What will be the wavelength and pitch of the note emitted by a closed organ pipe 32.4 cm long at `0^@C` if the velocity of sound in air at `0^@C` is `332 ms^(-1)`.

Answer

To solve the problem of finding the wavelength and pitch of the note emitted by a closed organ pipe, we will follow these steps:

### Step 1: Convert the length of the closed organ pipe to meters
The length of the closed organ pipe is given as 32.4 cm. We need to convert this into meters.
- \( L = 32.4 \, \text{cm} = \frac{32.4}{100} \, \text{m} = 0.324 \, \text{m} \)

### Step 2: Use the formula for frequency of a closed organ pipe
For a closed organ pipe, the fundamental frequency \( f \) is given by the formula:
\[ f = \frac{v}{4L} \]
where \( v \) is the velocity of sound and \( L \) is the length of the pipe.

### Step 3: Substitute the values into the frequency formula
We know:
- \( v = 332 \, \text{m/s} \)
- \( L = 0.324 \, \text{m} \)

Now substituting these values into the frequency formula:
\[ f = \frac{332 \, \text{m/s}}{4 \times 0.324 \, \text{m}} \]

### Step 4: Calculate the frequency
Calculating the denominator:
\[ 4 \times 0.324 = 1.296 \, \text{m} \]

Now substituting back:
\[ f = \frac{332}{1.296} \approx 256.2 \, \text{Hz} \]

### Step 5: Determine the pitch
The pitch of a sound is directly proportional to its frequency. Therefore, the pitch of the note emitted by the organ pipe is also:
\[ \text{Pitch} \approx 256.2 \, \text{Hz} \]

### Step 6: Calculate the wavelength
The wavelength \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]

### Step 7: Substitute the values into the wavelength formula
Substituting the values we have:
\[ \lambda = \frac{332 \, \text{m/s}}{256.2 \, \text{Hz}} \]

### Step 8: Calculate the wavelength
Calculating:
\[ \lambda \approx 1.296 \, \text{m} \]

### Final Answers
- **Wavelength (\( \lambda \))**: 1.296 m
- **Pitch**: 256.2 Hz

---