A tuning fork of frequency 200 Hz is in unison with a sonometer wire . Tension is the wire of sonometer is increased by 1% without any change in its length . Find the number of beats heard in 9 s.

Options

A. 9
B. 3
C. 6
D. 12

Answer

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

### Step-by-Step Solution:

1. **Understanding the Relationship Between Frequency and Tension**:
The frequency \( f \) of a sonometer wire is given by the formula:
\[ f \propto \sqrt{T} \]
where \( T \) is the tension in the wire. This means that the frequency is directly proportional to the square root of the tension.

2. **Calculating the Change in Frequency**:
When the tension is increased by 1%, we can express this mathematically as:
\[ \Delta T = 0.01T \]
The new tension \( T' \) can be written as:
\[ T' = T + \Delta T = T + 0.01T = 1.01T \]

3. **Finding the New Frequency**:
The new frequency \( f' \) can be calculated using the relationship:
\[ f' = k\sqrt{T'} = k\sqrt{1.01T} \]
where \( k \) is a constant of proportionality. Since \( f = k\sqrt{T} \), we can express the new frequency in terms of the old frequency:
\[ f' = f\sqrt{1.01} \]

4. **Calculating the Change in Frequency**:
The change in frequency \( \Delta f \) is given by:
\[ \Delta f = f' - f = f(\sqrt{1.01} - 1) \]
Using the approximation \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) for small \( x \):
\[ \sqrt{1.01} \approx 1 + \frac{0.01}{2} = 1 + 0.005 = 1.005 \]
Therefore:
\[ \Delta f \approx f(0.005) = 200 \times 0.005 = 1 \text{ Hz} \]

5. **Calculating the Beat Frequency**:
The beat frequency is equal to the change in frequency:
\[ \text{Beat frequency} = \Delta f = 1 \text{ Hz} \]

6. **Finding the Number of Beats in 9 Seconds**:
The number of beats heard in a given time \( t \) is given by:
\[ \text{Number of beats} = \text{Beat frequency} \times t \]
Substituting the values:
\[ \text{Number of beats} = 1 \text{ Hz} \times 9 \text{ s} = 9 \]

### Final Answer:
The number of beats heard in 9 seconds is **9**.