Two steel wires have equal volumes. Their diameters are in the ratio `2:1`. When the same force is applied on them, the elongation produced will be in the ratio of

Options

A. `1:8`
B. `8:1`
C. `1:16`
D. `16:1`

Explanation

To solve the problem, we need to find the ratio of elongation produced in two steel wires that have equal volumes and diameters in the ratio of 2:1 when the same force is applied to them.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Let the diameters of the two wires be \( D_1 \) and \( D_2 \) such that \( \frac{D_1}{D_2} = \frac{2}{1} \).
- Since the volumes of the wires are equal, we have \( V_1 = V_2 \).
- Both wires are made of steel, so their Young's moduli \( Y_1 \) and \( Y_2 \) are equal.

2. **Calculate the Areas of Cross-Section:**
- The area of cross-section \( A \) of a wire is given by the formula:
\[ A = \frac{\pi D^2}{4} \]
- Therefore, the areas for the two wires can be expressed as:
\[ A_1 = \frac{\pi D_1^2}{4}, \quad A_2 = \frac{\pi D_2^2}{4} \]
- To find the ratio of the areas:
\[ \frac{A_1}{A_2} = \frac{D_1^2}{D_2^2} \]
- Substituting \( D_1 = 2D_2 \):
\[ \frac{A_1}{A_2} = \frac{(2D_2)^2}{D_2^2} = \frac{4D_2^2}{D_2^2} = 4 \]

3. **Relate Lengths and Volumes:**
- Since the volumes are equal:
\[ V_1 = A_1 L_1 = A_2 L_2 \]
- From the ratio of areas:
\[ A_1 = 4A_2 \implies 4A_2 L_1 = A_2 L_2 \]
- Dividing both sides by \( A_2 \) (assuming \( A_2 \neq 0 \)):
\[ 4L_1 = L_2 \implies \frac{L_1}{L_2} = \frac{1}{4} \]

4. **Use Young's Modulus to Find Elongation:**
- Young's modulus is defined as:
\[ Y = \frac{F/A}{\Delta L/L} \]
- For both wires, since \( Y_1 = Y_2 \) and the same force \( F \) is applied:
\[ \frac{Y_1}{Y_2} = \frac{F/A_1 \cdot L_1/\Delta L_1}{F/A_2 \cdot L_2/\Delta L_2} \]
- Simplifying gives:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{A_2 L_1}{A_1 L_2} \]
- Substituting the ratios we found:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{A_2}{4A_2} \cdot \frac{L_1}{L_2} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

5. **Conclusion:**
- The ratio of elongation produced in the two wires is:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{1}{16} \]

### Final Answer:
The elongation produced in the two steel wires will be in the ratio \( \Delta L_1 : \Delta L_2 = 1 : 16 \).