The tension in a wire is decreased by `19%` the percentage decrease in frequency will be

Options

A. `19 %`
B. `10%`
C. `0.9%`
D. None of these

Explanation

To solve the problem of determining the percentage decrease in frequency when the tension in a wire is decreased by 19%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship Between Frequency and Tension**:
The frequency (F) of a vibrating wire is directly proportional to the square root of the tension (T) in the wire. This can be expressed as:
\[ F \propto \sqrt{T} \]

2. **Initial and Final Tension**:
Let the initial tension be \( T_1 \) and the final tension after a 19% decrease be \( T_2 \). The tension decreases by 19%, so:
\[ T_2 = T_1 - 0.19 T_1 = 0.81 T_1 \]

3. **Expressing the Frequencies**:
Let the initial frequency be \( F_1 \) and the final frequency be \( F_2 \). According to the proportionality, we can write:
\[ \frac{F_1}{F_2} = \sqrt{\frac{T_1}{T_2}} \]

4. **Substituting for \( T_2 \)**:
Substitute \( T_2 \) into the equation:
\[ \frac{F_1}{F_2} = \sqrt{\frac{T_1}{0.81 T_1}} = \sqrt{\frac{1}{0.81}} = \frac{1}{\sqrt{0.81}} = \frac{1}{0.9} \]
Therefore, we have:
\[ F_2 = 0.9 F_1 \]

5. **Calculating the Percentage Decrease in Frequency**:
The percentage decrease in frequency can be calculated using the formula:
\[ \text{Percentage Decrease} = \frac{F_1 - F_2}{F_1} \times 100 \]
Substituting \( F_2 \):
\[ \text{Percentage Decrease} = \frac{F_1 - 0.9 F_1}{F_1} \times 100 = \frac{0.1 F_1}{F_1} \times 100 = 10\% \]

### Final Answer:
The percentage decrease in frequency is **10%**.

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