The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, angle $\theta$ should be

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The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, angle $	heta$ should be

Aicharya · Accepted Answer

From the diagram,

$\mathrm{T}=\mathrm{Mg}$ and, For the system to remain in equilibrium, $\begin{array}{ll} & 2 \mathrm{~T} \cos \theta=\sqrt{2} \cdot \mathrm{Mg} \\ \therefore & 2 \mathrm{~T} \cos \theta=\mathrm{T} \sqrt{2} \\ \therefore & \cos \theta=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \\ \therefore & \theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \end{array}$

The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in…

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