Wire having tension 225 N produces six beats per second when it is tuned with a fork. When tension changes to 256 N, it is tuned with the same fork, the number of beats remain unchanged. The frequency of the fork will be

Options

A. 186 Hz
B. 225 Hz
C. 256 Hz
D. 280 Hz

Answer

To solve the problem, we need to analyze the relationship between tension, frequency, and the beats produced when the wire is tuned with a tuning fork.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Frequency and Tension**:
The frequency of a vibrating string is given by the formula:
\[ f \propto \sqrt{T} \]
where \( T \) is the tension in the wire. Therefore, the frequency can be expressed as:
\[ f = k \sqrt{T} \]
where \( k \) is a constant that depends on the length and mass per unit length of the wire.

2. **Setting Up the Frequencies**:
Let \( f_1 \) be the frequency of the wire under a tension of 225 N, and \( f_2 \) be the frequency under a tension of 256 N. According to the problem, the beat frequency is 6 beats per second, which means:
\[ |f_1 - f_0| = 6 \quad \text{and} \quad |f_2 - f_0| = 6 \]
where \( f_0 \) is the frequency of the tuning fork.

3. **Expressing Frequencies in Terms of Tension**:
From the relationship established earlier:
\[ f_1 = k \sqrt{225} = 15k \quad \text{and} \quad f_2 = k \sqrt{256} = 16k \]

4. **Setting Up the Beat Frequency Equations**:
Since \( f_1 \) is less than \( f_2 \):
\[ f_0 - f_1 = 6 \quad \text{(1)} \]
\[ f_2 - f_0 = 6 \quad \text{(2)} \]

5. **Substituting Frequencies into the Equations**:
From equation (1):
\[ f_0 = f_1 + 6 = 15k + 6 \]
From equation (2):
\[ f_0 = f_2 - 6 = 16k - 6 \]

6. **Setting the Two Expressions for \( f_0 \) Equal**:
Now we can set the two expressions for \( f_0 \) equal to each other:
\[ 15k + 6 = 16k - 6 \]

7. **Solving for \( k \)**:
Rearranging gives:
\[ 6 + 6 = 16k - 15k \]
\[ 12 = k \]

8. **Finding the Frequency of the Tuning Fork**:
Now substituting \( k \) back into either expression for \( f_0 \):
\[ f_0 = 15k + 6 = 15(12) + 6 = 180 + 6 = 186 \text{ Hz} \]

### Final Answer:
The frequency of the tuning fork is \( \boxed{186 \text{ Hz}} \).