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 instant radius R(t) is :

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the behavior of a uniformly charged spherical surface as it expands due to mutual repulsion. We'll derive the relationship between the speed \( V(r(t)) \) of the charge distribution and its instantaneous radius \( R(t) \).

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a uniform spherical surface charge density at a distance \( R_0 \) from the origin.
- The charge distribution is initially at rest and begins to expand due to the mutual electrostatic repulsion between the charges.

2. **Energy Conservation**:
- The total energy of the charge distribution remains constant during the expansion.
- The total energy \( E \) can be expressed as the sum of kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]

3. **Kinetic Energy**:
- The kinetic energy \( K \) of the charge distribution can be expressed as:
\[ K = \frac{1}{2} mv^2 \]
- Here, \( m \) is the mass of the charge distribution and \( v \) is the speed of the charge distribution.

4. **Potential Energy**:
- The potential energy \( U \) due to the electrostatic interaction can be given by:
\[ U = -\frac{kQ^2}{R} \]
- Where \( k \) is Coulomb's constant and \( Q \) is the total charge.

5. **Setting Up the Energy Equation**:
- At the initial state (when the radius is \( R_0 \)):
\[ E = \frac{1}{2} mv^2 + U \]
- At the final state (when the radius is \( R(t) \)):
\[ E = 0 + U = -\frac{kQ^2}{R_0} \]

6. **Equating Energies**:
- Since total energy is conserved, we can write:
\[ \frac{1}{2} mv^2 - \frac{kQ^2}{R} = -\frac{kQ^2}{R_0} \]
- Rearranging gives:
\[ \frac{1}{2} mv^2 = -\frac{kQ^2}{R} + \frac{kQ^2}{R_0} \]

7. **Solving for Velocity**:
- We can express the velocity \( v \) in terms of \( R \):
\[ mv^2 = 2kQ^2\left(\frac{1}{R_0} - \frac{1}{R}\right) \]
- Thus,
\[ v = \sqrt{\frac{2kQ^2}{m}\left(\frac{1}{R_0} - \frac{1}{R}\right)} \]

8. **Analyzing the Function**:
- As the radius \( R \) increases, the term \( \left(\frac{1}{R_0} - \frac{1}{R}\right) \) decreases, which means that \( v \) decreases as \( R \) increases.
- Therefore, \( v(R) \) is a decreasing function of \( R \).

9. **Choosing the Correct Graph**:
- From the analysis, we conclude that the speed \( v(R) \) decreases as the radius \( R \) increases.
- Among the given options, we look for a graph that shows a decreasing trend as \( R \) increases.

### Conclusion:
Based on the analysis, the correct option that represents the speed \( V(r(t)) \) as a function of the instantaneous radius \( R(t) \) is **Graph C**, which shows a decreasing trend.