A tuning fork of frequency 200Hz is in unison with a sonometer wire . The number of beats heard per second when the tension is increased by 1% will be

Options

A. 1
B. 2
C. 4
D. 0.5

Explanation

To solve the problem, we need to determine the number of beats heard per second when the tension in the sonometer wire is increased by 1%.

### Step-by-Step Solution:

1. **Understand the relationship between frequency and tension**:
The fundamental frequency \( f \) of a sonometer wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire, \( L \) is the length of the wire, and \( \mu \) is the linear mass density of the wire.

2. **Initial frequency**:
Since the tuning fork is in unison with the sonometer wire, the initial frequency \( f_1 \) is:
\[ f_1 = 200 \text{ Hz} \]

3. **Increase in tension**:
When the tension is increased by 1%, the new tension \( T' \) can be expressed as:
\[ T' = T + 0.01T = 1.01T \]

4. **Calculate the new frequency**:
The new frequency \( f_2 \) with the increased tension is:
\[ f_2 = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.01T}{\mu}} = \frac{1}{2L} \sqrt{1.01} \sqrt{\frac{T}{\mu}} = \sqrt{1.01} \cdot f_1 \]
Since \( f_1 = 200 \text{ Hz} \):
\[ f_2 = \sqrt{1.01} \cdot 200 \]

5. **Calculate \( \sqrt{1.01} \)**:
Using a calculator or approximation:
\[ \sqrt{1.01} \approx 1.005 \]
Therefore:
\[ f_2 \approx 1.005 \cdot 200 \approx 201 \text{ Hz} \]

6. **Determine the number of beats**:
The number of beats per second \( n \) is given by the absolute difference between the two frequencies:
\[ n = |f_2 - f_1| = |201 - 200| = 1 \text{ beat per second} \]

### Final Answer:
The number of beats heard per second when the tension is increased by 1% is **1 beat per second**.