question_answer

For the system shown in the 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) \[2mg\,\sin \theta \]  

Correct Answer: C

Solution :

From force diagram in figure [a], \[mg\sin \theta -T=ma.\]                                      (i) In force diagram in the figure [b], T = ma   (ii) From Eqs. (i) and (ii), \[a=\frac{g}{2}\sin \theta \]   \[\therefore \]\[T=\frac{mg\sin \theta }{2}\] Hence, the correction option is [c].