1. Potential of a uniformly charged solid sphere

For a uniformly charged solid sphere of radius $R$ and total charge $Q$:

• Outside the sphere $(r \ge R)$, $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$
• On the surface $(r=R)$, $V_0=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$

So outside, $V(r)=V_0\frac{R}{r}$

• Inside the sphere $(r\le R)$, $V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{2R}\left(3-\frac{r^2}{R^2}\right)$ Using $V_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$, $V(r)=\frac{V_0}{2}\left(3-\frac{r^2}{R^2}\right)$

Thus, $V_{\text{inside}}=\frac{V_0}{2}\left(3-\frac{r^2}{R^2}\right), \qquad V_{\text{outside}}=V_0\frac{R}{r}$

2. Find $R_1$ for potential $\dfrac{3V_0}{2}$

Since the maximum potential occurs at the center: $V(0)=\frac{V_0}{2}(3)=\frac{3V_0}{2}$ So the equipotential surface for $\dfrac{3V_0}{2}$ is just the center point.

Hence, $R_1=0$

3. Find $R_2$ for potential $\dfrac{5V_0}{4}$

This value is greater than $V_0$, so it lies inside the sphere.

Use $\frac{V_0}{2}\left(3-\frac{r^2}{R^2}\right)=\frac{5V_0}{4}$ Cancel $V_0$: $\frac{1}{2}\left(3-\frac{r^2}{R^2}\right)=\frac{5}{4}$ $3-\frac{r^2}{R^2}=\frac{5}{2}$ $\frac{r^2}{R^2}=\frac{1}{2}$ $R_2=\frac{R}{\sqrt{2}}$

4. Find $R_3$ for potential $\dfrac{3V_0}{4}$

This is less than $V_0$, so it lies outside the sphere.

Use $V_0\frac{R}{r}=\frac{3V_0}{4}$ $\frac{R}{r}=\frac{3}{4}$ $R_3=\frac{4R}{3}$

5. Find $R_4$ for potential $\dfrac{V_0}{4}$

Again outside the sphere: $V_0\frac{R}{r}=\frac{V_0}{4}$ $\frac{R}{r}=\frac{1}{4}$ $R_4=4R$

6. Check each option

Option A:

Claims: $R_1=0$ and $R_2<(R_4-R_3)$

We have $R_4-R_3=4R-\frac{4R}{3}=\frac{8R}{3}$ And $R_2=\frac{R}{\sqrt{2}}$ Clearly, $\frac{R}{\sqrt{2}}<\frac{8R}{3}$ So Option A is correct.

Option B:

Claims: $2R<R_4$ Since $R_4=4R$ we get $2R<4R$ So Option B is correct.

Option C:

Claims: $R_1=0$ and $R_2>(R_4-R_3)$ But we found $\frac{R}{\sqrt{2}}<\frac{8R}{3}$ So Option C is false.

Option D:

Claims: $R_1\ne 0$ and $(R_2-R_1)>(R_4-R_3)$ But $R_1=0$, so first part itself is false. Also, $R_2-R_1=\frac{R}{\sqrt{2}}$ which is not greater than $\frac{8R}{3}$. So Option D is false.

7. Final answer

The correct options are: $\boxed{A,\ B}$

8. Comparison with stored answer

Stored correct answer: A, B

My derived answer: A, B

They match.