A cylindrical tube open at both the ends has a
fundamental frequency of 390 Hz in air. If `1/4` th of the tube is immesed vertically in water the fundamental
frequency of air column is

Options

A. (a)260 Hz
B. (b)130 Hz
C. (c)390 Hz
D. (d)520 Hz

Answer

To solve the problem step by step, we will follow the principles of sound waves in tubes and the relationship between frequency, wavelength, and length of the tube.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a cylindrical tube that is open at both ends.
- The fundamental frequency of the tube in air is given as \( f = 390 \, \text{Hz} \).
- When \( \frac{1}{4} \) of the tube is immersed in water, we need to find the new fundamental frequency of the air column.

2. **Determine the Length of the Tube**:
- The fundamental frequency of a tube open at both ends is given by the formula:
\[ f = \frac{v}{2L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the tube.
- Rearranging the formula gives:
\[ L = \frac{v}{2f} \]

3. **Calculate the Length of the Tube**:
- We need the speed of sound in air, which is approximately \( v = 343 \, \text{m/s} \) (at room temperature).
- Substituting the values into the length formula:
\[ L = \frac{343 \, \text{m/s}}{2 \times 390 \, \text{Hz}} = \frac{343}{780} \approx 0.440 \, \text{m} \]

4. **Determine the New Length of the Air Column**:
- When \( \frac{1}{4} \) of the tube is immersed in water, the length of the air column becomes:
\[ L' = L - \frac{1}{4}L = \frac{3}{4}L \]
- Therefore:
\[ L' = \frac{3}{4} \times 0.440 \, \text{m} = 0.330 \, \text{m} \]

5. **Calculate the New Fundamental Frequency**:
- The new fundamental frequency \( f' \) for the air column is given by:
\[ f' = \frac{v}{2L'} \]
- Substituting \( L' \):
\[ f' = \frac{343 \, \text{m/s}}{2 \times 0.330 \, \text{m}} = \frac{343}{0.660} \approx 519.7 \, \text{Hz} \]

6. **Final Calculation**:
- Rounding off, the new fundamental frequency is approximately:
\[ f' \approx 520 \, \text{Hz} \]

### Conclusion:
The fundamental frequency of the air column when \( \frac{1}{4} \) of the tube is immersed in water is approximately **520 Hz**.