A tuning fork is used to produce resonance in glass tuve. The length of the air column in the tube can be adjusted by a variable piston. At room temperature of `27^@C` two succesive resonance are produced at 20 cm and 73 cm column length. If the frequency of the tuning fork is 320 Hz. the velocity of sound is air at `27^(@)C` is

Answer

To solve the problem, we need to find the velocity of sound in air at a temperature of 27°C using the information about the resonant lengths of the air column in the tube.

### Step-by-Step Solution:

1. **Identify the Lengths of Resonance:**
- The first resonance occurs at \( L_1 = 20 \, \text{cm} \).
- The second resonance occurs at \( L_2 = 73 \, \text{cm} \).

2. **Understand the Resonance in a Closed Tube:**
- In a closed tube, the resonating lengths correspond to odd multiples of \( \frac{\lambda}{4} \) (where \( \lambda \) is the wavelength).
- The first resonance (20 cm) corresponds to \( \frac{\lambda}{4} \).
- The second resonance (73 cm) corresponds to \( \frac{3\lambda}{4} \).

3. **Set Up the Equation:**
- The difference between the two lengths gives us:
\[ L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2} \]
- Substitute the lengths:
\[ 73 \, \text{cm} - 20 \, \text{cm} = \frac{\lambda}{2} \]
- This simplifies to:
\[ 53 \, \text{cm} = \frac{\lambda}{2} \]

4. **Calculate the Wavelength:**
- To find \( \lambda \):
\[ \lambda = 2 \times 53 \, \text{cm} = 106 \, \text{cm} = 1.06 \, \text{m} \]

5. **Use the Frequency to Find Velocity:**
- The formula for the velocity of sound is given by:
\[ V = f \cdot \lambda \]
- Where \( f = 320 \, \text{Hz} \) and \( \lambda = 1.06 \, \text{m} \):
\[ V = 320 \, \text{Hz} \times 1.06 \, \text{m} = 339.2 \, \text{m/s} \]

### Final Answer:
The velocity of sound in air at \( 27^\circ C \) is approximately \( 339.2 \, \text{m/s} \).