An organ pipe of cross-sectional area 100 `cm^(2)` resonates with a tuning fork of frequency 1000 Hz in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is
(Take velocity of wave = 320 m `s^(-1)`)

Options

A. `800 cm^(3)`
B. `1200 cm^(3)`
C. `1600 cm^(3)`
D. `2000 cm^(3)`

Explanation

To solve the problem step by step, we need to determine the lengths of the air column for the fundamental frequency and the first overtone in a closed organ pipe, and then calculate the volume of water that needs to be drained to achieve this.

### Step 1: Understand the relationship between frequency, velocity, and length of the air column.
For a closed organ pipe, the fundamental frequency (first harmonic) is given by:
\[ f_1 = \frac{V}{4L_1} \]
Where:
- \( V \) is the velocity of sound in air (320 m/s),
- \( L_1 \) is the length of the air column for the fundamental frequency.

### Step 2: Calculate the length of the air column for the fundamental frequency.
Rearranging the formula to find \( L_1 \):
\[ L_1 = \frac{V}{4f_1} \]
Substituting the values:
\[ L_1 = \frac{320 \, \text{m/s}}{4 \times 1000 \, \text{Hz}} \]
\[ L_1 = \frac{320}{4000} \]
\[ L_1 = 0.08 \, \text{m} = 8 \, \text{cm} \]

### Step 3: Calculate the length of the air column for the first overtone.
The frequency for the first overtone (third harmonic) is given by:
\[ f_2 = \frac{3V}{4L_2} \]
Where \( L_2 \) is the length of the air column for the first overtone.

Rearranging the formula to find \( L_2 \):
\[ L_2 = \frac{3V}{4f_2} \]
Substituting the values:
\[ L_2 = \frac{3 \times 320 \, \text{m/s}}{4 \times 1000 \, \text{Hz}} \]
\[ L_2 = \frac{960}{4000} \]
\[ L_2 = 0.24 \, \text{m} = 24 \, \text{cm} \]

### Step 4: Determine the difference in lengths of the air columns.
The difference in lengths of the air columns is:
\[ \Delta L = L_2 - L_1 = 24 \, \text{cm} - 8 \, \text{cm} = 16 \, \text{cm} \]

### Step 5: Calculate the volume of water to be drained.
The volume of water drained can be calculated using the cross-sectional area of the pipe and the difference in lengths:
\[ \text{Volume} = \text{Area} \times \Delta L \]
Given the area is 100 cm²:
\[ \text{Volume} = 100 \, \text{cm}^2 \times 16 \, \text{cm} = 1600 \, \text{cm}^3 \]

### Final Answer:
The minimum volume of water to be drained so that the pipe again resonates with the same tuning fork is **1600 cm³**.

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