On vibrating an air column at `627^(@)C` and a tuning fork simultaneously, `6` beats/sec are heard. The frequency of fork is less than that of air column. No beats are heard at `-48^(@)C`.The frequency of fork is

Options

A. `3Hz`
B. `6Hz`
C. `10Hz`
D. `15Hz`

Explanation

To solve the problem, we need to determine the frequency of the tuning fork based on the information given about the air column and the beats heard.

### Step-by-Step Solution:

1. **Understanding Beats**:
When two sound waves of slightly different frequencies interfere, they produce a phenomenon called "beats." The number of beats per second is equal to the absolute difference in their frequencies. In this case, we hear 6 beats per second.

2. **Assigning Variables**:
Let the frequency of the tuning fork be \( n \) Hz. The frequency of the air column at \( 627^\circ C \) is higher than that of the tuning fork since it is stated that the tuning fork's frequency is less than that of the air column.

3. **Frequency Relationship**:
Since we hear 6 beats per second, we can express the relationship between the frequencies as:
\[ f_{air} - n = 6 \]
where \( f_{air} \) is the frequency of the air column.

4. **Calculating Frequency of Air Column**:
The frequency of sound in air increases with temperature. The formula to calculate the frequency of sound in air is:
\[ f = 331 + 0.6 \times T \]
where \( T \) is the temperature in Celsius. For \( T = 627^\circ C \):
\[ f_{air} = 331 + 0.6 \times 627 = 331 + 376.2 = 707.2 \text{ Hz} \]

5. **Substituting into the Beats Equation**:
Now, substituting \( f_{air} \) into the beats equation:
\[ 707.2 - n = 6 \]
Rearranging gives:
\[ n = 707.2 - 6 = 701.2 \text{ Hz} \]

6. **Final Answer**:
The frequency of the tuning fork is approximately \( 701.2 \text{ Hz} \).

### Summary:
The frequency of the tuning fork is \( 701.2 \text{ Hz} \).