For definite length of wire, if the weight used for applying tension is immersed in water , then frequency will

Options

A. become less
B. become more
C. remain equal
D. become zero

Answer

To solve the problem, we need to analyze how immersing the weight in water affects the tension in the wire and subsequently the frequency of the vibrating wire.

### Step-by-step Solution:

1. **Understanding Tension in the Wire**:
- Initially, the tension \( T \) in the wire is given by the weight of the mass \( m \) hanging from it, which is \( T = mg \), where \( g \) is the acceleration due to gravity.

2. **Effect of Immersing the Weight in Water**:
- When the weight is immersed in water, a buoyant force acts on it. The buoyant force \( F_b \) is equal to the weight of the water displaced by the weight, which can be expressed as:
\[ F_b = \text{Volume} \times \rho_{\text{water}} \times g \]
- Here, \( \rho_{\text{water}} \) is the density of water.

3. **New Tension Calculation**:
- The new tension \( T' \) in the wire when the weight is submerged is given by:
\[ T' = mg - F_b \]
- This indicates that the tension in the wire decreases because the buoyant force opposes the weight of the mass.

4. **Frequency of the Wire**:
- The frequency \( f \) of the vibrating wire can be expressed using the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
- Where \( L \) is the length of the wire and \( \mu \) is the mass per unit length of the wire.

5. **Substituting the New Tension**:
- Substituting the new tension \( T' \) into the frequency formula gives:
\[ f' = \frac{1}{2L} \sqrt{\frac{mg - F_b}{\mu}} \]
- Since \( F_b \) is positive, it follows that \( T' < mg \), which means:
\[ f' < f \]
- Therefore, the frequency decreases when the weight is immersed in water.

6. **Conclusion**:
- The frequency of the wire decreases when the weight used to apply tension is immersed in water.

### Final Answer:
The frequency will **decrease** when the weight is immersed in water.