A tuning fork of frequency 100 Hz emits a wave of wavelength 4m. What is the temperature of the atmosphere, if the velocity of sound at `27^(@)C` is 330m/s?

Options

A. `0^(@)C`
B. `100^(@)C`
C. `167^(@)C`
D. `227^(@)C`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Frequency of the tuning fork, \( F = 100 \, \text{Hz} \)
- Wavelength of the wave, \( \lambda = 4 \, \text{m} \)
- Velocity of sound at \( 27^\circ C \), \( V_2 = 330 \, \text{m/s} \)
- Temperature at which \( V_2 \) is measured, \( T_2 = 27^\circ C = 300 \, \text{K} \)

### Step 2: Calculate the velocity of sound emitted by the tuning fork
Using the formula for the velocity of a wave:
\[ V_1 = F \times \lambda \]
Substituting the values:
\[ V_1 = 100 \, \text{Hz} \times 4 \, \text{m} = 400 \, \text{m/s} \]

### Step 3: Use the relationship between velocity and temperature
The velocity of sound is directly proportional to the square root of the temperature:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
Rearranging this equation gives:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \implies \left(\frac{V_1}{V_2}\right)^2 = \frac{T_1}{T_2} \]

### Step 4: Substitute known values into the equation
Substituting \( V_1 = 400 \, \text{m/s} \), \( V_2 = 330 \, \text{m/s} \), and \( T_2 = 300 \, \text{K} \):
\[ \left(\frac{400}{330}\right)^2 = \frac{T_1}{300} \]

### Step 5: Calculate \( T_1 \)
Calculating the left side:
\[ \frac{400}{330} \approx 1.2121 \]
Squaring this value:
\[ (1.2121)^2 \approx 1.4747 \]
Now substituting back to find \( T_1 \):
\[ 1.4747 = \frac{T_1}{300} \implies T_1 = 1.4747 \times 300 \approx 442.41 \, \text{K} \]

### Step 6: Convert Kelvin to Celsius
To convert Kelvin to Celsius:
\[ T_1 = 442.41 \, \text{K} - 273.15 \approx 169.26^\circ C \]

### Step 7: Round to the nearest option
Rounding \( 169.26^\circ C \) gives approximately \( 167^\circ C \).

### Final Answer
Thus, the temperature of the atmosphere is approximately:
\[ \boxed{167^\circ C} \]

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