Assertion: In case of charged spherical shells, E-r graph is discontinuous while V-r graph is continuous
Reason: According to Gauss's theorem only the charge inside a closed surface can produce electric field at some point.

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, and then determine their validity based on the principles of electrostatics, particularly focusing on charged spherical shells.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that in the case of charged spherical shells, the electric field (E) versus radius (r) graph is discontinuous, while the potential (V) versus radius (r) graph is continuous.
- For a charged spherical shell, inside the shell (r < R, where R is the radius of the shell), the electric field is zero. Outside the shell (r > R), the electric field behaves as if all the charge were concentrated at the center (E = kQ/r²).
- Thus, the electric field graph shows a jump from 0 (inside) to kQ/R² (on the surface) and then follows the kQ/r² curve outside, indicating a discontinuity at r = R.

2. **Analyzing the Potential**:
- The electric potential inside the shell remains constant (V = kQ/R) since the electric field is zero. At the surface, the potential is also kQ/R, and outside the shell, it decreases as V = kQ/r.
- Therefore, the potential graph is continuous across the boundary at r = R, as it does not have a jump; it remains constant inside and matches the value at the surface.

3. **Understanding the Reason**:
- The reason states that according to Gauss's theorem, only the charge inside a closed surface can produce an electric field at some point.
- This statement is partially true but misleading. While it is correct that only the enclosed charge contributes to the electric field at a point inside the Gaussian surface, charges outside the surface can still influence the electric field at points outside the surface.

4. **Conclusion**:
- The assertion is true because the E-r graph is indeed discontinuous while the V-r graph is continuous.
- The reason is false because it incorrectly implies that charges outside a closed surface cannot produce an electric field at points outside that surface.

### Final Answer:
- The assertion is true, and the reason is false.