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 solve the problem step by step, we need to determine the unknown frequency of the tuning fork before it is loaded with wax.

### Step 1: Understand the Beat Frequency
The beat frequency is the absolute difference between the frequencies of two tuning forks. The formula for beat frequency is given by:
\[ f_{\text{beat}} = |f_1 - f_2| \]
where \( f_1 \) is the frequency of the unknown tuning fork and \( f_2 \) is the frequency of the known tuning fork (254 Hz in this case).

### Step 2: Set Up the Equation
From the problem, we know that the beat frequency is 4 Hz. Therefore, we can write:
\[ |f_1 - 254| = 4 \]

### Step 3: Solve the Absolute Value Equation
This absolute value equation can be split into two cases:
1. \( f_1 - 254 = 4 \)
2. \( f_1 - 254 = -4 \)

#### Case 1:
\[ f_1 - 254 = 4 \]
\[ f_1 = 254 + 4 = 258 \, \text{Hz} \]

#### Case 2:
\[ f_1 - 254 = -4 \]
\[ f_1 = 254 - 4 = 250 \, \text{Hz} \]

So, the two possible frequencies for the unknown tuning fork are 258 Hz and 250 Hz.

### Step 4: Analyze the Effect of Wax
The problem states that the same tuning fork produces 4 beats per second when loaded with wax. When the tuning fork is loaded with wax, its frequency decreases.

- If \( f_1 = 250 \, \text{Hz} \), after loading with wax, it would decrease further, leading to a beat frequency greater than 4 Hz (for example, if it decreases to 248 Hz, the beat frequency would be 6 Hz).
- If \( f_1 = 258 \, \text{Hz} \), after loading with wax, it could decrease to 254 Hz, maintaining a beat frequency of 4 Hz.

### Step 5: Conclusion
Thus, the only viable solution is:
\[ f_1 = 258 \, \text{Hz} \]

### Final Answer
The unknown frequency before loading with wax is **258 Hz**.

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