Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $$d \, (d \ll l)$$ 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,

Let each sphere have mass $$m$$ and (instantaneous) charge $$q$$. The two light strings of equal length $$l$$ make an angle $$\theta$$ with the vertical when the spheres are separated by a horizontal distance $$x$$. Because $$d\ll l$$, the angles are small and the system can be treated as quasistatic (the spheres readjust their positions continuously as the charge leaks).

Force balance at any instant
For one sphere, the horizontal component of the string tension balances the electrostatic repulsion while the vertical component balances the weight:
$$T\sin\theta = \frac{k\,q^{2}}{x^{2}}, \qquad T\cos\theta = mg$$
Dividing the two equations gives
$$\tan\theta = \frac{k\,q^{2}}{mg\,x^{2}} \; -(1)$$

Small-angle geometry
With both strings making the same angle, the separation is $$x = 2l\sin\theta \approx 2l\theta$$, so $$\theta \approx \dfrac{x}{2l}$$. For small angles $$\tan\theta \approx \theta$$, hence from (1)

$$\frac{x}{2l} = \frac{k\,q^{2}}{mg\,x^{2}} \;\;\Longrightarrow\;\; q^{2} = \frac{mg}{2lk}\,x^{3}$$

Therefore $$q = C\,x^{3/2}, \quad\text{where}\quad C = \sqrt{\frac{mg}{2lk}} = \text{constant}$$

Relating leak rate to approach speed
Charge leaks at a constant rate, so $$\dfrac{dq}{dt} = -\lambda$$ with $$\lambda$$ positive and constant.
Differentiate $$q = Cx^{3/2}$$ with respect to time:

$$\frac{dq}{dt} = \frac{3}{2}\,C\,x^{1/2}\,\frac{dx}{dt}$$

The spheres are approaching, so we deal with the magnitude of velocity $$v = \left|\dfrac{dx}{dt}\right|$$. Using $$\left|\dfrac{dq}{dt}\right| = \lambda$$:

$$\lambda = \frac{3}{2}\,C\,x^{1/2}\,v \;\;\Longrightarrow\;\; v = \frac{2\lambda}{3C}\,x^{-1/2}$$

Thus the speed varies inversely with the square root of the instantaneous separation:

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

Hence, the correct choice is Option D.