A uniformly charged solid sphere of radius $\mathrm{R}$ has potential $\mathrm{V}_0$ (measured with respect to infinity) on its surface. For this sphere the equipotential surfaces with potentials $\dfrac{3 \mathrm{~V}_0}{2}, \dfrac{\mathrm{V}_0}{1}, \dfrac{3 \mathrm{~V}_0}{4}$ and $\dfrac{\mathrm{V}_0}{4}$ have radius $\mathrm{R}_1, \mathrm{R}_2, \mathrm{R}_3$ and $\mathrm{R}_4$ and respectively, then

The potential $V(r)$ of a uniformly charged solid sphere of radius $R$ and total charge $Q$ is given by:

$V(r) = \dfrac{kQ}{R} = V_0$ for $r \le R$

$V(r) = \dfrac{kQ}{r}$ for $r > R$

Given $V(R) = V_0$, we have $V_0 = \dfrac{kQ}{R}$.

For $r \le R$, the potential is constant at $V_0$. Since $\dfrac{3V_0}{2} > V_0$, no point inside or on the surface has this potential. Thus, $R_1$ does not exist or is not defined in the physical sense of the sphere, but the option $R_1=0$ is often used in such contexts to denote the impossibility of the condition. However, checking the potential values: $V_0 = V_0$ at $r=R$, so $R_2 = R$.

For $r > R$, $V(r) = \dfrac{V_0 R}{r}$.

Equating potentials:

For $V = \dfrac{3V_0}{2}$: Since $\dfrac{3V_0}{2} > V_0$, this potential is only possible if $r < R$, but $V(r) = V_0$ for all $r \le R$. Thus $R_1$ is not a valid radius for this potential, often represented as $R_1=0$.

For $V = V_0$: $R_2 = R$.

For $V = \dfrac{3V_0}{4}$: $\dfrac{V_0 R}{R_3} = \dfrac{3V_0}{4} \Rightarrow R_3 = \dfrac{4R}{3}$.

For $V = \dfrac{V_0}{4}$: $\dfrac{V_0 R}{R_4} = \dfrac{V_0}{4} \Rightarrow R_4 = 4R$.

Now evaluate the condition $R_2 < (R_4 - R_3)$:

$R_2 = R$

$R_4 - R_3 = 4R - \dfrac{4R}{3} = \dfrac{8R}{3}$.

Since $R < \dfrac{8R}{3}$, the condition $R_2 < (R_4 - R_3)$ holds true.

Answer: $R_1=0 \text { and } R_2&lt;\left(R_4-R_3\right)$