There is a uniform spherically symmetric surface charge density at a distance `R_(0)` from the origin . The charge distribution is intially at rest and starts expanding because of mutual repulsion , the figure that repeesents best the speed V (r(t)) of the distribution as a function of its intantanradius R(t) is :

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the expansion of a uniform spherically symmetric surface charge density due to mutual repulsion. We'll derive the relationship between the speed \( V(r(t)) \) of the distribution and its instantaneous radius \( R(t) \).

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a uniform spherically symmetric surface charge density at an initial radius \( R_0 \).
- The charge distribution is initially at rest and starts expanding due to mutual repulsion among the charges.

2. **Applying Work-Energy Theorem**:
- Since there are no external forces acting on the system, the work done by external forces is zero.
- According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy:
\[ W_{\text{total}} = \Delta KE \]

3. **Identifying Forces**:
- The only forces doing work are the electric forces due to the surface charge. The work done by the electric field can be expressed as:
\[ W = U_{\text{final}} - U_{\text{initial}} \]
- Here, \( U \) is the electric potential energy.

4. **Calculating Electric Potential Energy**:
- The potential energy \( U \) for a uniformly charged sphere can be expressed as:
\[ U = \frac{KQ^2}{2R} \]
- Where \( K \) is Coulomb's constant, \( Q \) is the total charge, and \( R \) is the radius.

5. **Setting Up the Energy Equation**:
- Initially, at radius \( R_0 \):
\[ U_{\text{initial}} = \frac{KQ^2}{2R_0} \]
- At an instantaneous radius \( R \):
\[ U_{\text{final}} = \frac{KQ^2}{2R} \]
- The change in kinetic energy is given by:
\[ KE = \frac{1}{2} MV^2 \]

6. **Equating Energies**:
- Setting the work done equal to the change in kinetic energy:
\[ \frac{KQ^2}{2R} - \frac{KQ^2}{2R_0} = \frac{1}{2} MV^2 \]
- Rearranging gives:
\[ \frac{KQ^2}{2R} - \frac{KQ^2}{2R_0} = \frac{1}{2} MV^2 \]

7. **Solving for Speed \( V \)**:
- Rearranging the equation, we find:
\[ MV^2 = KQ^2 \left( \frac{1}{R} - \frac{1}{R_0} \right) \]
- Thus, the speed \( V \) can be expressed as:
\[ V = \sqrt{\frac{KQ^2}{M} \left( \frac{1}{R_0} - \frac{1}{R} \right)} \]

8. **Differentiating with Respect to Radius**:
- To understand how \( V \) changes with \( R \), we differentiate:
\[ \frac{dV}{dR} = \frac{KQ^2}{M} \cdot \frac{1}{2V} \cdot \left( -\frac{1}{R^2} \right) \]
- This shows that \( \frac{dV}{dR} \) is negative, indicating that as \( R \) increases, \( V \) decreases.

9. **Conclusion**:
- The speed \( V \) as a function of the instantaneous radius \( R \) decreases with increasing \( R \). This means that the graph of \( V \) versus \( R \) will have a negative slope.

### Final Answer:
The correct figure that represents the speed \( V(r(t)) \) of the distribution as a function of its instantaneous radius \( R(t) \) is option **third**.