Two identical charged spheres suspended from a common point by two massless strings of length l, are initially at a distance d 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:

Let the charge on each sphere be q. The force of repulsion between the spheres is given by Coulomb's law:
F = kq²/x²
where k is Coulomb's constant and x is the distance between the spheres.
The horizontal component of the tension in the strings is Tsinθ, where θ is the angle between the string and the vertical. Since the spheres are in equilibrium, the horizontal component of the tension must balance the force of repulsion:
2Tsinθ = kq²/x²
Since the spheres are identical and the strings are massless, the angle θ is small, and we can approximate sinθ ≈ tanθ = x/(2l). Thus,
2T(x/(2l)) = kq²/x²
T = klq²/x³
The spheres are approaching each other with velocity v, so dx/dt = -v (negative because x is decreasing).
The rate of change of charge on each sphere is dq/dt = -c, where c is a constant.
The force of repulsion is given by:
F = kq²/x²
The acceleration of each sphere is given by:
a = F/m = kq²/mx²
Since v = dx/dt, we have a = dv/dt = (dv/dx)(dx/dt) = -v(dv/dx).
Thus, -v(dv/dx) = kq²/mx²
Integrating both sides, we get:
∫v dv = -∫kq²/mx² dx
v²/2 = kq²/mx + C
Initially, when x = d, v = 0, so C = -kq²/md.
Thus, v²/2 = kq²/mx - kq²/md
v² = 2kq²/m(1/x - 1/d)
Since dq/dt = -c, we have q = q₀ - ct, where q₀ is the initial charge.
As the charges leak, the distance between the spheres decreases, and the velocity increases.
The velocity v varies as x⁻¹/².
Therefore, v∝x⁻¹/²
The correct option is v∝x⁻¹/².