Two identical wires are stretched by the same tension of `101 N` and each emits a note of frequency `202 Hz`. If the tension in one wire is increased by `1N`, then the beat frequency is

Options

A. `2 Hz`
B. `(1)/(2)Hz`
C. `1Hz`
D. none of these

Explanation

To solve the problem, we need to determine the beat frequency produced when the tension in one of the wires is increased by 1 N.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Frequency and Tension**:
The frequency of a vibrating string (or wire) is given by the formula:
\[ f \propto \sqrt{T} \]
where \( f \) is the frequency and \( T \) is the tension in the wire.

2. **Initial Conditions**:
We have two identical wires, both under a tension of \( T_1 = 101 \, \text{N} \) and both producing a frequency of \( f_1 = 202 \, \text{Hz} \).

3. **Change in Tension**:
When the tension in one wire is increased by \( 1 \, \text{N} \), the new tension becomes:
\[ T_2 = T_1 + 1 = 101 + 1 = 102 \, \text{N} \]

4. **Calculating the New Frequency**:
Using the proportionality of frequency to the square root of tension, we can express the new frequency \( f_2 \) as:
\[ f_2 = k \sqrt{T_2} \]
where \( k \) is a constant that relates frequency and tension. Since we know that \( f_1 = k \sqrt{T_1} \), we can find \( k \):
\[ k = \frac{f_1}{\sqrt{T_1}} = \frac{202}{\sqrt{101}} \]
Now substituting \( T_2 \):
\[ f_2 = \frac{202}{\sqrt{101}} \cdot \sqrt{102} \]

5. **Calculating the New Frequency**:
To find \( f_2 \):
\[ f_2 = 202 \cdot \sqrt{\frac{102}{101}} \]

6. **Finding the Beat Frequency**:
The beat frequency \( f_b \) is given by the absolute difference between the two frequencies:
\[ f_b = |f_2 - f_1| \]
Substituting \( f_1 \) and \( f_2 \):
\[ f_b = \left| 202 \cdot \sqrt{\frac{102}{101}} - 202 \right| \]
Factoring out \( 202 \):
\[ f_b = 202 \left| \sqrt{\frac{102}{101}} - 1 \right| \]

7. **Calculating the Value**:
To simplify \( \sqrt{\frac{102}{101}} - 1 \), we can use the approximation for small changes:
\[ \sqrt{\frac{102}{101}} \approx 1 + \frac{1}{2} \left(\frac{1}{101}\right) = 1 + \frac{0.5}{101} \]
Thus,
\[ \sqrt{\frac{102}{101}} - 1 \approx \frac{0.5}{101} \]
Therefore,
\[ f_b \approx 202 \cdot \frac{0.5}{101} = 1 \]

### Final Answer:
The beat frequency is approximately \( 1 \, \text{Hz} \).