Question

Two metallic spheres each of mass $M$ are suspended by two strings each of length $L$ . The distance between the upper ends of strings is $L.$ The angle which the strings will make with the vertical due to mutual attraction of the spheres is

• A. $t a n^{- 1}\left[\frac{G M}{g L}\right] \, \, $
• B. $t a n^{- 1}\left[\frac{G M}{2 g L}\right]$
• C. $t a n^{- 1}\left[\frac{G M}{g L^{2}}\right]$
• D. $t a n^{- 1}\left[\frac{2 G M}{g L^{2}}\right]$

Answer 1

Answer: (C)

$ F =\frac{ GM \cdot M }{ L ^2} $
due to mutual attraction
Gravitational force
$ \theta=\text { ? } $
distance $\approx L$ (approx)
because $\rightarrow$ Grat. Force very very less strong
$ \Rightarrow T \sin \theta= F $
$\therefore T \cos \theta= Mg$
$ \tan \theta=\frac{ F }{ Mg } $
$\therefore \tan \theta=\frac{\frac{ GM ^2}{ L ^2}}{ Mg }$
$ \Rightarrow \theta=\tan ^{-1}\left(\frac{ GM }{ L ^2 g }\right) $