The length of two wires are in the ratio `3:4`. Ratio of the determeters is `1:2,` young's modulus of the wires are in the ratio `3:2`. If they are subjected to same tensile force, the ratio of the elogation produced is

Options

A. `1:1`
B. `1:2`
C. `2:3`
D. `2:1`

Explanation

To solve the problem, we need to find the ratio of the elongation produced in two wires subjected to the same tensile force. We will use the relationship between Young's modulus, force, area, length, and elongation.

### Given Data:
1. Length of wires (L1:L2) = 3:4
2. Diameter of wires (D1:D2) = 1:2
3. Young's modulus (Y1:Y2) = 3:2

### Step-by-Step Solution:

1. **Identify the relationship for elongation**:
The elongation (ΔL) produced in a wire can be expressed using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( F \) = tensile force
- \( L \) = original length of the wire
- \( A \) = cross-sectional area
- \( Y \) = Young's modulus

2. **Express the areas in terms of diameters**:
The area \( A \) of a wire can be calculated using the formula:
\[ A = \frac{\pi D^2}{4} \]
Therefore, the areas for the two wires can be expressed as:
\[ A_1 = \frac{\pi D_1^2}{4}, \quad A_2 = \frac{\pi D_2^2}{4} \]
Given the ratio of diameters \( D_1:D_2 = 1:2 \), we have:
\[ A_1 = \frac{\pi (1)^2}{4} = \frac{\pi}{4}, \quad A_2 = \frac{\pi (2)^2}{4} = \frac{4\pi}{4} = \pi \]
Thus, the ratio of areas \( A_1:A_2 = \frac{\pi/4}{\pi} = \frac{1}{4} \).

3. **Set up the elongation ratio**:
Using the elongation formula, we can write:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{F \cdot L_1 / (A_1 \cdot Y_1)}{F \cdot L_2 / (A_2 \cdot Y_2)} = \frac{L_1 \cdot A_2 \cdot Y_2}{L_2 \cdot A_1 \cdot Y_1} \]
Since the force \( F \) cancels out.

4. **Substitute the known ratios**:
Now we substitute the known values:
- Length ratio \( \frac{L_1}{L_2} = \frac{3}{4} \)
- Area ratio \( \frac{A_2}{A_1} = 4 \)
- Young's modulus ratio \( \frac{Y_2}{Y_1} = \frac{2}{3} \)

Therefore:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{3}{4} \cdot 4 \cdot \frac{2}{3} \]

5. **Simplify the expression**:
Simplifying this gives:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{3 \cdot 4 \cdot 2}{4 \cdot 3} = 2 \]

6. **Final Ratio**:
Thus, the ratio of elongations is:
\[ \Delta L_1 : \Delta L_2 = 2 : 1 \]

### Conclusion:
The ratio of the elongation produced in the two wires is \( 2:1 \).