A uniformly charged solid sphere of radius `R` has potential `V_(0)` (measured with respect to `oo`) on its surface. For this sphere the equipotential surfaces with potentials `(3V_(0))/(2) , (5 V_(0))/(4), (3V_(0))/(4)` and `(V_(0))/(4)` have radius `R_(1), R_(2), R_(3)` and `R_(4)` respecatively. Then

Options

A. `R_1=0 and R_2 gt (R_4 - R_3)`
B. `R_(1) ne 0 and (R_(2) - R_(1)) gt (R_(4) - R_(3))`
C. `R_1 =0 and R_2 lt(R_4 - R_3)`
D. `2R gtR_4`

Explanation

To solve the problem, we need to find the radii \( R_1, R_2, R_3, \) and \( R_4 \) corresponding to the given potentials \( \frac{3V_0}{2}, \frac{5V_0}{4}, \frac{3V_0}{4}, \) and \( \frac{V_0}{4} \) respectively for a uniformly charged solid sphere of radius \( R \) and surface potential \( V_0 \).

### Step-by-Step Solution:

1. **Understanding the Potential Inside and Outside the Sphere:**
The potential \( V \) at a distance \( r \) from the center of a uniformly charged solid sphere is given by:
\[ V(r) = \begin{cases} \frac{kQ}{R} & \text{for } r \geq R \\ \frac{kQ}{R} - \frac{kQ}{2R} + \frac{kQ}{3R} & \text{for } r < R \end{cases} \]
where \( k \) is Coulomb's constant and \( Q \) is the total charge of the sphere.

2. **Setting the Surface Potential:**
The potential at the surface of the sphere (at \( r = R \)) is given as \( V_0 \):
\[ V_0 = \frac{kQ}{R} \]

3. **Finding the Radius for Each Given Potential:**
We will use the potential formula to find the corresponding radius for each potential.

- **For \( V = \frac{3V_0}{2} \):**
\[ \frac{3V_0}{2} = V_0 \left( \frac{3}{2} - \frac{r_1^2}{R^2} \right) \]
Simplifying gives:
\[ \frac{3}{2} = \frac{3}{2} - \frac{r_1^2}{R^2} \implies r_1^2 = 0 \implies r_1 = 0 \]

- **For \( V = \frac{5V_0}{4} \):**
\[ \frac{5V_0}{4} = V_0 \left( \frac{3}{2} - \frac{r_2^2}{R^2} \right) \]
Simplifying gives:
\[ \frac{5}{4} = \frac{3}{2} - \frac{r_2^2}{R^2} \implies \frac{5}{4} = \frac{6}{4} - \frac{r_2^2}{R^2} \implies \frac{r_2^2}{R^2} = \frac{1}{4} \implies r_2 = \frac{R}{2} \]

- **For \( V = \frac{3V_0}{4} \):**
\[ \frac{3V_0}{4} = V_0 \left( \frac{3}{2} - \frac{r_3^2}{R^2} \right) \]
Simplifying gives:
\[ \frac{3}{4} = \frac{3}{2} - \frac{r_3^2}{R^2} \implies \frac{3}{4} = \frac{6}{4} - \frac{r_3^2}{R^2} \implies \frac{r_3^2}{R^2} = \frac{3}{4} \implies r_3 = \frac{\sqrt{3}}{2}R \]

- **For \( V = \frac{V_0}{4} \):**
\[ \frac{V_0}{4} = V_0 \left( \frac{3}{2} - \frac{r_4^2}{R^2} \right) \]
Simplifying gives:
\[ \frac{1}{4} = \frac{3}{2} - \frac{r_4^2}{R^2} \implies \frac{1}{4} = \frac{6}{4} - \frac{r_4^2}{R^2} \implies \frac{r_4^2}{R^2} = \frac{5}{4} \implies r_4 = \frac{R\sqrt{5}}{2} \]

4. **Final Results:**
- \( r_1 = 0 \)
- \( r_2 = \frac{R}{2} \)
- \( r_3 = \frac{\sqrt{3}}{2}R \)
- \( r_4 = \frac{R\sqrt{5}}{2} \)