The length of a sonometer wire tuned to a frequency of 250 Hz is 0.60 metre . The frequency of tuning fork with which the vibrating wire will be in tune when the length is made 0.40 metre is

Options

A. 250 Hz
B. 375 Hz
C. 56 Hz
D. 384 Hz

Answer

To solve the problem, we will use the relationship between the frequency of a vibrating wire and its length. The fundamental frequency of a vibrating wire is inversely proportional to its length. This can be expressed mathematically as:

\[ n_1 \cdot l_1 = n_2 \cdot l_2 \]

Where:
- \( n_1 \) = frequency of the first tuning fork (250 Hz)
- \( l_1 \) = length of the wire in the first case (0.60 m)
- \( n_2 \) = frequency of the second tuning fork (unknown)
- \( l_2 \) = length of the wire in the second case (0.40 m)

### Step-by-Step Solution:

1. **Identify the known values:**
- \( n_1 = 250 \, \text{Hz} \)
- \( l_1 = 0.60 \, \text{m} \)
- \( l_2 = 0.40 \, \text{m} \)

2. **Set up the equation using the relationship:**
\[ n_1 \cdot l_1 = n_2 \cdot l_2 \]

3. **Substitute the known values into the equation:**
\[ 250 \cdot 0.60 = n_2 \cdot 0.40 \]

4. **Calculate the left side of the equation:**
\[ 250 \cdot 0.60 = 150 \]
So, we have:
\[ 150 = n_2 \cdot 0.40 \]

5. **Solve for \( n_2 \):**
\[ n_2 = \frac{150}{0.40} \]

6. **Calculate \( n_2 \):**
\[ n_2 = 375 \, \text{Hz} \]

### Final Answer:
The frequency of the tuning fork when the length of the wire is made 0.40 meters is **375 Hz**.