When the streching force of a wire is increased by `2.5kg`, the frequency of the note emitted is changed in the ratio `3//2`. Calculate the original stretching force

Options

A. `3kg`
B. `2kg`
C. `1.5kg`
D. `1kg`

Explanation

To solve the problem, we need to calculate the original stretching force (T) of the wire based on the information provided about the change in frequency when the stretching force is increased.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let the original stretching force be \( T \) (in Newtons).
- The increase in stretching force is given as \( 2.5 \, \text{kg} \), which is equivalent to \( 2.5 \times 9.8 \, \text{N} = 24.5 \, \text{N} \) (using \( g \approx 9.8 \, \text{m/s}^2 \)).
- The new stretching force after the increase is \( T + 24.5 \).

2. **Frequency Relation:**
- The frequency of the wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( L \) is the length of the wire and \( \mu \) is the linear mass density.
- Let the initial frequency be \( f_1 \) and the new frequency after the increase be \( f_2 \).

3. **Express Frequencies:**
- The initial frequency \( f_1 \) is:
\[ f_1 = \sqrt{\frac{T}{\mu}} \]
- The new frequency \( f_2 \) is:
\[ f_2 = \sqrt{\frac{T + 24.5}{\mu}} \]

4. **Set Up the Ratio:**
- According to the problem, the ratio of the new frequency to the original frequency is given as:
\[ \frac{f_2}{f_1} = \frac{3}{2} \]
- Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{\sqrt{\frac{T + 24.5}{\mu}}}{\sqrt{\frac{T}{\mu}}} = \frac{3}{2} \]

5. **Simplify the Equation:**
- This simplifies to:
\[ \frac{\sqrt{T + 24.5}}{\sqrt{T}} = \frac{3}{2} \]
- Squaring both sides gives:
\[ \frac{T + 24.5}{T} = \left(\frac{3}{2}\right)^2 \]
\[ \frac{T + 24.5}{T} = \frac{9}{4} \]

6. **Cross Multiply:**
- Cross multiplying gives:
\[ 4(T + 24.5) = 9T \]
- Expanding this:
\[ 4T + 98 = 9T \]

7. **Rearranging the Equation:**
- Rearranging gives:
\[ 9T - 4T = 98 \]
\[ 5T = 98 \]

8. **Solve for T:**
- Dividing both sides by 5:
\[ T = \frac{98}{5} = 19.6 \, \text{N} \]

9. **Convert to kg:**
- To convert this to kg, we use \( T = 19.6 \, \text{N} \div 9.8 \, \text{m/s}^2 = 2 \, \text{kg} \).

### Final Answer:
The original stretching force is \( 19.6 \, \text{N} \) or \( 2 \, \text{kg} \).