Tuning fork A of frequency 258 Hz gives 8 beats with a tuning fork B. When the tuning fork A is filed and again A and B are sounded the number of beats heard decreases. The frequency of B is

Options

A. 250 Hz
B. 266 Hz
C. 258 Hz
D. 242 Hz

Answer

To solve the problem, we need to determine the frequency of tuning fork B based on the information provided about tuning fork A and the beats produced when both forks are sounded together.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Frequency of tuning fork A (FA) = 258 Hz
- Number of beats heard when A and B are sounded together = 8 beats

2. **Understanding Beats:**
- The number of beats heard is given by the absolute difference between the frequencies of the two tuning forks:
\[ \text{Number of beats} = |FA - FB| \]
- Therefore, we can write:
\[ |258 - FB| = 8 \]

3. **Setting Up the Equations:**
- This absolute value equation can lead to two possible cases:
1. Case 1: \( 258 - FB = 8 \)
2. Case 2: \( FB - 258 = 8 \)

4. **Solving Case 1:**
- From Case 1:
\[ 258 - FB = 8 \implies FB = 258 - 8 = 250 \text{ Hz} \]

5. **Solving Case 2:**
- From Case 2:
\[ FB - 258 = 8 \implies FB = 258 + 8 = 266 \text{ Hz} \]

6. **Analyzing the Situation:**
- Initially, tuning fork A produces 8 beats with tuning fork B. This means that the frequencies are either 8 Hz apart.
- When tuning fork A is filed (which means its frequency is decreased), the number of beats heard decreases. This indicates that tuning fork B must have a frequency that is higher than tuning fork A after filing.
- Since we found two possible frequencies for tuning fork B (250 Hz and 266 Hz), we need to determine which one is valid based on the condition that the number of beats decreases after filing.

7. **Conclusion:**
- Since filing tuning fork A (258 Hz) would decrease its frequency, the only valid solution that satisfies the condition of fewer beats is:
\[ FB = 266 \text{ Hz} \]

### Final Answer:
The frequency of tuning fork B is **266 Hz**.

---