To solve the problem, we will follow these steps:
### Step 1: Understand the setup
We have a resonance column (a tube) with a diameter of 10 cm and a length of 70 cm. The first resonance occurs when the tube has 55 cm of water. The speed of sound in the air at the given temperature is 330 m/s.
### Step 2: Identify the parameters
- Diameter of the tube, \( D = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Length of the tube, \( L = 70 \, \text{cm} = 0.7 \, \text{m} \)
- Water level in the tube, \( h = 55 \, \text{cm} = 0.55 \, \text{m} \)
- Speed of sound, \( v = 330 \, \text{m/s} \)
### Step 3: Calculate the end correction
The end correction \( e \) can be calculated using the formula:
\[
e = 0.3 \times D
\]
Substituting the diameter:
\[
e = 0.3 \times 0.1 \, \text{m} = 0.03 \, \text{m}
\]
### Step 4: Determine the effective length of the air column
The effective length of the air column \( L_{\text{eff}} \) is given by:
\[
L_{\text{eff}} = L - h + e
\]
Substituting the values:
\[
L_{\text{eff}} = 0.7 \, \text{m} - 0.55 \, \text{m} + 0.03 \, \text{m} = 0.7 - 0.55 + 0.03 = 0.18 \, \text{m}
\]
### Step 5: Relate the effective length to the wavelength
For the first resonance in a tube that is closed at one end, the relationship between the effective length and the wavelength \( \lambda \) is given by:
\[
L_{\text{eff}} = \frac{\lambda}{4}
\]
Thus,
\[
\lambda = 4 \times L_{\text{eff}} = 4 \times 0.18 \, \text{m} = 0.72 \, \text{m}
\]
### Step 6: Calculate the frequency of the tuning fork
Using the formula for wave speed:
\[
v = f \cdot \lambda
\]
We can rearrange this to find the frequency \( f \):
\[
f = \frac{v}{\lambda}
\]
Substituting the values:
\[
f = \frac{330 \, \text{m/s}}{0.72 \, \text{m}} \approx 458.33 \, \text{Hz}
\]
### Step 7: Round to the nearest option
The closest option to 458.33 Hz is 460 Hz.
### Final Answer
The frequency of the tuning fork is approximately **460 Hz**.
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