A steel wire fo length `5m` and cross-sectional area `2xx10^(-6)m^(2)` streches by the same amount as a copper wire of length `4m` and cross sectional area of `3xx10^(-6) m^(2)` under a given load. The ratio of young's mouduls of steel to that of copper is

Options

A. `8:15`
B. `15:8`
C. `5:3`
D. `3:5`

Explanation

To find the ratio of Young's modulus of steel to that of copper, we can use the relationship between Young's modulus (Y), length (L), cross-sectional area (A), and the amount of stretch (ΔL) for the two wires.

### Step-by-step Solution:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress to strain. Mathematically, it can be expressed as:
\[ Y = \frac{F/A}{\Delta L/L} \]
where:
- \( F \) = Force applied
- \( A \) = Cross-sectional area
- \( \Delta L \) = Change in length (stretch)
- \( L \) = Original length

2. **Given Data**:
- For the steel wire:
- Length (\( L_1 \)) = 5 m
- Cross-sectional area (\( A_1 \)) = \( 2 \times 10^{-6} \, m^2 \)
- For the copper wire:
- Length (\( L_2 \)) = 4 m
- Cross-sectional area (\( A_2 \)) = \( 3 \times 10^{-6} \, m^2 \)

3. **Using the Given Conditions**:
Since both wires stretch by the same amount under the same load, we can set up the following relationship:
\[ \frac{Y_{\text{steel}}}{Y_{\text{copper}}} = \frac{F/A_1}{\Delta L/L_1} \div \frac{F/A_2}{\Delta L/L_2} \]
Since \( F \) and \( \Delta L \) are the same for both wires, they cancel out, leading to:
\[ \frac{Y_{\text{steel}}}{Y_{\text{copper}}} = \frac{L_1 \cdot A_2}{L_2 \cdot A_1} \]

4. **Substituting the Values**:
Now substituting the given values into the equation:
\[ \frac{Y_{\text{steel}}}{Y_{\text{copper}}} = \frac{5 \cdot (3 \times 10^{-6})}{4 \cdot (2 \times 10^{-6})} \]

5. **Calculating the Ratio**:
Simplifying the right-hand side:
\[ = \frac{5 \cdot 3}{4 \cdot 2} = \frac{15}{8} \]

6. **Final Result**:
Thus, the ratio of Young's modulus of steel to that of copper is:
\[ \frac{Y_{\text{steel}}}{Y_{\text{copper}}} = \frac{15}{8} \]

### Conclusion:
The ratio of Young's modulus of steel to that of copper is \( \frac{15}{8} \).