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 a weight in water affects the tension in a wire and subsequently the frequency of the wire. Here’s a step-by-step breakdown:

### Step 1: Understand the Setup
We have a wire of length \( L \) with a mass \( M \) attached to it. The weight of the mass \( M \) creates tension in the wire.

**Hint:** Identify the forces acting on the mass when it is submerged in water.

### Step 2: Write the Equilibrium Condition
In equilibrium, the tension \( T \) in the wire is equal to the weight of the mass \( M \) minus the buoyant force \( F_B \) acting on it when submerged in water.

The weight of the mass is given by:
\[ W = mg \]
where \( m \) is the mass and \( g \) is the acceleration due to gravity.

The buoyant force \( F_B \) is given by:
\[ F_B = \rho V g \]
where \( \rho \) is the density of water, \( V \) is the volume of the mass submerged, and \( g \) is the acceleration due to gravity.

Thus, the equilibrium condition can be expressed as:
\[ T + F_B = mg \]
or rearranging gives:
\[ T = mg - F_B \]

**Hint:** Remember that the buoyant force reduces the effective weight of the mass when it is submerged.

### Step 3: Express Tension in Terms of Buoyant Force
Substituting the expression for buoyant force into the tension equation:
\[ T = mg - \rho V g \]

**Hint:** This shows how the tension in the wire changes when the mass is immersed in water.

### Step 4: Relate Tension to Frequency
The frequency \( f \) of the wire can be expressed as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( \mu \) is the linear mass density of the wire (mass per unit length).

**Hint:** The frequency is directly proportional to the square root of the tension.

### Step 5: Analyze the Effect of Buoyancy on Frequency
Since the tension \( T \) decreases when the mass is submerged in water (due to the buoyant force), we can conclude that:
- If \( T \) decreases, then \( f \) must also decrease.

**Hint:** Compare the initial tension (when the mass is in air) to the new tension (when the mass is submerged) to see the effect on frequency.

### Conclusion
Therefore, when the weight used for applying tension is immersed in water, the frequency of the wire will **decrease**.

**Final Answer:** The frequency will become less.