Two identical charged spheres suspended from a common point by two massless strings of lengths $l$, are initially at a distance $d\left(d\ll l\right)$ 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 $\mathrm{\upsilon }$. Then $\mathrm{\upsilon }$ varies as a function of the distance $x$ between the spheres, as

• A

  $v\propto x^{\frac{1}{2}}$
• B

  $v\propto x$
• Correct answer
  C

  $v\propto x^{-\frac{1}{2}}$
• D

  $v\propto x^{-1}$

Solution

$\sin \theta =$ $\theta =\frac{x}{2l}$, $\cos \theta \approx 1$
$f_e\cos \theta =m\mathrm{gsin}\theta$
$f_e=mg.\left(\frac{x}{2l}\right)$
$\frac{k{q}^2}{x^2}=\frac{mgx}{2l}$
$k{q}^2=\frac{mg}{2l}{x}^3$
$q\propto x^{\frac{3}{2}}$
$\frac{dq}{dt}\propto \frac{3}{2} x^{\frac{1}{2}}.\frac{dx}{dt}$
$\Rightarrow x^{\frac{1}{2}}.v=constant$
$v\propto x^{-\frac{1}{2}}$

Hence, the correct answer is option: 'C'.