Two spheres of silver of same radii, one solid and the other hollow are charged to the same potential, then:

The correct option is: both spheres will have same charge.

Let us understand the physical principles that lead to this result step-by-step.

Step 1: Charge distribution on conductors
Silver is a highly conducting metal. In electrostatics, any excess charge given to a conductor distributes itself entirely on its outer surface. This is because like charges repel each other and move as far apart as possible, which is the outer boundary. Consequently, there is no electrostatic charge inside the body of a conductor under static conditions. This holds true whether the conducting sphere is solid or hollow.

Step 2: Relate potential to charge and radius
For a conducting sphere of radius $R$ carrying a charge $Q$, the electrostatic potential $V$ at its surface (and everywhere inside) is given by the relation:

$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$

where ${\epsilon }_0$ is the permittivity of free space.

Step 3: Analyze the given conditions
We are given two spheres made of the same material (silver):
1. One sphere is solid and the other is hollow.
2. They have the same radius, so $R_{\text{solid}}=R_{\text{hollow}}=R$.
3. They are charged to the same potential, so $V_{\text{solid}}=V_{\text{hollow}}=V$.

Rearranging the potential formula to solve for charge $Q$, we get:

$Q=4\pi {\epsilon }_{0}RV$

Since both $R$ and $V$ are identical for both the solid and hollow spheres, their charges must also be identical:
$Q_{\text{solid}}=Q_{\text{hollow}}$.

Thus, both spheres will acquire the exact same amount of charge.