When two tuning forks A and B are sounded together x `beats//s` are heard. Frequency A is `n`. Now when one prong of B is loaded with a little wax, the number of beats/s decreases. The frequency of fork B is

Options

A. `n+x`
B. `n-x`
C. `n-x^2`
D. `n-2x`

Explanation

To solve the problem, we need to analyze the situation with the two tuning forks A and B, where the frequency of A is given as \( n \) and the number of beats produced when both forks are sounded together is \( x \) beats per second.

### Step-by-step Solution:

1. **Understand the Beat Frequency**:
The beat frequency is defined as the absolute difference between the frequencies of the two tuning forks. Therefore, we can express this as:
\[ |f_A - f_B| = x \]
where \( f_A \) is the frequency of fork A and \( f_B \) is the frequency of fork B.

2. **Assign Known Values**:
From the problem, we know:
- Frequency of fork A, \( f_A = n \)
- The number of beats per second, \( x \)

Thus, we can write:
\[ |n - f_B| = x \]

3. **Consider Two Cases**:
This absolute value equation can be split into two cases:
- Case 1: \( n - f_B = x \)
- Case 2: \( f_B - n = x \)

From Case 1, we can rearrange to find:
\[ f_B = n - x \]
From Case 2, we can rearrange to find:
\[ f_B = n + x \]

4. **Effect of Loading Wax on Fork B**:
When one prong of fork B is loaded with wax, the frequency of fork B decreases. This means that the new frequency of fork B will be less than its original frequency.

5. **Analyze the Change in Beats**:
Since the number of beats per second decreases when the frequency of fork B is reduced, we can conclude that the original frequency \( f_B \) must have been \( n + x \) (Case 2). If it was \( n - x \), then loading wax would have increased the beat frequency, which contradicts the given information.

6. **Final Expression for Frequency of Fork B**:
Therefore, the original frequency of fork B before loading wax is:
\[ f_B = n + x \]

### Conclusion:
The frequency of fork B is \( n + x \).