Two tuning forks having frequency 256 Hz (A) and 262 Hz (B) tuning fork. A produces some beats per second with unknown tuning fork, same unknown tuning fork produce double beats per second from B tuning fork then the frequency of unknown tuning fork is :-

Options

A. 262
B. 260
C. 250
D. 300

Explanation

To solve the problem, we need to find the frequency of the unknown tuning fork (let's denote it as \( f_U \)). We know the frequencies of the two tuning forks: \( f_A = 256 \, \text{Hz} \) and \( f_B = 262 \, \text{Hz} \).

### Step-by-Step Solution:

1. **Understanding Beats**: The number of beats produced when two tuning forks are sounded together is equal to the absolute difference of their frequencies.
\[ \text{Beats} = |f_1 - f_2| \]

2. **Beats with Tuning Fork A**: When tuning fork A (256 Hz) is played with the unknown tuning fork, let the number of beats produced be \( n \).
\[ n = |256 - f_U| \]

3. **Beats with Tuning Fork B**: When tuning fork B (262 Hz) is played with the unknown tuning fork, it produces double the beats compared to when it was played with tuning fork A. Therefore, the number of beats produced in this case is \( 2n \):
\[ 2n = |262 - f_U| \]

4. **Setting Up the Equations**: From the above, we have two equations:
- \( n = |256 - f_U| \)
- \( 2n = |262 - f_U| \)

5. **Substituting for n**: Substitute \( n \) from the first equation into the second:
\[ 2|256 - f_U| = |262 - f_U| \]

6. **Case Analysis**: We need to analyze the absolute values. There are two cases to consider based on the value of \( f_U \):

**Case 1**: \( f_U < 256 \)
- Then \( |256 - f_U| = 256 - f_U \) and \( |262 - f_U| = 262 - f_U \)
- The equation becomes:
\[ 2(256 - f_U) = 262 - f_U \]
Simplifying this gives:
\[ 512 - 2f_U = 262 - f_U \implies 512 - 262 = f_U \implies f_U = 250 \, \text{Hz} \]

**Case 2**: \( 256 \leq f_U < 262 \)
- Then \( |256 - f_U| = f_U - 256 \) and \( |262 - f_U| = 262 - f_U \)
- The equation becomes:
\[ 2(f_U - 256) = 262 - f_U \]
Simplifying this gives:
\[ 2f_U - 512 = 262 - f_U \implies 3f_U = 774 \implies f_U = 258 \, \text{Hz} \]

**Case 3**: \( f_U \geq 262 \) (not applicable since it contradicts the problem statement of producing beats with A and B)

7. **Conclusion**: The only valid solutions are \( f_U = 250 \, \text{Hz} \) and \( f_U = 258 \, \text{Hz} \). However, since the unknown tuning fork produces double beats with B, we conclude that the unknown tuning fork must be \( 258 \, \text{Hz} \).

### Final Answer:
The frequency of the unknown tuning fork is \( f_U = 258 \, \text{Hz} \).