Assertion : Resonance frequency will decrease in `L-C-R` series circuit if a dielectric slab is inserted in between the plates of the capacitor.
Reason : By doing so, capacity of capacitor will increase.

Options

A. If both Assertion and Reason are true and the Reason is correct explanation of the 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 will analyze the assertion and reason provided, and then derive the relationship between capacitance, resonance frequency, and the effect of inserting a dielectric slab into a capacitor.

### Step-by-Step Solution:

1. **Understanding Resonance in L-C-R Circuit**:
- In an L-C-R series circuit, resonance occurs when the inductive reactance (X_L) equals the capacitive reactance (X_C).
- Mathematically, this is expressed as:
\[ X_L = X_C \]
- Where:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]

2. **Deriving the Resonance Frequency**:
- Setting \(X_L\) equal to \(X_C\):
\[ \omega L = \frac{1}{\omega C} \]
- Rearranging gives:
\[ \omega^2 = \frac{1}{LC} \]
- Therefore, the resonance frequency (\(f_0\)) is given by:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

3. **Effect of Inserting a Dielectric**:
- When a dielectric slab is inserted between the plates of a capacitor, the capacitance increases. The new capacitance (\(C'\)) can be expressed as:
\[ C' = kC \]
- Where \(k\) is the dielectric constant (greater than 1) of the material.

4. **Analyzing the Change in Resonance Frequency**:
- Substituting the new capacitance into the resonance frequency formula:
\[ f_0' = \frac{1}{2\pi\sqrt{L C'}} = \frac{1}{2\pi\sqrt{L (kC)}} = \frac{1}{2\pi\sqrt{kLC}} \]
- Since \(k > 1\), it follows that:
\[ \sqrt{k} > 1 \quad \Rightarrow \quad f_0' < f_0 \]
- Thus, the resonance frequency decreases when a dielectric slab is inserted.

5. **Conclusion**:
- The assertion is correct: the resonance frequency will decrease when a dielectric is inserted.
- The reason is also correct: inserting a dielectric increases the capacitance of the capacitor.

### Final Answer:
Both the assertion and reason are true, and the reason is the correct explanation of the assertion.