Q. 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) $\tan^{- 1}\left\lbrack \frac{GM}{gL} \right\rbrack\ \ $
B) $\tan^{- 1}\left\lbrack \frac{GM}{2gL} \right\rbrack$
C) $\tan^{- 1}\left\lbrack \frac{GM}{{gL}^{2}} \right\rbrack$
D) $\tan^{- 1}\left\lbrack \frac{2GM}{{gL}^{2}} \right\rbrack$
Explanation
(c) The metallic spheres will be at positions as shown in the figure.
[image:8_image59.emf]
$T\sin{\theta = F = \frac{GM \times M}{L^{2}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{{GM}^{2}}{L^{2}}$
$\ \ \ \ \ \ \ \ T\cos{\theta = Mg}$
$\therefore\ \ \ \ \ \ \ \ \tan{\theta = \frac{GM}{gL^{2}}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta = \tan^{- 1}\left( \frac{GM}{gL^{2}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.