text{4Ma + 2Mg sin}\theta$......(3)
On adding equation (2) and (3)
$\text{2Mg}=\text{6Ma + 2Mg sin}\theta$
(Given $\theta ={30}^{\circ }$)
$\Rightarrow 6\text{Ma}+2\text{Mg sin}{30}^{\circ }-2\text{Mg}=0$
$\Rightarrow 6\text{Ma = Mg}\Rightarrow a=\frac{g}{6}$.
Acceleration of mass $M$ is $2a=2\times \frac{g}{6}=\frac{g}{3}$ up the plane.