To find the ratio of the elongations produced in two steel wires under the same stretching force, we can use the formula for elongation in a wire, which is given by:
\[
\Delta L = \frac{F \cdot L}{A \cdot Y}
\]
where:
- \(\Delta L\) is the elongation,
- \(F\) is the force applied,
- \(L\) is the original length of the wire,
- \(A\) is the cross-sectional area of the wire,
- \(Y\) is the Young's modulus of the material (which is the same for both wires since they are made of steel).
Let's denote:
- For wire 1: \(L_1 = 100 \, \text{cm} = 1 \, \text{m}\), \(r_1 = 6 \, \text{mm} = 0.006 \, \text{m}\)
- For wire 2: \(L_2 = 200 \, \text{cm} = 2 \, \text{m}\), \(r_2 = 3 \, \text{mm} = 0.003 \, \text{m}\)
### Step 1: Calculate the cross-sectional areas of the wires
The cross-sectional area \(A\) of a wire can be calculated using the formula:
\[
A = \pi r^2
\]
For wire 1:
\[
A_1 = \pi (0.006)^2 = \pi \times 0.000036 \, \text{m}^2
\]
For wire 2:
\[
A_2 = \pi (0.003)^2 = \pi \times 0.000009 \, \text{m}^2
\]
### Step 2: Write the expressions for elongation for both wires
Using the elongation formula:
\[
\Delta L_1 = \frac{F \cdot L_1}{A_1 \cdot Y}
\]
\[
\Delta L_2 = \frac{F \cdot L_2}{A_2 \cdot Y}
\]
### Step 3: Find the ratio of elongations
Now, we can find the ratio of the elongations:
\[
\frac{\Delta L_1}{\Delta L_2} = \frac{F \cdot L_1}{A_1 \cdot Y} \div \frac{F \cdot L_2}{A_2 \cdot Y}
\]
This simplifies to:
\[
\frac{\Delta L_1}{\Delta L_2} = \frac{L_1 \cdot A_2}{L_2 \cdot A_1}
\]
### Step 4: Substitute the values
Now we substitute the values we calculated:
- \(L_1 = 1 \, \text{m}\)
- \(L_2 = 2 \, \text{m}\)
- \(A_1 = \pi \times 0.000036 \, \text{m}^2\)
- \(A_2 = \pi \times 0.000009 \, \text{m}^2\)
Thus:
\[
\frac{\Delta L_1}{\Delta L_2} = \frac{1 \cdot (A_2)}{2 \cdot (A_1)} = \frac{1 \cdot \pi \times 0.000009}{2 \cdot \pi \times 0.000036}
\]
The \(\pi\) cancels out:
\[
\frac{\Delta L_1}{\Delta L_2} = \frac{0.000009}{2 \cdot 0.000036} = \frac{0.000009}{0.000072} = \frac{1}{8}
\]
### Final Result
The ratio of the elongations produced in the two wires is:
\[
\Delta L_1 : \Delta L_2 = 1 : 8
\]
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