A stretched wire of stone length under a tension is vibrating with its fundamental frequency . Its length is decreased by `45 %` and tension is increased by `21 %` . Now fundamental frequency

Options

A. increases by `50 %`
B. increases by `100 %`
C. decreases by `50 %`
D. decreases by `25 %`

Explanation

To solve the problem, we need to determine how the fundamental frequency of a stretched wire changes when its length is decreased by 45% and its tension is increased by 21%.

### Step-by-Step Solution:

1. **Understand the Formula for Fundamental Frequency**:
The fundamental frequency \( n \) of a stretched wire is given by the formula:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{m}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( m \) = mass per unit length of the wire

2. **Define Initial Conditions**:
Let's denote the initial length and tension as:
- \( L_1 = L \)
- \( T_1 = T \)

The initial fundamental frequency is:
\[ n_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{m}} \]

3. **Calculate New Length and Tension**:
- The length is decreased by 45%, so the new length \( L_2 \) is:
\[ L_2 = L_1 - 0.45L_1 = 0.55L_1 = 0.55L \]

- The tension is increased by 21%, so the new tension \( T_2 \) is:
\[ T_2 = T_1 + 0.21T_1 = 1.21T_1 = 1.21T \]

4. **Calculate New Fundamental Frequency**:
The new fundamental frequency \( n_2 \) is given by:
\[ n_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{m}} \]
Substituting \( L_2 \) and \( T_2 \):
\[ n_2 = \frac{1}{2(0.55L)} \sqrt{\frac{1.21T}{m}} \]

5. **Relate \( n_2 \) to \( n_1 \)**:
To find the ratio \( \frac{n_2}{n_1} \):
\[ \frac{n_2}{n_1} = \frac{\frac{1}{2(0.55L)} \sqrt{\frac{1.21T}{m}}}{\frac{1}{2L} \sqrt{\frac{T}{m}}} \]
Simplifying this gives:
\[ \frac{n_2}{n_1} = \frac{L}{0.55L} \cdot \sqrt{\frac{1.21T}{T}} = \frac{1}{0.55} \cdot \sqrt{1.21} \]

6. **Calculate the Values**:
- \( \sqrt{1.21} = 1.1 \)
- Therefore:
\[ \frac{n_2}{n_1} = \frac{1.1}{0.55} = 2 \]

7. **Conclusion**:
Thus, we find:
\[ n_2 = 2n_1 \]
This means the new fundamental frequency is double the original frequency, indicating an increase of 100%.

### Final Answer:
The fundamental frequency is increased by 100%.