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 based on the lengths of the sonometer wire and the number of beats produced.

### Step-by-Step Solution:

1. **Identify the Lengths of the Sonometer Wire:**
- Let the first length \( L_1 = 95 \, \text{cm} \)
- Let the second length \( L_2 = 100 \, \text{cm} \)

2. **Define the Frequencies:**
- Let the frequency of the tuning fork be \( \nu \).
- The frequency of the wire at length \( L_1 \) (95 cm) is \( \nu_1 = \nu + 4 \) (since it produces 4 beats).
- The frequency of the wire at length \( L_2 \) (100 cm) is \( \nu_2 = \nu - 4 \) (since it produces 4 beats in the opposite direction).

3. **Use the Relationship Between Frequencies and Lengths:**
- The relationship between the frequencies and lengths of the sonometer wire is given by:
\[ \frac{\nu_1}{\nu_2} = \frac{L_2}{L_1} \]
- Substituting the expressions for \( \nu_1 \) and \( \nu_2 \):
\[ \frac{\nu + 4}{\nu - 4} = \frac{100}{95} \]

4. **Cross-Multiply to Solve for \( \nu \):**
- Cross-multiplying gives:
\[ 95(\nu + 4) = 100(\nu - 4) \]
- Expanding both sides:
\[ 95\nu + 380 = 100\nu - 400 \]

5. **Rearranging the Equation:**
- Rearranging the equation to isolate \( \nu \):
\[ 100\nu - 95\nu = 380 + 400 \]
\[ 5\nu = 780 \]

6. **Solving for \( \nu \):**
- Dividing both sides by 5:
\[ \nu = \frac{780}{5} = 156 \, \text{Hz} \]

### Final Answer:
The frequency of the tuning fork is \( 156 \, \text{Hz} \).