question_answer

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance \[d(d<<l)\] apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance \[x\] between them   AIEEE  Solved  Paper-2011

A) \[v\propto x\]                     

B) \[v\propto {{x}^{-\frac{1}{2}}}\]

C) \[v\propto {{x}^{-1}}\]                     

D) \[v\propto {{x}^{\frac{1}{2}}}\]

Correct Answer: B

Solution :

             \[\tan \theta =\frac{F}{mg}\] or \[\frac{x}{2l}=\frac{k{{q}^{2}}}{mg{{x}^{2}}}\] \[\frac{{{x}^{3}}}{2l}=\frac{k{{q}^{2}}}{mg}\] \[\frac{3{{x}^{2}}\frac{dx}{dt}}{2l}=\frac{2kq\frac{dq}{dt}}{mg}\]              Also, \[q\propto {{x}^{3/2}}\] \[\Rightarrow \,\,\frac{dx}{dt}\propto \frac{{{x}^{3/2}}}{{{x}^{2}}},i.e.,\,v\propto {{x}^{-1/2}}\]