Question

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 the tension in the string will be:

• A. $\frac{2}{3}Mg\sin \theta$
• B. $\frac{3}{2}Mg\sin \theta$
• C. $\frac{Mg\sin \theta }{2}$
• D. $2Mg\sin \theta$

Answer 1

Answer: (C)

$\frac{Mg\sin \theta }{2}$

Here do the following
1)draw the free body diagram of the blocks.
2) The acceleration of the block lying on the inclined plane will be in a downward direction.
3)block kept on the flat surface having acceleration in the leftward direction.
now writing the equation for both after resolving the forces
$mg\sin \theta -T=ma$
$T=ma$
hence $ma=T$
on putting in equation (1) we get
$T=\frac{mgsin\theta }{2}$