Question

Two blocks, each having a mass $M$, rest on frictionless surface 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. $2.3 Mg \sin \theta$
• B. $1.3 Mg \sin \theta$
• C. $\frac{ Mg }{2} \sin \theta$
• D. $2 Mg \sin 0$

Answer 1

Answer: (C)

$a=\frac{M g \sin \theta}{(M+M)}=\frac{g}{2} \sin \theta$
$\therefore T=M a=\frac{M g}{2} \sin \theta$