Two blocks each having mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then tension in the string will be:

A.\[\dfrac{2}{3}Mg\sin \theta \]
B.\[\dfrac{3}{2}Mg\sin \theta \]
C.\[\dfrac{{Mg\sin \theta }}{2}\]
D.\[2Mg\sin \theta \]

Two blocks of each mass \[M\] rest on a frictionless surface.

Draw the free body diagram of the blocks.


In the above free-body diagram of the blocks, \[\theta \] is the angle of inclination of the inclined plane, \[Mg\] is the weight of the blocks and \[T\] is the tension in the string. The directions of X and Y axes for the forces on both the blocks are shown in the diagram.

Apply Newton’s second law of motion to the block on inclined plane in horizontal direction.
\[Mg\sin \theta - T = Ma\] - (Eq 2)

Apply Newton’s second law of motion to the block on horizontal plane in horizontal direction.
\[T = Ma\]

Substitute \[Ma\] for \[T\] in equation (2).
\[Mg\sin \theta - Ma = Ma\]
\[ \Rightarrow Mg\sin \theta = 2Ma\]
\[ \Rightarrow a = \dfrac{{g\sin \theta }}{2}\]

Substitute \[\dfrac{{g\sin \theta }}{2}\] for \[a\] in equation (2).
\[Mg\sin \theta - T = M\dfrac{{g\sin \theta }}{2}\]

Rearrange the above equation for the tension \[T\] in the string.
\[T = Mg\sin \theta - \dfrac{{Mg\sin \theta }}{2}\]
\[ \Rightarrow T = \dfrac{{Mg\sin \theta }}{2}\]

Therefore, the tension in the string is \[\dfrac{{Mg\sin \theta }}{2}\].

Hence, the correct option is C.

Note: There is no need to apply Newton’s law in the vertical direction as the required tension in the string is along the horizontal direction.