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

Answer

To solve the problem, we need to determine the frequency of tuning fork \( F_2 \) before it was loaded with wax, given that it produces 6 beats per second with tuning fork \( F_1 \) which has a frequency of 256 Hz.

### 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. If \( F_1 \) has a frequency of 256 Hz and it produces 6 beats per second with \( F_2 \), we can express this as:
\[ |f_1 - f_2| = 6 \text{ Hz} \]

2. **Setting Up the Equation**: Since \( f_1 = 256 \) Hz, we can set up two possible equations based on the beat frequency:
\[ f_2 = f_1 + 6 \quad \text{or} \quad f_2 = f_1 - 6 \]
This gives us:
\[ f_2 = 256 + 6 = 262 \text{ Hz} \quad \text{(1)} \]
\[ f_2 = 256 - 6 = 250 \text{ Hz} \quad \text{(2)} \]

3. **Considering the Effect of Loading with Wax**: When \( F_2 \) is loaded with wax, its frequency decreases. Let’s denote the frequency of \( F_2 \) after loading as \( f_2' \). Since it still produces 6 beats per second with \( F_1 \), we have:
\[ |f_1 - f_2'| = 6 \text{ Hz} \]
This means:
\[ f_2' = f_1 + 6 \quad \text{or} \quad f_2' = f_1 - 6 \]

4. **Determining the New Frequency**: Since \( f_2' < f_2 \) (because loading with wax decreases the frequency), we can conclude:
\[ f_2' = 256 - 6 = 250 \text{ Hz} \]

5. **Finalizing the Frequency of \( F_2 \)**: Now we know that before loading, \( F_2 \) must have been the higher frequency, which is:
\[ f_2 = 262 \text{ Hz} \]

Thus, the frequency of \( F_2 \) before loading was **262 Hz**.

### Answer:
The frequency of \( F_2 \) before loading was **262 Hz**.