Question

Two blocks, each having a mass $\text{M}$, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and $\text{M}$ on the incline is allowed to move down along the incline, 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

The correct option is C $\frac{Mg\sin \theta }{2}$

By the free body diagram of the blocks,

Applying Newton's second law,
$Mg\sin \theta -T=Ma$...$\left(i\right)$

Now by the free body diagram of the other block,

We get $T=Ma$ ...$\left(ii\right)$

Substituting $T$ in equation $\left(i\right)$ we get

$a=\frac{Mg\sin \theta }{2M}$ and

$T=Ma=\frac{Mg\sin \theta }{2}$

Hence option C is the correct answer.