Tuning fork `F_1` has a frequency of 256 Hz and it is observed to produce `6 beats//second` with another tuning fork `F_2`. When `F_2` is loaded with wax, it still produces `6 beats//sec` with `F_1`. The frequency of `F_2` before loading was

Options

A. `253Hz`
B. `262Hz`
C. `250Hz`
D. `259Hz`

Explanation

To solve the problem, we need to find the frequency of tuning fork \( F_2 \) before it was loaded with wax. We know the following:

- Frequency of tuning fork \( F_1 \) (\( N_A \)) = 256 Hz
- Number of beats produced = 6 beats/second
- When \( F_2 \) is loaded with wax, it still produces 6 beats/second with \( F_1 \).

### Step-by-step Solution:

1. **Understanding Beats**:
The number of beats per second is given by the absolute difference in frequencies of the two tuning forks. Therefore, we can write:
\[ |N_A - N_B| = 6 \]
where \( N_B \) is the frequency of tuning fork \( F_2 \).

2. **Setting Up the Equation**:
Since \( N_A = 256 \) Hz, we can set up two possible equations based on the beat frequency:
\[ N_A - N_B = 6 \quad \text{(1)} \]
or
\[ N_B - N_A = 6 \quad \text{(2)} \]

3. **Solving Equation (1)**:
From equation (1):
\[ 256 - N_B = 6 \]
Rearranging gives:
\[ N_B = 256 - 6 = 250 \text{ Hz} \]

4. **Solving Equation (2)**:
From equation (2):
\[ N_B - 256 = 6 \]
Rearranging gives:
\[ N_B = 256 + 6 = 262 \text{ Hz} \]

5. **Analyzing the Effect of Wax**:
When \( F_2 \) is loaded with wax, its frequency decreases. Therefore, if \( N_B \) was initially 262 Hz, after loading with wax, it would produce a lower frequency, which would still result in 6 beats with \( F_1 \) (256 Hz). This means that the only feasible solution for \( N_B \) before loading with wax is:
\[ N_B = 262 \text{ Hz} \]

### Final Answer:
The frequency of tuning fork \( F_2 \) before loading with wax was **262 Hz**.