If two tuning fork A and B are sounded together they produce 4 beats per second. A is then slightly loaded with wax, they produce 2 beats when sounded again. The frequency of A is 256. The frequency of B will be

Options

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

Explanation

To solve the problem, we need to find the frequency of tuning fork B (fB) given the frequency of tuning fork A (fA) and the number of beats produced in two different scenarios.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency of tuning fork A (fA) = 256 Hz
- Beats per second when A and B are sounded together = 4 beats/s
- Beats per second after loading A with wax = 2 beats/s

2. **Use the Beat Frequency Formula**:
The beat frequency is given by the absolute difference between the frequencies of the two tuning forks:
\[ |fA - fB| = \text{Number of beats per second} \]
For the first scenario:
\[ |256 - fB| = 4 \]
This gives us two possible equations:
\[ fB = 256 + 4 \quad \text{or} \quad fB = 256 - 4 \]
Thus:
\[ fB = 260 \quad \text{or} \quad fB = 252 \]

3. **Consider the Effect of Loading A with Wax**:
When tuning fork A is loaded with wax, its frequency decreases. Let's denote the new frequency of A as fA'. Since the frequency decreases, we have:
\[ fA' < 256 \text{ Hz} \]

4. **Apply the Beat Frequency Formula Again**:
Now, we know that after loading A, the beat frequency is 2 beats/s:
\[ |fA' - fB| = 2 \]
Since fA' is less than 256 Hz, we can analyze the two cases for fB:
- If \( fB = 260 \):
\[ |fA' - 260| = 2 \]
This would imply that \( fA' = 262 \) or \( fA' = 258 \), both of which are greater than 256, which contradicts our earlier conclusion that \( fA' < 256 \).

- If \( fB = 252 \):
\[ |fA' - 252| = 2 \]
This gives us:
\[ fA' = 254 \quad \text{or} \quad fA' = 250 \]
Both of these values are less than 256, which is consistent with our understanding that loading A with wax decreases its frequency.

5. **Conclusion**:
Since the only consistent solution is when \( fB = 252 \) Hz, we conclude that:
\[ \text{Frequency of tuning fork B (fB)} = 252 \text{ Hz} \]

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