To solve this problem, we need to analyze the forces and motion of the charged spheres.$\\$Given:$\\$• Two identical charged spheres suspended from a common point by massless strings of length $l. \\$• Initial separation between the spheres is $d$ (where $d \ll l). \\$• Charges leak at a constant rate, causing the spheres to approach each other with velocity $v. \\$We need to find how $v$ varies as a function of the distance $x$ between the spheres.$\\$Step 1: Analyze the forces acting on the spheres.$\\$The electrostatic force between the spheres is given by Coulomb's law:$\\ F = \frac{k q^2}{x^2} \\$where $k$ is Coulomb's constant and $q$ is the charge on each sphere.$\\$Step 2: Consider the equilibrium of forces.$\\$At equilibrium, the electrostatic force is balanced by the tension in the strings.$\\$For small angles, the horizontal component of tension $T$ is approximately $T \sin \theta \approx T \frac{x}{2l}. \\$Thus, $\frac{k q^2}{x^2} = T \frac{x}{2l}. \\$Step 3: Relate the rate of change of charge to velocity.$\\$As the charges leak, the force decreases, causing the spheres to move closer.$\\$The rate of change of force is related to the velocity $v$ by:$\\ \frac{d}{dt} \left( \frac{k q^2}{x^2} \right) = \frac{d}{dt} \left( T \frac{x}{2l} \right). \\$Step 4: Simplify the expression.$\\$Assuming $q$ decreases at a constant rate, $\frac{dq}{dt} = -C$ (a constant).$\\$Thus, $\frac{d}{dt} \left( \frac{k q^2}{x^2} \right) = \frac{2k q}{x^2} \frac{dq}{dt} - \frac{2k q^2}{x^3} \frac{dx}{dt}. \\$Substitute $\frac{dq}{dt} = -C$ and $\frac{dx}{dt} = v: \\ \frac{2k q (-C)}{x^2} - \frac{2k q^2 v}{x^3} = 0. \\$Step 5: Solve for $v. \\ \frac{2k q^2 v}{x^3} = \frac{2k q C}{x^2} \\ v = \frac{C x}{q}. \\$Since $q$ is decreasing linearly with time, $q \propto x. \\$Thus, $v \propto x^{-1/2}. \\$Therefore, the velocity $v$ varies as $x^{-1/2},$ which corresponds to Option 1.