A cylindrical tube (L = 125 cm) is resonant with a tuning fork of frequency 330 Hz. If it is filling by water then to get resonance again, minimum length of water column is (`V_air` = 330 m/s) -

Options

A. 50 cm
B. 60 cm
C. 25 cm
D. 20 cm

Explanation

To solve the problem, we need to find the minimum length of the water column in a cylindrical tube that allows for resonance with a tuning fork of frequency 330 Hz, given that the speed of sound in air is 330 m/s.

### Step-by-Step Solution:

1. **Identify the Length of the Tube**:
The total length of the cylindrical tube (L) is given as 125 cm, which we convert to meters:
\[ L = 125 \text{ cm} = 1.25 \text{ m} \]

2. **Calculate the Wavelength**:
We know the relationship between speed (V), frequency (f), and wavelength (λ):
\[ V = f \cdot \lambda \]
Rearranging gives us:
\[ \lambda = \frac{V}{f} \]
Substituting the given values:
\[ \lambda = \frac{330 \text{ m/s}}{330 \text{ Hz}} = 1 \text{ m} \]

3. **Determine Resonance Conditions**:
For a closed pipe (one end closed), the resonance occurs at odd multiples of quarter wavelengths. The lengths at which resonance occurs are given by:
\[ L_n = \frac{(2n + 1) \lambda}{4} \]
where \(n = 0, 1, 2, \ldots\)

The first resonance (n=0) occurs at:
\[ L_0 = \frac{1 \text{ m}}{4} = 0.25 \text{ m} = 25 \text{ cm} \]

The second resonance (n=1) occurs at:
\[ L_1 = \frac{3 \cdot 1 \text{ m}}{4} = 0.75 \text{ m} = 75 \text{ cm} \]

4. **Calculate the Length of the Air Column**:
The length of the air column at the second resonance is 75 cm. Since the total length of the tube is 125 cm, the length of the water column (L_water) can be calculated as:
\[ L_{\text{water}} = L - L_{\text{air}} = 125 \text{ cm} - 75 \text{ cm} = 50 \text{ cm} \]

5. **Conclusion**:
The minimum length of the water column required to achieve resonance again is:
\[ \boxed{50 \text{ cm}} \]