Statement I: When air between the plates of a parallel plate condenser is replaced by an insulating dielectric medium, its capacity decreases.
Statement II: Electric field intensity between the plates with dielectric in between is reduced.

Options

A. Statement I is true, Statement II is True, Statement II is a correct explanation for statement I.
B. Statement I is true, Statement II is True, Statement II is Not a correct explanation for statement I.
C. Statement I is True, Statement II is False.
D. Statement I is False, Statement II is True.

Explanation

To analyze the given statements, we will break down the concepts of capacitance and electric field intensity in a parallel plate capacitor.

### Step 1: Understanding Capacitance
The capacitance \( C \) of a parallel plate capacitor in a vacuum (or air) is given by the formula:
\[ C_v = \frac{A \epsilon_0}{d} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( d \) is the separation between the plates.

When a dielectric medium with permittivity \( \epsilon \) is introduced between the plates, the capacitance becomes:
\[ C_m = \frac{A \epsilon}{d} = \frac{A \epsilon_0 k}{d} \]
where \( k \) is the dielectric constant of the material, and \( k \geq 1 \).

### Step 2: Analyzing Statement I
Statement I claims that replacing air with a dielectric medium decreases the capacitance.

From the formulas above, we see that:
\[ C_m = k C_v \]
Since \( k \) is always greater than or equal to 1, it follows that:
\[ C_m \geq C_v \]
Thus, the capacitance increases when a dielectric is introduced. Therefore, Statement I is **false**.

### Step 3: Understanding Electric Field Intensity
The electric field intensity \( E \) between the plates of a capacitor without a dielectric is given by:
\[ E_v = \frac{\sigma}{\epsilon_0} \]
where \( \sigma \) is the surface charge density.

When a dielectric is present, the electric field intensity becomes:
\[ E_m = \frac{\sigma}{\epsilon} = \frac{\sigma}{\epsilon_0 k} \]
This shows that:
\[ E_m = \frac{E_v}{k} \]
Since \( k > 1 \), it follows that:
\[ E_m < E_v \]
Thus, the electric field intensity is reduced when a dielectric is introduced. Therefore, Statement II is **true**.

### Conclusion
- **Statement I** is false: The capacitance increases when a dielectric is introduced.
- **Statement II** is true: The electric field intensity decreases with the introduction of a dielectric.

### Final Answer
- Statement I: False
- Statement II: True