Two identical charged spheres suspended from a common point by tw… - Vidyadip

MCQ

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

A
$v \propto x$
✓
$v \propto {x^{ - \frac{1}{2}}}$
C
$\;v \propto {x^{ - 1}}$
D
$\;v \propto {x^{\frac{1}{2}}}$

✓

Answer

Correct option: B.

$v \propto {x^{ - \frac{1}{2}}}$

b
$\text { From figure, } T \cos \theta=m g.........(i)$

$T \sin \theta=\frac{k q^{2}}{x^{2}}.........(ii)$

From eqns. $(i)$ and $(ii)$, $tan \theta=\frac{k q^{2}}{x^{2} m g}$

since $\theta$ is small, $\therefore \tan \theta \approx \sin \theta = \frac{x}{{2l}}$

$\therefore \quad \frac{x}{2 l}=\frac{k q^{2}}{x^{2} m g} \Rightarrow q^{2}=x^{3} \frac{m g}{2 l k}$ or $q \propto x^{3 / 2}$

$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} \frac{d x}{d t}=\frac{3}{2} \sqrt{x} v$

since, $\frac{d q}{d t}=$ constant

$\therefore v \propto \frac{1}{\sqrt{x}}$

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