Two identical wires have a fundamental frequency of 100Hz, when kept under the same tension . If the tension of one of the wires is increased by 21% , the number of beats produced is

Options

A. 11
B. 10
C. 9
D. 8

Explanation

To solve the problem step by step, we will follow the principles of wave frequency and tension in strings.

### Step-by-Step Solution:

1. **Identify the Fundamental Frequency**:
The fundamental frequency of both wires is given as \( F_1 = 100 \, \text{Hz} \).

2. **Understand the Relationship Between Frequency and Tension**:
The frequency of a wire is related to its tension by the formula:
\[ F \propto \sqrt{T} \]
This means that if the tension increases, the frequency will also increase.

3. **Calculate the New Tension**:
If the tension of one wire is increased by 21%, the new tension \( T_2 \) can be expressed as:
\[ T_2 = T + 0.21T = 1.21T \]

4. **Determine the New Frequency**:
Using the proportionality of frequency to the square root of tension, we can express the new frequency \( F_2 \) as:
\[ \frac{F_1}{F_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \( T_1 = T \) and \( T_2 = 1.21T \):
\[ \frac{F_1}{F_2} = \sqrt{\frac{T}{1.21T}} = \sqrt{\frac{1}{1.21}} = \frac{1}{\sqrt{1.21}} \]

5. **Calculate the New Frequency \( F_2 \)**:
Rearranging gives us:
\[ F_2 = F_1 \cdot \sqrt{1.21} \]
Substituting \( F_1 = 100 \, \text{Hz} \):
\[ F_2 = 100 \cdot \sqrt{1.21} = 100 \cdot 1.1 = 110 \, \text{Hz} \]

6. **Calculate the Number of Beats**:
The number of beats produced is the difference between the two frequencies:
\[ \text{Number of beats} = |F_2 - F_1| = |110 - 100| = 10 \, \text{beats per second} \]

### Final Answer:
The number of beats produced is **10 beats per second**.