Physics - NLM Question with Solution | TestHub

Let $T$ be the tension in the string.

For the hanging masses $M$ to be in equilibrium, the tension in the string must balance their weight.

$T=Mg$

For the central mass $\sqrt{2}M$ to be in equilibrium, the net vertical force must be zero. The upward components of the tension from both sides of the string balance the downward weight of the central mass.

From the given figure, the angle between the upward vertical and the string is $\theta$. By symmetry, the string on the other side also makes an angle $\theta$ with the vertical.

The total upward force is $2T\cos \theta$.

Equating the upward force to the downward weight of the central mass:

$2T\cos \theta =\sqrt{2}Mg$

Substituting $T=Mg$ into the equation:

$2(Mg)\cos \theta =\sqrt{2}Mg$

$2\cos \theta =\sqrt{2}$

$\cos \theta =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

$\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Answer: ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$