A tuning fork of unknown frequency produces 4 beats per second when sounded with another tuning fork of frequency 254 Hz. It gives the same number of beats per second when unknown tuning fork loaded with wax . The unknown frequency before loading with wax is

Options

A. 258
B. 254
C. 250
D. Can't be determined

Answer

To find the unknown frequency of the tuning fork before it was loaded with wax, we can follow these steps:

### Step 1: Understand the concept of beats
When two tuning forks of different frequencies are sounded together, they produce beats. The number of beats per second is equal to the absolute difference between their frequencies.

### Step 2: Set up the equation for beats
Let the unknown frequency of the tuning fork be \( f \). According to the problem, the known frequency is \( 254 \, \text{Hz} \) and the number of beats produced is \( 4 \, \text{beats/second} \). This gives us the equation:
\[ |f - 254| = 4 \]

### Step 3: Solve for the unknown frequency
From the equation \( |f - 254| = 4 \), we can derive two possible cases:
1. \( f - 254 = 4 \)
2. \( f - 254 = -4 \)

Solving these equations:
1. For \( f - 254 = 4 \):
\[ f = 254 + 4 = 258 \, \text{Hz} \]
2. For \( f - 254 = -4 \):
\[ f = 254 - 4 = 250 \, \text{Hz} \]

Thus, the possible frequencies for the unknown tuning fork are \( 258 \, \text{Hz} \) and \( 250 \, \text{Hz} \).

### Step 4: Analyze the effect of loading with wax
When the tuning fork is loaded with wax, its frequency decreases. The problem states that the number of beats remains the same (4 beats per second) even after loading with wax. This implies that the frequency of the loaded tuning fork must still produce 4 beats with the 254 Hz fork.

### Step 5: Determine the frequency after loading with wax
If we assume the unknown frequency was \( 258 \, \text{Hz} \) before loading with wax, after loading, the frequency decreases. The new frequency can be represented as \( f' \). Thus:
\[ |f' - 254| = 4 \]
Since \( f' < 258 \), we can conclude:
\[ f' - 254 = -4 \implies f' = 250 \, \text{Hz} \]

### Step 6: Verify the other case
If we assume the unknown frequency was \( 250 \, \text{Hz} \) before loading with wax, then after loading, the frequency would decrease further, leading to:
\[ |f' - 254| = 4 \]
This would imply that \( f' \) could be \( 246 \, \text{Hz} \) or \( 254 \, \text{Hz} \), which would not maintain the same beat frequency of 4 beats per second.

### Conclusion
Thus, the only consistent solution is that the unknown frequency before loading with wax is:
\[ \boxed{258 \, \text{Hz}} \]