A tuning fork produces `6` beats/sec with sonometer wire when its tensions are either `169N` or `196N`. The frequency of that fork is

Options

A. `162Hz`
B. `190Hz`
C. `200Hz`
D. `80Hz`

Explanation

To solve the problem, we need to find the frequency of the tuning fork based on the information given about the beats produced with the sonometer wire at two different tensions.

### Step-by-Step Solution:

1. **Understanding the Concept of Beats**:
When two frequencies are close to each other, they produce beats. The number of beats per second is equal to the absolute difference between the two frequencies. In this case, the tuning fork produces 6 beats/sec with the sonometer wire.

2. **Identifying the Frequencies**:
Let the frequency of the tuning fork be \( f \). The frequencies of the sonometer wire when the tensions are \( 169N \) and \( 196N \) can be calculated using the formula for the frequency of a vibrating string:
\[ f_n = k \sqrt{T} \]
where \( T \) is the tension and \( k \) is a constant that depends on the length and mass per unit length of the wire.

3. **Setting Up the Equations**:
Let \( f_1 \) be the frequency corresponding to \( T_1 = 169N \) and \( f_2 \) be the frequency corresponding to \( T_2 = 196N \). We can express these as:
\[ f_1 = k \sqrt{169} = 13k \]
\[ f_2 = k \sqrt{196} = 14k \]

4. **Using the Beat Frequency**:
According to the problem, the beat frequency is 6 beats/sec. Therefore, we can write:
\[ |f - f_1| = 6 \quad \text{or} \quad |f - f_2| = 6 \]

5. **Setting Up the Two Cases**:
This gives us two cases to consider:
- Case 1: \( f - f_1 = 6 \) or \( f - f_2 = -6 \)
- Case 2: \( f - f_2 = 6 \) or \( f - f_1 = -6 \)

From Case 1:
- \( f = f_1 + 6 \) or \( f = f_2 - 6 \)

From Case 2:
- \( f = f_2 + 6 \) or \( f = f_1 - 6 \)

6. **Substituting the Frequencies**:
Substitute \( f_1 \) and \( f_2 \):
- From \( f = 13k + 6 \) and \( f = 14k - 6 \)
- From \( f = 14k + 6 \) and \( f = 13k - 6 \)

7. **Solving for \( k \)**:
Equating the two expressions from Case 1:
\[ 13k + 6 = 14k - 6 \]
Rearranging gives:
\[ k = 12 \]

8. **Finding the Frequency of the Tuning Fork**:
Substitute \( k \) back into either frequency equation:
\[ f_1 = 13k = 13 \times 12 = 156 \text{ Hz} \]
\[ f_2 = 14k = 14 \times 12 = 168 \text{ Hz} \]

Now, we can find \( f \):
- Using \( f = 156 + 6 = 162 \) Hz
- Using \( f = 168 - 6 = 162 \) Hz

Thus, the frequency of the tuning fork is **162 Hz**.

### Final Answer:
The frequency of the tuning fork is **162 Hz**.