MHT CET · Physics · Laws of Motion
The pulleys and strings shown in figure are smooth and of negligible mass. For the system to remain in equilibrium, angle \(\theta\) should be

- A \(\cos ^{-1}(1)\)
- B \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
- C \(\cos ^{-1}\left(\frac{1}{2}\right)\)
- D \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(D) \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
From the diagram,

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

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