In a resonance tube experiment, a closed organ pipe of diameter 10 cm and of length 120 cm resonates, when tuned with a tuning fork of frequency 340 Hz. If water is poured into the pipe, then (speed of sound in air `340ms^(-1)`).

Options

A. minimum length of water column to have resonance is 50 cm
B. the distance between two successive nodes is 50 cm
C. the maximum length of water column to create resonance is 98 cm
D. the minimum length of water column to have resonance is 45 cm

Explanation

To solve the problem step by step, we will analyze the resonance tube experiment involving a closed organ pipe and the effects of pouring water into it.

### Step 1: Understand the relationship between speed, frequency, and wavelength
The speed of sound in air is given as \( V = 340 \, \text{m/s} \) and the frequency of the tuning fork is \( f = 340 \, \text{Hz} \). We can use the formula:
\[ V = f \cdot \lambda \]
where \( \lambda \) is the wavelength.

### Step 2: Calculate the wavelength
Substituting the values into the equation:
\[ 340 \, \text{m/s} = 340 \, \text{Hz} \cdot \lambda \]
Dividing both sides by \( 340 \, \text{Hz} \):
\[ \lambda = \frac{340 \, \text{m/s}}{340 \, \text{Hz}} = 1 \, \text{m} \]

### Step 3: Determine the length of the water column for resonance
In a closed organ pipe, the fundamental frequency (first harmonic) corresponds to a quarter of the wavelength fitting into the length of the pipe. Thus, the length of the air column \( L \) can be expressed as:
\[ L = \frac{\lambda}{4} \]
Substituting the value of \( \lambda \):
\[ L = \frac{1 \, \text{m}}{4} = 0.25 \, \text{m} = 25 \, \text{cm} \]

### Step 4: Calculate the maximum length of the water column
The total length of the pipe is \( 120 \, \text{cm} \). When water is poured into the pipe, the maximum length of the air column that can resonate is:
\[ L_{\text{max}} = 120 \, \text{cm} - 25 \, \text{cm} = 95 \, \text{cm} \]

### Step 5: Determine the next harmonic length
The next harmonic (third harmonic) corresponds to three-quarters of the wavelength fitting into the length of the pipe:
\[ L = \frac{3\lambda}{4} \]
Substituting the value of \( \lambda \):
\[ L = \frac{3 \cdot 1 \, \text{m}}{4} = 0.75 \, \text{m} = 75 \, \text{cm} \]

### Step 6: Calculate the minimum length of the water column
The minimum length of the air column that can resonate when the water is poured is:
\[ L_{\text{min}} = 120 \, \text{cm} - 75 \, \text{cm} = 45 \, \text{cm} \]

### Step 7: Determine the distance between two nodes
The distance between two consecutive nodes in a standing wave is given by:
\[ \text{Distance between nodes} = \frac{\lambda}{2} \]
Substituting the value of \( \lambda \):
\[ \text{Distance} = \frac{1 \, \text{m}}{2} = 0.5 \, \text{m} = 50 \, \text{cm} \]

### Final Summary
- Maximum length of the water column for resonance: \( 95 \, \text{cm} \)
- Minimum length of the water column for resonance: \( 45 \, \text{cm} \)
- Distance between two nodes: \( 50 \, \text{cm} \)