At a distance of `5 cm and 10 cm` outward from the surface of a uniformly charged solid sphere, the potentials are `100 V and 75 V`, repectively. Then.

Options

A. potential at its surface is `150 V`
B. the charge on the shre is `50/3xx10^-10C`
C. the electric field on the surface is `1500 V//m`
D. the electric potential at its centre is `25 V`

Explanation

To solve the problem, we need to find the radius of a uniformly charged solid sphere given the potentials at two points outside the sphere. We will denote the radius of the sphere as \( R \) and the charge on the sphere as \( Q \).

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given the potentials at distances of \( 5 \, \text{cm} \) and \( 10 \, \text{cm} \) from the surface of the sphere. Therefore, the distances from the center of the sphere are:
- Distance 1: \( r_1 = R + 5 \, \text{cm} \)
- Distance 2: \( r_2 = R + 10 \, \text{cm} \)

2. **Using the Formula for Electric Potential:**
The electric potential \( V \) at a distance \( r \) from the center of a uniformly charged sphere is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant.

3. **Setting Up the Equations:**
From the problem, we have:
- At \( r_1 = R + 5 \):
\[ V_1 = \frac{kQ}{R + 5} = 100 \, \text{V} \]
- At \( r_2 = R + 10 \):
\[ V_2 = \frac{kQ}{R + 10} = 75 \, \text{V} \]

4. **Dividing the Two Equations:**
We can divide the two equations to eliminate \( kQ \):
\[ \frac{V_1}{V_2} = \frac{\frac{kQ}{R + 5}}{\frac{kQ}{R + 10}} \implies \frac{100}{75} = \frac{R + 10}{R + 5} \]
Simplifying gives:
\[ \frac{4}{3} = \frac{R + 10}{R + 5} \]

5. **Cross Multiplying:**
Cross-multiplying gives:
\[ 4(R + 5) = 3(R + 10) \]
Expanding both sides:
\[ 4R + 20 = 3R + 30 \]

6. **Solving for \( R \):**
Rearranging the equation:
\[ 4R - 3R = 30 - 20 \implies R = 10 \, \text{cm} \]

7. **Conclusion:**
The radius of the uniformly charged solid sphere is \( R = 10 \, \text{cm} \).