In two similar wires of tension 16 N and T, 3 beats are heard, then T=? if wire having tension 16N has a frequency of 4 Hz

Options

A. 49 N
B. 25 N
C. 64 N
D. 80 N

Answer

To solve the problem, we need to find the tension \( T \) in the second wire given that the first wire has a tension of 16 N and a frequency of 4 Hz, and that 3 beats are heard between the two wires.

### Step-by-Step Solution:

1. **Understand the relationship between tension and frequency**:
The frequency of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the wire, \( T \) is the tension, and \( \mu \) is the mass per unit length. Since both wires are similar, \( L \) and \( \mu \) are constant for both wires.

2. **Establish the frequency relationship**:
Since the frequencies \( f_1 \) and \( f_2 \) of the two wires are related to their tensions \( T_1 \) and \( T_2 \) as follows:
\[ \frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}} \]
We know \( T_1 = 16 \, \text{N} \) and \( f_1 = 4 \, \text{Hz} \). Let \( f_2 \) be the frequency of the second wire with tension \( T \).

3. **Express \( f_2 \) in terms of \( T \)**:
Rearranging the frequency relationship gives:
\[ f_2 = f_1 \cdot \sqrt{\frac{T_2}{T_1}} = 4 \cdot \sqrt{\frac{T}{16}} = 4 \cdot \frac{\sqrt{T}}{4} = \sqrt{T} \]

4. **Use the beat frequency condition**:
The beat frequency is given by the absolute difference between the two frequencies:
\[ |f_2 - f_1| = 3 \, \text{Hz} \]
Substituting \( f_2 \) and \( f_1 \):
\[ |\sqrt{T} - 4| = 3 \]

5. **Solve the absolute value equation**:
This gives us two cases to consider:
- Case 1: \( \sqrt{T} - 4 = 3 \)
\[ \sqrt{T} = 7 \implies T = 49 \, \text{N} \]
- Case 2: \( \sqrt{T} - 4 = -3 \)
\[ \sqrt{T} = 1 \implies T = 1 \, \text{N} \]

6. **Determine the valid solution**:
Since the tension in a wire cannot be negative or zero, the valid solution is:
\[ T = 49 \, \text{N} \]

### Final Answer:
The tension \( T \) in the second wire is \( 49 \, \text{N} \).