Two identical charged spheres suspended form a common point by two massless strings of lengths l, are initially at a distance `d(d gt gt l)` apart because of their mutual repulsion. The charges brgin 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

Options

A. `v prop x^(-1)`
B. `v prop x^((1)/(2))`
C. `v prop x`
D. `v prop x^(-(1)/(2))`

Answer

To solve the problem step by step, we will analyze the situation of two identical charged spheres that are leaking charge and approaching each other.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have two identical charged spheres suspended from a common point by two massless strings of length \( L \).
- Initially, the distance between the spheres is \( d \) (where \( d \gg L \)).
- The spheres repel each other due to their charges.

**Hint**: Visualize the setup and understand the forces acting on the spheres due to their charges.

2. **Force Between the Spheres**:
- The electrostatic force \( F \) between the spheres is given by Coulomb's law:
\[ F = \frac{k Q^2}{d^2} \]
- As the charges leak from the spheres, the charge \( Q \) decreases over time.

**Hint**: Remember that as the charge decreases, the force of repulsion will also decrease.

3. **Change in Distance**:
- As the charges leak, the spheres will move closer together, reducing the distance from \( d \) to \( x \).
- The new force \( F' \) when the distance is \( x \) becomes:
\[ F' = \frac{k Q^2}{x^2} \]

**Hint**: Keep track of how \( Q \) changes as the spheres approach each other.

4. **Equilibrium of Forces**:
- The tension \( T \) in the strings can be resolved into components:
- \( T \cos \theta = mg \) (vertical component)
- \( T \sin \theta = F' \) (horizontal component)
- For small angles, \( \tan \theta \approx \sin \theta \approx \theta \).

**Hint**: Use trigonometric identities for small angles to simplify your calculations.

5. **Relating Tension and Force**:
- From the above, we can write:
\[ \tan \theta = \frac{F'}{mg} = \frac{k Q^2}{mg x^2} \]
- Also, for small angles:
\[ \tan \theta \approx \frac{x/2}{L} \Rightarrow \theta \approx \frac{x}{2L} \]

**Hint**: Set the two expressions for \( \tan \theta \) equal to each other to find a relationship between \( Q \), \( x \), and \( mg \).

6. **Finding the Charge Relation**:
- Equating the two expressions gives:
\[ \frac{k Q^2}{mg x^2} = \frac{x}{2L} \]
- Rearranging leads to:
\[ k Q^2 = \frac{mg x^3}{2L} \]
- Thus, we can express \( Q \) in terms of \( x \):
\[ Q = \sqrt{\frac{mg x^3}{2kL}} \]

**Hint**: This step is crucial as it relates the charge to the distance between the spheres.

7. **Rate of Change of Charge**:
- Since the charge is leaking at a constant rate, we denote this rate as \( \frac{dQ}{dt} = -k \) (where \( k \) is a positive constant).
- Using the chain rule:
\[ \frac{dQ}{dt} = \frac{dQ}{dx} \cdot \frac{dx}{dt} \]

**Hint**: Remember that \( \frac{dx}{dt} \) is the velocity \( v \) of the spheres approaching each other.

8. **Substituting and Solving for Velocity**:
- Substitute \( \frac{dQ}{dx} \) from the expression for \( Q \):
\[ \frac{dQ}{dx} = \frac{3mg x^2}{4kL \sqrt{\frac{mg x^3}{2kL}}} \]
- Setting this equal to \( -k \cdot v \) gives:
\[ -k v = \frac{3mg x^2}{4kL \sqrt{\frac{mg x^3}{2kL}}} \]

**Hint**: Simplify this equation to isolate \( v \).

9. **Final Expression for Velocity**:
- After simplification, we find:
\[ v \propto x^{-1/2} \]
- Thus, the velocity \( v \) varies inversely with the square root of the distance \( x \).

**Hint**: This relationship indicates that as the spheres get closer, their velocity increases.

### Conclusion:
The final result is that the velocity \( v \) of the spheres as a function of the distance \( x \) between them is given by:
\[ v \propto x^{-1/2} \]