A sonometer wire resonates with a given tuning fork forming standing waves with three antinodes between the two bridges when a mass of 16 Kg is suspended from the wire. When this mass is replaced by a mass .9 kg. the wire resonates with the same tuning fork for .p. antinodes for the same positions of the bridges. Then .p. is

Answer

To solve the problem, we need to analyze the relationship between the mass suspended from the sonometer wire and the number of antinodes formed when the wire resonates with a tuning fork.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a sonometer wire that resonates with a tuning fork, forming standing waves with 3 antinodes (P1 = 3) when a mass of 16 kg (M1 = 16 kg) is suspended from it.
- When this mass is replaced with a mass of 9 kg (M2 = 9 kg), we need to find the new number of antinodes (P2 = P).

2. **Using the Formula for Frequency**:
- The frequency of the standing wave in the wire is given by the formula:
\[ f = \frac{P}{2L} \sqrt{\frac{T}{\mu}} \]
- Here, \(T\) is the tension in the wire, \(\mu\) is the mass per unit length of the wire, and \(L\) is the length of the wire.

3. **Setting Up the Tension**:
- The tension in the wire due to the suspended mass is given by:
\[ T = M \cdot g \]
- For the first case (with 16 kg):
\[ T1 = M1 \cdot g = 16g \]
- For the second case (with 9 kg):
\[ T2 = M2 \cdot g = 9g \]

4. **Equating Frequencies**:
- Since the frequency must be the same for both cases (as the same tuning fork is used), we can set up the equation:
\[ \frac{P1}{2L} \sqrt{\frac{T1}{\mu}} = \frac{P2}{2L} \sqrt{\frac{T2}{\mu}} \]
- Canceling \(2L\) and \(\mu\) from both sides, we have:
\[ P1 \sqrt{T1} = P2 \sqrt{T2} \]

5. **Substituting Values**:
- Substitute \(P1 = 3\), \(T1 = 16g\), and \(T2 = 9g\):
\[ 3 \sqrt{16g} = P \sqrt{9g} \]
- Simplifying gives:
\[ 3 \cdot 4 \sqrt{g} = P \cdot 3 \sqrt{g} \]

6. **Solving for P**:
- Dividing both sides by \(3\sqrt{g}\):
\[ 12 = P \]
- Thus, \(P = 4\).

### Final Answer:
The value of \(P\) is 4.