Length of a sonometer wire is either 95 cm or 100 cm. In both the cases a tuning fork produces 4 beats then the frequency of tuning fork is :-

Options

A. 152
B. 156
C. 160
D. 164

Explanation

To solve the problem, we need to find the frequency of the tuning fork (N) based on the information given about the sonometer wire lengths and the number of beats produced.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two lengths of a sonometer wire: 95 cm and 100 cm. A tuning fork produces 4 beats per second when used with these wires. The frequency of the tuning fork (N) differs from the frequencies produced by the sonometer wires.

2. **Frequency of the Sonometer Wire**:
The frequency of a vibrating string (or sonometer wire) is given by the formula:
\[ f = \frac{n}{L} \]
where \( n \) is a constant depending on the tension and mass per unit length of the wire, and \( L \) is the length of the wire.

3. **Calculate Frequencies for Both Lengths**:
- For the 95 cm wire:
\[ f_1 = \frac{n}{0.95} \]
- For the 100 cm wire:
\[ f_2 = \frac{n}{1.00} \]

4. **Beat Frequency**:
The beat frequency is given by the absolute difference between the frequencies of the tuning fork and the sonometer wire. Since the tuning fork produces 4 beats, we can write:
\[ |N - f_1| = 4 \quad \text{(1)} \]
\[ |N - f_2| = 4 \quad \text{(2)} \]

5. **Setting Up Equations**:
From equation (1):
\[ N - f_1 = 4 \quad \text{or} \quad f_1 - N = 4 \]
From equation (2):
\[ N - f_2 = 4 \quad \text{or} \quad f_2 - N = 4 \]

6. **Solving for N**:
We can express \( f_1 \) and \( f_2 \) in terms of \( N \):
- From \( N - f_1 = 4 \):
\[ N = f_1 + 4 \]
- From \( N - f_2 = 4 \):
\[ N = f_2 + 4 \]

Setting these equal to each other gives:
\[ f_1 + 4 = f_2 + 4 \]
Simplifying this, we find:
\[ f_1 = f_2 \]

7. **Using the Ratio of Frequencies**:
Since the frequencies are inversely proportional to their lengths:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{100}{95} \]

8. **Substituting Values**:
Let \( f_1 = N - 4 \) and \( f_2 = N + 4 \):
\[ \frac{N - 4}{N + 4} = \frac{100}{95} \]

9. **Cross Multiplying**:
Cross-multiplying gives:
\[ 95(N - 4) = 100(N + 4) \]
Expanding both sides:
\[ 95N - 380 = 100N + 400 \]
Rearranging gives:
\[ 5N = 780 \]
Thus:
\[ N = \frac{780}{5} = 156 \text{ Hz} \]

### Final Answer:
The frequency of the tuning fork is **156 Hz**.