(A ): The surface densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal.
(R ): Surface density is equal to charge per unit area.

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 problem, we need to analyze the statements given:

**Statement (A)**: The surface densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal.

**Statement (R)**: Surface density is equal to charge per unit area.

### Step-by-Step Solution:

1. **Understanding Surface Charge Density**:
- Surface charge density (σ) is defined as the charge (Q) per unit area (A) on the surface of a conductor.
- Mathematically, it is expressed as:
\[ \sigma = \frac{Q}{A} \]

2. **Electric Field Due to a Charged Sphere**:
- For a spherical conductor, the electric field (E) just outside its surface can be derived from Gauss's law. The electric field intensity near the surface of a charged sphere is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r^2} \]
- Here, \( r \) is the radius of the sphere.

3. **Relating Surface Charge Density to Electric Field**:
- The electric field just outside the surface can also be expressed in terms of surface charge density:
\[ E = \frac{\sigma}{\epsilon_0} \]
- Therefore, if the surface charge densities of two spheres are equal (σ1 = σ2), the electric fields near their surfaces can be expressed as:
\[ E_1 = \frac{\sigma_1}{\epsilon_0} \quad \text{and} \quad E_2 = \frac{\sigma_2}{\epsilon_0} \]

4. **Comparing Electric Fields**:
- If σ1 = σ2, then:
\[ E_1 = E_2 \]
- Thus, the electric fields near the surfaces of the two spherical conductors are equal.

5. **Conclusion for Statement (A)**:
- Since the electric fields near the surfaces of the two spherical conductors are equal when their surface charge densities are equal, Statement (A) is **True**.

6. **Analyzing Statement (R)**:
- The statement (R) claims that surface density is equal to charge per unit area, which is indeed correct as per the definition of surface charge density (σ = Q/A).
- Therefore, Statement (R) is also **True**.

### Final Assessment:
- Since both statements (A) and (R) are true, the correct answer is that both statements are true.