A steel wire of length 5.0 m and cross-section `3.0xx10^(-5) m^(2)` stretches by the same amount as a copper wire of length 3.0 m and cross -section `4.0xx10^(-5)m^(2)` under a given load. What is the ratio of Young's modulus of steel to that of copper?

Answer

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

### Step 1: Write the formula for Young's modulus
Young's modulus (Y) is defined as the ratio of stress to strain. The formula for Young's modulus is given by:

\[ Y = \frac{F/A}{\Delta L/L} \]

Where:
- \( F \) = Load applied
- \( A \) = Cross-sectional area
- \( \Delta L \) = Extension (change in length)
- \( L \) = Original length

### Step 2: Set up the equations for steel and copper
For the steel wire:
- Length \( L_s = 5.0 \, m \)
- Cross-sectional area \( A_s = 3.0 \times 10^{-5} \, m^2 \)

The Young's modulus for steel \( Y_s \) can be expressed as:

\[ Y_s = \frac{F}{A_s} \cdot \frac{L_s}{\Delta L} = \frac{F}{3.0 \times 10^{-5}} \cdot \frac{5.0}{\Delta L} \]

For the copper wire:
- Length \( L_c = 3.0 \, m \)
- Cross-sectional area \( A_c = 4.0 \times 10^{-5} \, m^2 \)

The Young's modulus for copper \( Y_c \) can be expressed as:

\[ Y_c = \frac{F}{A_c} \cdot \frac{L_c}{\Delta L} = \frac{F}{4.0 \times 10^{-5}} \cdot \frac{3.0}{\Delta L} \]

### Step 3: Set up the ratio of Young's moduli
Since the extension \( \Delta L \) is the same for both wires under the same load, we can set up the ratio of Young's moduli:

\[ \frac{Y_s}{Y_c} = \frac{\frac{F}{3.0 \times 10^{-5}} \cdot \frac{5.0}{\Delta L}}{\frac{F}{4.0 \times 10^{-5}} \cdot \frac{3.0}{\Delta L}} \]

### Step 4: Simplify the ratio
The \( F \) and \( \Delta L \) cancel out:

\[ \frac{Y_s}{Y_c} = \frac{5.0}{3.0} \cdot \frac{4.0 \times 10^{-5}}{3.0 \times 10^{-5}} \]

This simplifies to:

\[ \frac{Y_s}{Y_c} = \frac{5.0 \cdot 4.0}{3.0 \cdot 3.0} = \frac{20.0}{9.0} \]

### Step 5: Final result
Thus, the ratio of Young's modulus of steel to that of copper is:

\[ \frac{Y_s}{Y_c} = \frac{20}{9} \]