NEET Solved Paper 2016 Question 9
Question: Two identical charged spheres suspended from a common point by two massless strings of lengths $ l, $ are initially at a distance $ d,(d«l) $ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:
Options:
A) $ v\propto {x^{\frac{1}{2}}} $
B) $ v\propto x $
C) $ v\propto {x^{-\frac{1}{2}}} $
D) $ v\propto {x^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tan \theta =\frac{F _{e}}{mg}\simeq \theta $ $ \frac{Kq^{2}}{x^{2}mg}=\frac{x}{2\ell } $ or $ $ ..(1)
or $ {x^{3/2}}\propto q $ ..(2)
differentiate eq.(i) w. r .t. time $ 3x^{2}\frac{dx}{dt}\propto 2q\frac{dq}{dt} $
but $ \frac{dq}{dt} $ is constant so x $ x^{2}(v)\propto q $ replace q from eq. (2) $ x^{2}(v)\propto {x^{3/2}} $
BETA
