Q.
For the system shown in figure, the pulleys are light and frictionless. The tension in the string will be
A) $\frac{2}{3}mg\sin\theta$
B) $\frac{3}{2}mg\sin\theta$
C) $\frac{1}{2}mg\sin\theta$
D) 2$mg\sin\theta$
Explanation
(c) $mg\sin{\theta - T = ma}$
$T = ma$
[image:5_image198.emf]
$\therefore\ mg\sin{\theta - T = T}$
or $2T = mg\sin\theta$ or $T = \frac{1}{2}mg\sin\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.