Question

Two blocks,each having a mass $M$,rest on frictionless surfaces as shown in the figure. If the pulley 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

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

Ref. image 1

Make free body diagrams first:

According to Newton's law:

$Mg\sin \theta =\left(2M\right)a$

$a=\frac{g\sin \theta }{2}$

$Mg\sin \theta -T=Ma$

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

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