question_answer

For the given system as shown in figure, the pulley are light and frictionless. The tension in the string is

A)  \[2\,\,Mg\,\,sin\,\theta \]             

B)  \[\frac{1}{2}\,\,Mg\,\,sin\,\theta \]

C)  \[\frac{2}{3}\,Mg \sin \,\theta \]             

D)  \[\frac{3}{2}\,Mg \sin \,\theta \]

Correct Answer: B

Solution :

The FBD of both the block is shown For 1st block, \[\operatorname{T}\,\,=\,\,Ma\]                                       ... (i) For 2nd block, \[\operatorname{Mg}\,\,sin\,\,\theta -T=Ma\]                     ... (ii) From Eqs. (i) and (ii) \[2T=Mg\,\sin \,\theta \,\,\,\Rightarrow \,\,\,\,T=\frac{Mg\,\sin \,\theta }{2}\]