(A ): Sharper is the curvature of spot on a charged body lesser will be the surface density of charge at that point.
(R ): Electric field is zero inside a charged conductor.

Options

A. Both .A. and .R. are true and .R. is the correct explanation of .A.
B. Both .A. and .R. are tru and .R. is not the correct explanation of .A.
C. A. is true and .R. is false
D. A. is false and .R. is true

Answer

To solve the given question, we need to analyze both the assertion (A) and the reason (R) provided.

### Step-by-Step Solution:

1. **Understanding the Assertion (A)**:
- The assertion states: "Sharper is the curvature of a spot on a charged body, lesser will be the surface density of charge at that point."
- Surface charge density (σ) is defined as the amount of charge (Q) per unit area (A) on the surface:
\[ \sigma = \frac{Q}{A} \]
- If the curvature of a spot is sharper, it implies that the area (A) at that point is smaller.

2. **Analyzing the Relationship**:
- Since surface charge density is inversely proportional to the area:
- If the area (A) is smaller (due to sharper curvature), the surface charge density (σ) must be greater, not lesser.
- Therefore, the assertion is incorrect because it states the opposite.

3. **Understanding the Reason (R)**:
- The reason states: "Electric field is zero inside a charged conductor."
- According to electrostatic principles, when a conductor is charged, the charges redistribute themselves on the surface of the conductor.
- Inside the conductor, the electric field (E) is indeed zero because the charges balance out any internal electric fields.

4. **Conclusion**:
- Assertion (A) is false because it incorrectly states the relationship between curvature and surface charge density.
- Reason (R) is true as it accurately describes the behavior of electric fields inside a charged conductor.
- Therefore, the correct conclusion is that A is false and R is true.

### Final Answer:
- Assertion (A) is false.
- Reason (R) is true.