In a resonance tube with tuning fork of frequency `512 Hz`, first resonance occurs at water level equal to `30.3cm` and second resonance ocuurs at `63.7cm`. The maximum possible error in the speed of sound is

Options

A. 51.2 cm/s
B. 102.4 cm/s
C. 204.8 cm/s
D. 153.6 cm/s

Answer

To solve the problem, we need to determine the maximum possible error in the speed of sound based on the given resonance tube measurements.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of the tuning fork, \( f = 512 \, \text{Hz} \)
- First resonance length, \( L_1 = 30.3 \, \text{cm} = 0.303 \, \text{m} \)
- Second resonance length, \( L_2 = 63.7 \, \text{cm} = 0.637 \, \text{m} \)

2. **Calculate the Difference in Lengths:**
\[ \Delta L = L_2 - L_1 = 0.637 \, \text{m} - 0.303 \, \text{m} = 0.334 \, \text{m} \]

3. **Relate the Length Difference to the Speed of Sound:**
The difference in lengths corresponds to half a wavelength:
\[ \Delta L = \frac{V}{2f} \]
Rearranging gives:
\[ V = 2f \Delta L \]

4. **Substitute the Values:**
\[ V = 2 \times 512 \, \text{Hz} \times 0.334 \, \text{m} = 343.488 \, \text{m/s} \]

5. **Determine the Maximum Possible Error in Length Measurements:**
The maximum possible error in the length measurements is given as \( \Delta L_1 = \Delta L_2 = 0.1 \, \text{cm} = 0.001 \, \text{m} \).

6. **Calculate the Error in Speed of Sound:**
The error in speed \( \Delta V \) can be calculated using the formula:
\[ \Delta V = \frac{dV}{dL} \Delta L \]
Since \( V = 2f \Delta L \), the derivative \( \frac{dV}{dL} = 2f \).
Therefore:
\[ \Delta V = 2f (\Delta L_1 + \Delta L_2) = 2 \times 512 \times (0.001 + 0.001) = 2 \times 512 \times 0.002 = 2.048 \, \text{m/s} \]

### Final Result:
The maximum possible error in the speed of sound is approximately:
\[ \Delta V \approx 2.048 \, \text{m/s} \]