Frequency of tuning fork A is 256 Hz. It produces 4 beats/second with tuning fork B. When wax is applied at tuning fork B then 6 beats/second are heard. Frequency of B is :-

Options

A. 250 Hz
B. 260 Hz
C. 252 Hz
D. (A) & (C) both may possible

Explanation

To find the frequency of tuning fork B, we can follow these steps:

### Step 1: Understand the beat frequency
The beat frequency is the difference in frequencies between two tuning forks. When tuning fork A (with a frequency of 256 Hz) produces 4 beats per second with tuning fork B, it means the frequency of tuning fork B can either be higher or lower than that of A.

### Step 2: Set up the equations
Let the frequency of tuning fork B be \( f_B \). The relationship can be expressed as:
- \( f_B = 256 \, \text{Hz} + 4 \, \text{Hz} \) (if B is higher)
- or \( f_B = 256 \, \text{Hz} - 4 \, \text{Hz} \) (if B is lower)

This gives us two potential frequencies for tuning fork B:
- \( f_B = 260 \, \text{Hz} \) (higher)
- \( f_B = 252 \, \text{Hz} \) (lower)

### Step 3: Analyze the effect of wax
When wax is applied to tuning fork B, its frequency decreases. The problem states that after applying wax, 6 beats per second are heard. This means the new frequency of tuning fork B (after waxing) can be expressed as:
- \( f_B' = f_B - \Delta f \)
Where \( \Delta f \) is the change in frequency due to the wax.

### Step 4: Set up the new beat frequency equation
Now, we know that the new beat frequency is 6 Hz:
- If \( f_B = 260 \, \text{Hz} \):
- \( f_B' = 260 - \Delta f \)
- The beat frequency with A will be \( |256 - (260 - \Delta f)| = 6 \)
- This simplifies to \( |256 - 260 + \Delta f| = 6 \)
- Which gives us two cases:
1. \( \Delta f - 4 = 6 \) → \( \Delta f = 10 \)
2. \( \Delta f + 4 = 6 \) → \( \Delta f = 2 \) (not possible since frequency can’t increase)

- If \( f_B = 252 \, \text{Hz} \):
- \( f_B' = 252 - \Delta f \)
- The beat frequency with A will be \( |256 - (252 - \Delta f)| = 6 \)
- This simplifies to \( |256 - 252 + \Delta f| = 6 \)
- Which gives us two cases:
1. \( \Delta f + 4 = 6 \) → \( \Delta f = 2 \)
2. \( \Delta f - 4 = 6 \) → \( \Delta f = 10 \) (not possible since frequency can’t increase)

### Step 5: Conclusion
Since the only feasible solution is when \( f_B = 252 \, \text{Hz} \) and \( \Delta f = 2 \, \text{Hz} \), we conclude that the frequency of tuning fork B is:
- **Frequency of tuning fork B = 252 Hz**

### Final Answer
The frequency of tuning fork B is **252 Hz**.