A steel wire of length 4.5 m and cross-sectional area `3 xx 10^(-5) m^(2)` stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of `4 x 10^(-5) m^(2)` under a given load. The ratio of the Young's modulus of steel to that of copper is

Options

A. `1.3`
B. `1.5`
C. `1.7`
D. `1.9`

Explanation

To find the ratio of the Young's modulus of steel to that of copper, we can follow these steps:

### Step 1: Understand the given information
We have two wires:
- Steel wire: Length \( L_s = 4.5 \, m \), Cross-sectional area \( A_s = 3 \times 10^{-5} \, m^2 \)
- Copper wire: Length \( L_c = 3.5 \, m \), Cross-sectional area \( A_c = 4 \times 10^{-5} \, m^2 \)

Both wires stretch by the same amount under the same load.

### Step 2: Write the formula for Young's modulus
The Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} \]
where:
- \( F \) is the force applied,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the extension (stretch),
- \( L \) is the original length of the wire.

### Step 3: Set up the equations for both wires
For the steel wire:
\[ Y_s = \frac{F}{A_s} \cdot \frac{L_s}{\Delta L} \]

For the copper wire:
\[ Y_c = \frac{F}{A_c} \cdot \frac{L_c}{\Delta L} \]

### Step 4: Relate the two Young's moduli
Since both wires stretch by the same amount (\( \Delta L \)) and are subjected to the same force (\( F \)), we can set up the ratio:
\[ \frac{Y_s}{Y_c} = \frac{A_c}{A_s} \cdot \frac{L_s}{L_c} \]

### Step 5: Substitute the values
Substituting the values we have:
- \( A_c = 4 \times 10^{-5} \, m^2 \)
- \( A_s = 3 \times 10^{-5} \, m^2 \)
- \( L_s = 4.5 \, m \)
- \( L_c = 3.5 \, m \)

Now substituting these into the ratio:
\[ \frac{Y_s}{Y_c} = \frac{4 \times 10^{-5}}{3 \times 10^{-5}} \cdot \frac{4.5}{3.5} \]

### Step 6: Simplify the expression
The \( 10^{-5} \) cancels out:
\[ \frac{Y_s}{Y_c} = \frac{4}{3} \cdot \frac{4.5}{3.5} \]

Calculating \( \frac{4.5}{3.5} \):
\[ \frac{4.5}{3.5} = \frac{45}{35} = \frac{9}{7} \]

Now substituting back:
\[ \frac{Y_s}{Y_c} = \frac{4}{3} \cdot \frac{9}{7} = \frac{36}{21} = \frac{12}{7} \approx 1.714 \]

### Step 7: Conclusion
Thus, the ratio of the Young's modulus of steel to that of copper is approximately:
\[ \frac{Y_s}{Y_c} \approx 1.7 \]

### Final Answer
The ratio of the Young's modulus of steel to that of copper is \( \approx 1.7 \).

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