A tuning fork of frequency 500Hz when sounded on the mouth of resonance tube the lengths of air column for first and second resonance are recorded as 10cm and 40cm respectively. The room temperature is 20°C. The velocity of sound at the room temperature obtained from the explanation is

Options

A. `300 m//s`
B. `332 m//s`
C. `290 m//s`
D. `310 m//s`

Answer

To find the velocity of sound using the given data, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of the tuning fork, \( f = 500 \, \text{Hz} \)
- Length of air column for first resonance, \( L_1 = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Length of air column for second resonance, \( L_2 = 40 \, \text{cm} = 0.4 \, \text{m} \)

2. **Understand Resonance in the Tube:**
- At the first resonance, the length of the air column corresponds to \( \frac{\lambda}{4} \).
- At the second resonance, the length of the air column corresponds to \( \frac{3\lambda}{4} \).

3. **Set Up the Equation for Resonances:**
- From the first resonance:
\[ L_1 = \frac{\lambda}{4} \implies \lambda = 4L_1 = 4 \times 0.1 = 0.4 \, \text{m} \]
- From the second resonance:
\[ L_2 = \frac{3\lambda}{4} \implies \lambda = \frac{4L_2}{3} = \frac{4 \times 0.4}{3} \approx 0.533 \, \text{m} \]

4. **Calculate the Wavelength:**
- The difference in lengths between the first and second resonance gives:
\[ L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2} \]
- Therefore:
\[ L_2 - L_1 = 0.4 - 0.1 = 0.3 \, \text{m} \implies \frac{\lambda}{2} = 0.3 \implies \lambda = 0.6 \, \text{m} \]

5. **Calculate the Velocity of Sound:**
- The velocity of sound \( v \) can be calculated using the formula:
\[ v = f \cdot \lambda \]
- Substituting the known values:
\[ v = 500 \, \text{Hz} \times 0.6 \, \text{m} = 300 \, \text{m/s} \]

### Final Answer:
The velocity of sound at room temperature is \( 300 \, \text{m/s} \).