An air column in a pipe, when is closed at one end, is in resonance with a vibrating tuning fork of frequency `264 H_(Z)`. If `upsilon = 330 m//s`, the length of the column in cm is (are)

Options

A. 31.25
B. 62.5
C. 93.75
D. 125

Answer

To find the length of the air column in a pipe closed at one end that is in resonance with a vibrating tuning fork, we can use the formula for the fundamental frequency of a closed organ pipe.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency (f) = 264 Hz
- Speed of sound in air (v) = 330 m/s

2. **Use the Formula for the Fundamental Frequency:**
The fundamental frequency (f) of a closed organ pipe is given by the formula:
\[ f = \frac{v}{4L} \]
where \(L\) is the length of the air column.

3. **Rearranging the Formula:**
To find the length \(L\), we can rearrange the formula:
\[ L = \frac{v}{4f} \]

4. **Substituting the Values:**
Now substitute the values of \(v\) and \(f\) into the equation:
\[ L = \frac{330 \, \text{m/s}}{4 \times 264 \, \text{Hz}} \]

5. **Calculating the Length:**
Calculate the denominator:
\[ 4 \times 264 = 1056 \]
Now, calculate \(L\):
\[ L = \frac{330}{1056} \approx 0.3125 \, \text{m} \]

6. **Convert to Centimeters:**
To convert the length from meters to centimeters, multiply by 100:
\[ L = 0.3125 \times 100 = 31.25 \, \text{cm} \]

7. **Conclusion:**
The length of the air column in the pipe is \(31.25 \, \text{cm}\).

### Additional Options:
- To find other possible lengths for higher harmonics, we can use the odd harmonics (n = 1, 3, 5, ...):
- For n = 3:
\[ L = \frac{v}{4 \times 3f} = \frac{330}{4 \times 792} \approx 0.09375 \, \text{m} = 93.75 \, \text{cm} \]
- For n = 5, similar calculations can be done.

### Final Answer:
The correct options for the length of the column are:
- \(31.25 \, \text{cm}\) (for n = 1)
- \(93.75 \, \text{cm}\) (for n = 3)