A column of air at `51^(@) C` and a tuning fork produce `4` beats per second when sounded together. As the temperature of the air column is decreased, the number of beats per second tends to decrease and when the temperature is `16^(@) C` the two produce `1` beat per second. Find the frequency of the tuning fork.

Options

A. `100 Hz `
B. 75Hz
C. 150 Hz
D. 50 Hz

Answer

To find the frequency of the tuning fork, we will use the concept of beats and the relationship between the frequency of sound in air and temperature. Here’s a step-by-step solution:

### Step 1: Understand the relationship between frequency and temperature
The frequency of sound in air is directly proportional to the square root of the absolute temperature (in Kelvin). The formula can be expressed as:
\[ f \propto \sqrt{T} \]
where \( T \) is the absolute temperature in Kelvin.

### Step 2: Convert temperatures to Kelvin
We need to convert the given temperatures from Celsius to Kelvin:
- For \( T_1 = 51^\circ C \):
\[ T_1 = 51 + 273 = 324 \, K \]
- For \( T_2 = 16^\circ C \):
\[ T_2 = 16 + 273 = 289 \, K \]

### Step 3: Set up the equations for beats
Let \( f \) be the frequency of the tuning fork and \( f_1 \) be the frequency of the air column at \( 51^\circ C \). The number of beats produced is given by the difference in frequencies:
- At \( 51^\circ C \):
\[ f_1 - f = 4 \quad \Rightarrow \quad f_1 = f + 4 \]
- At \( 16^\circ C \):
\[ f_2 - f = 1 \quad \Rightarrow \quad f_2 = f + 1 \]

### Step 4: Use the relationship of frequencies and temperatures
Using the proportionality of frequencies and the square root of temperatures:
\[ \frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \( T_1 \) and \( T_2 \):
\[ \frac{f + 4}{f + 1} = \sqrt{\frac{324}{289}} = \frac{18}{17} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 17(f + 4) = 18(f + 1) \]
Expanding both sides:
\[ 17f + 68 = 18f + 18 \]
Rearranging gives:
\[ 68 - 18 = 18f - 17f \quad \Rightarrow \quad f = 50 \, Hz \]

### Step 6: Verify with the second case
Now, let’s verify using the second case where \( f_2 = f - 1 \):
\[ \frac{f + 4}{f - 1} = \frac{18}{17} \]
Cross-multiplying gives:
\[ 17(f + 4) = 18(f - 1) \]
Expanding both sides:
\[ 17f + 68 = 18f - 18 \]
Rearranging gives:
\[ 68 + 18 = 18f - 17f \quad \Rightarrow \quad f = 86 \, Hz \]

### Conclusion
We have two possible frequencies for the tuning fork: \( 50 \, Hz \) and \( 86 \, Hz \). Since only \( 50 \, Hz \) is provided in the options, the frequency of the tuning fork is:
\[ \boxed{50 \, Hz} \]