If two tuning forks A & B give 4 beats/sec. with each other, on loading A with wax , 2 beats/sec. are given. If frequency of A is 256 Hz, then frequency of B is-

Options

A. 250 Hz
B. 252 Hz
C. 260Hz
D. 262Hz

Explanation

To solve the problem step by step, we need to analyze the information given and apply the concept of beats in sound waves.

### Step-by-Step Solution:

1. **Understanding Beats**: The number of beats per second (beat frequency) is given by the absolute difference in frequencies of two tuning forks. If two tuning forks A and B produce 4 beats per second, we can express this mathematically as:
\[ |f_A - f_B| = 4 \text{ Hz} \]
where \( f_A \) is the frequency of tuning fork A and \( f_B \) is the frequency of tuning fork B.

2. **Given Frequency of A**: We know that the frequency of tuning fork A, \( f_A \), is 256 Hz. Therefore, we can write:
\[ |256 - f_B| = 4 \]

3. **Finding Possible Frequencies for B**: This equation can be solved to find the possible frequencies of tuning fork B:
- Case 1: \( 256 - f_B = 4 \)
\[ f_B = 256 - 4 = 252 \text{ Hz} \]
- Case 2: \( f_B - 256 = 4 \)
\[ f_B = 256 + 4 = 260 \text{ Hz} \]

Thus, the possible frequencies for tuning fork B are 252 Hz and 260 Hz.

4. **Effect of Loading A with Wax**: When tuning fork A is loaded with wax, its frequency decreases. After loading, it produces 2 beats per second with tuning fork B. This means:
\[ |f_A' - f_B| = 2 \text{ Hz} \]
where \( f_A' \) is the new frequency of tuning fork A after loading with wax.

5. **Calculating New Frequency of A**: Since loading with wax decreases the frequency, we can denote the new frequency of A as:
\[ f_A' = 256 - x \quad (\text{where } x \text{ is the decrease in frequency}) \]
We need to find \( f_A' \) to determine the new beat frequency with B.

6. **Considering the Possible Frequencies**: We can analyze both cases for \( f_B \):
- If \( f_B = 252 \text{ Hz} \):
\[ |f_A' - 252| = 2 \]
This gives two scenarios:
- \( f_A' = 252 + 2 = 254 \text{ Hz} \) (not possible since A's frequency decreases)
- \( f_A' = 252 - 2 = 250 \text{ Hz} \) (possible)
- If \( f_B = 260 \text{ Hz} \):
\[ |f_A' - 260| = 2 \]
This gives two scenarios:
- \( f_A' = 260 + 2 = 262 \text{ Hz} \) (not possible since A's frequency decreases)
- \( f_A' = 260 - 2 = 258 \text{ Hz} \) (possible)

7. **Conclusion**: Since the only valid scenario after loading A with wax is when \( f_B = 252 \text{ Hz} \) and \( f_A' = 250 \text{ Hz} \), we conclude that the frequency of tuning fork B is:
\[ f_B = 252 \text{ Hz} \]

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