Two tuning forks when sounded together produced `4beats//sec`. The frequency of one fork is 256 Hz. The number of beats heard increases when the fork of frequency 256 Hz is loaded with wax. The frequency of the other fork is

Options

A. `504 Hz`
B. `520 Hz`
C. `260 Hz`
D. `252 Hz`

Explanation

To solve the problem, we need to determine the frequency of the second tuning fork (let's call it Fork B) given that Fork A has a frequency of 256 Hz and they produce 4 beats per second when sounded together.

### Step-by-Step Solution:

1. **Understanding Beats**:
The number of beats per second (beats frequency) is given by the absolute difference between the frequencies of the two tuning forks. Mathematically, this can be expressed as:
\[ \text{Beats Frequency} = |f_A - f_B| \]
where \( f_A \) is the frequency of Fork A (256 Hz) and \( f_B \) is the frequency of Fork B.

2. **Setting Up the Equation**:
Since the problem states that the beat frequency is 4 beats per second, we can set up the equation:
\[ |256 - f_B| = 4 \]

3. **Solving the Absolute Value Equation**:
This absolute value equation can be split into two cases:
- Case 1: \( 256 - f_B = 4 \)
- Case 2: \( 256 - f_B = -4 \)

**Case 1**:
\[ 256 - f_B = 4 \implies f_B = 256 - 4 = 252 \text{ Hz} \]

**Case 2**:
\[ 256 - f_B = -4 \implies f_B = 256 + 4 = 260 \text{ Hz} \]

4. **Considering the Effect of Wax**:
The problem states that when Fork A (256 Hz) is loaded with wax, the number of beats increases. This implies that the frequency of Fork A decreases due to the loading of wax.

Since loading with wax decreases the frequency of Fork A, the new frequency of Fork A will be less than 256 Hz. Therefore, the frequency of Fork B must be higher than 256 Hz to increase the beat frequency.

5. **Determining the Frequency of Fork B**:
From our calculations, we have two possible frequencies for Fork B: 252 Hz and 260 Hz. Since the frequency of Fork A decreases when wax is added, the only feasible frequency for Fork B that satisfies the condition of increasing beats is:
\[ f_B = 260 \text{ Hz} \]

### Final Answer:
The frequency of the other tuning fork (Fork B) is **260 Hz**.