IF two tuning forks A and B are sounded together, they produce 4 beats per second. A is then slightly loaded with wax, they produce two 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 step by step, we can follow these instructions:

### Step 1: Understand the Concept of Beats
When two tuning forks are sounded together, the number of beats per second is equal to the absolute difference between their frequencies. This can be expressed mathematically as:
\[ \text{Number of beats} = |\nu_A - \nu_B| \]

### Step 2: Set Up the Initial Condition
From the problem, we know that:
- The frequency of tuning fork A, \(\nu_A\), is 256 Hz.
- The two tuning forks produce 4 beats per second when sounded together.

Using the beats formula:
\[ |\nu_A - \nu_B| = 4 \]
This gives us two possible equations:
1. \(\nu_A - \nu_B = 4\)
2. \(\nu_B - \nu_A = 4\)

### Step 3: Solve for \(\nu_B\)
Substituting the known value of \(\nu_A\) into the first equation:
\[ 256 - \nu_B = 4 \]
Rearranging gives:
\[ \nu_B = 256 - 4 = 252 \text{ Hz} \]

### Step 4: Consider the Effect of Loading Fork A
When fork A is slightly loaded with wax, its frequency decreases. The problem states that the new condition produces 2 beats per second. Thus, we can set up a new equation:
\[ |\nu_A' - \nu_B| = 2 \]
Where \(\nu_A'\) is the new frequency of fork A after loading. Since \(\nu_A'\) is less than \(\nu_A\), we can express this as:
\[ \nu_A' - \nu_B = -2 \quad \text{(since \(\nu_A' < \nu_B\))} \]

### Step 5: Substitute and Solve
Substituting \(\nu_B = 252\) into the equation:
\[ \nu_A' - 252 = -2 \]
This implies:
\[ \nu_A' = 252 - 2 = 250 \text{ Hz} \]

### Conclusion
The frequency of tuning fork B is:
\[ \nu_B = 252 \text{ Hz} \]

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

---