An organ pipe filled with a gas at `27^(@) C` resonates at `400 Hz` in its fundamental mode. If it is filled with the same gas at `90^(@) C`, the resonance frequency will be

Options

A. `420 Hz`
B. `440 Hz`
C. `484 Hz`
D. `512 Hz`

Explanation

To solve the problem, we need to find the new resonance frequency of an organ pipe filled with gas when the temperature changes from \(27^\circ C\) to \(90^\circ C\). We will use the relationship between frequency and temperature.

### Step-by-step Solution:

1. **Understand the relationship between frequency and temperature**:
The frequency of sound in a gas is directly proportional to the square root of the absolute temperature of the gas. This can be expressed mathematically as:
\[ \frac{F_1}{F_2} = \sqrt{\frac{T_1}{T_2}} \]
where \(F_1\) and \(F_2\) are the frequencies at temperatures \(T_1\) and \(T_2\) respectively.

2. **Convert temperatures from Celsius to Kelvin**:
The absolute temperature in Kelvin can be calculated by adding 273 to the Celsius temperature.
- For \(T_1 = 27^\circ C\):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \(T_2 = 90^\circ C\):
\[ T_2 = 90 + 273 = 363 \, K \]

3. **Substitute known values into the frequency ratio equation**:
We know that \(F_1 = 400 \, Hz\) and we need to find \(F_2\):
\[ \frac{400}{F_2} = \sqrt{\frac{300}{363}} \]

4. **Calculate the square root of the temperature ratio**:
First, calculate the ratio:
\[ \frac{300}{363} \approx 0.825 \]
Now, take the square root:
\[ \sqrt{0.825} \approx 0.908 \]

5. **Rearranging the equation to find \(F_2\)**:
\[ F_2 = \frac{400}{0.908} \approx 440 \, Hz \]

6. **Conclusion**:
The resonance frequency of the organ pipe when filled with gas at \(90^\circ C\) will be approximately \(440 \, Hz\).

### Final Answer:
The resonance frequency at \(90^\circ C\) is \(440 \, Hz\).