Assertion: Electric potential on the surface of a charged sphere of radius R is V. Then electric field at distance `r=R/2` from centre is `V/(2R)`. Charge is distributed uniformly over the volume.
Reason: From centre to surface, electric field varies linearly with r. Here r is distance from centre.

Options

A. If both Assertion and Reason are true but Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. If Assertion is true, but the Reason is false.
D. If Assertion is false but the Reason is true.

Explanation

To solve the question, we need to analyze the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that the electric potential \( V \) on the surface of a uniformly charged sphere of radius \( R \) leads to an electric field at a distance \( r = \frac{R}{2} \) from the center being \( \frac{V}{2R} \).

2. **Electric Potential on the Surface**:
- The electric potential \( V \) on the surface of a charged sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant and \( Q \) is the total charge on the sphere.

3. **Using Gauss's Law**:
- To find the electric field \( E \) inside the sphere at a distance \( r = \frac{R}{2} \), we apply Gauss's Law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- The charge enclosed \( Q_{\text{enc}} \) within a radius \( r \) (where \( r < R \)) is given by:
\[ Q_{\text{enc}} = \rho \cdot \text{Volume} = \rho \cdot \frac{4}{3}\pi r^3 \]
- The volume charge density \( \rho \) can be expressed as:
\[ \rho = \frac{Q}{\frac{4}{3}\pi R^3} \]
- Substituting this into the expression for \( Q_{\text{enc}} \):
\[ Q_{\text{enc}} = \frac{Q}{\frac{4}{3}\pi R^3} \cdot \frac{4}{3}\pi r^3 = Q \cdot \frac{r^3}{R^3} \]

4. **Calculating Electric Field**:
- Now, substituting \( Q_{\text{enc}} \) into Gauss's Law:
\[ E \cdot 4\pi r^2 = \frac{Q \cdot \frac{r^3}{R^3}}{\epsilon_0} \]
- Rearranging gives:
\[ E = \frac{Q}{4\pi \epsilon_0 R^3} \cdot r \]
- This shows that the electric field \( E \) varies linearly with \( r \).

5. **Finding Electric Field at \( r = \frac{R}{2} \)**:
- Substituting \( r = \frac{R}{2} \):
\[ E\left(\frac{R}{2}\right) = \frac{Q}{4\pi \epsilon_0 R^3} \cdot \frac{R}{2} = \frac{Q}{8\pi \epsilon_0 R^2} \]

6. **Relating Electric Field to Potential**:
- From the potential \( V = \frac{kQ}{R} \), we can express \( Q \) as:
\[ Q = \frac{VR}{k} \]
- Substituting this into our expression for \( E \):
\[ E\left(\frac{R}{2}\right) = \frac{V}{2R} \]

7. **Conclusion**:
- The assertion is true: the electric field at \( r = \frac{R}{2} \) is \( \frac{V}{2R} \).
- The reason provided states that the electric field varies linearly with \( r \), which is also true, but it does not directly explain the assertion.

### Final Answer:
- The assertion is true, and the reason is true but does not correctly explain the assertion.