Assertion : An electric bulb is first connected to a dc source and then to a ac source having the same brightness in both the cases.
Reason : The peak value of voltage for an A.C. source is `sqrt(2)` times the root mean square voltage.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is True
D. Statement-1 is False , Statement-2 is True

Answer

To solve the question, we need to analyze both the assertion and the reason provided.

### Step 1: Understand the Assertion
The assertion states that an electric bulb connected to a DC source and then to an AC source has the same brightness in both cases.

**Analysis**:
- Brightness of a bulb is proportional to the power it dissipates, which depends on the voltage across it.
- For a DC source, the voltage is constant, say \( V_{DC} \).
- For an AC source, the effective voltage (RMS voltage) is used to calculate power, which is given as \( V_{RMS} \).

### Step 2: Understand the Reason
The reason states that the peak value of voltage for an AC source is \( \sqrt{2} \) times the root mean square (RMS) voltage.

**Analysis**:
- This is a known fact in AC circuits: \( V_{peak} = \sqrt{2} \times V_{RMS} \).
- If the AC source has the same RMS voltage as the DC source, the peak voltage of the AC source will be higher than the DC voltage.

### Step 3: Compare Brightness in Both Cases
- For a DC voltage of \( V_{DC} \), the power \( P_{DC} \) can be expressed as:
\[ P_{DC} = \frac{V_{DC}^2}{R} \]

- For an AC voltage with the same RMS value \( V_{AC} = V_{RMS} \):
\[ P_{AC} = \frac{V_{RMS}^2}{R} \]

- If \( V_{DC} = V_{RMS} \), then:
\[ P_{AC} = \frac{V_{DC}^2}{R} \]

- However, the peak voltage \( V_{peak} \) for the AC source will be:
\[ V_{peak} = \sqrt{2} \times V_{RMS} = \sqrt{2} \times V_{DC} \]

### Step 4: Conclusion
Since the peak voltage for the AC source is higher than the DC voltage, the power (and thus brightness) of the bulb when connected to the AC source will be greater than when connected to the DC source. Therefore, the assertion is false.

### Final Answer
- **Assertion**: False
- **Reason**: True
- The correct option is D: Statement 1 is false and statement 2 is true.