Two exactly similar wire of steel and copper are stretched by equal force. If the difference in their elongations is 0.5 cm, the elongation `(l)` of each wire is
`Y_(s)("steel") =2.0xx10^(11) N//m^(2)`
`Y_(c)("copper")=1.2xx10^(11)N//m^(2)`

Options

A. `l_(s)=0.75 cm,l_(c)=1.25 cm`
B. `l_(s)=1.25 cm,l_(c)=0.75 cm`
C. `l_(s)=0.25 cm,l_(c)=0.75 cm`
D. `l_(s)=0.75cm,l_(c)=0.25 cm`

Answer

To solve the problem, we need to find the elongation of each wire (steel and copper) given the difference in their elongations and their respective Young's moduli.

### Step 1: Understand the relationship between Young's modulus, stress, and strain
Young's modulus (Y) is defined as the ratio of stress to strain. For two wires of the same length (L) and cross-sectional area (A), the elongation (ΔL) can be expressed as:

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

Where:
- \( Y \) = Young's modulus
- \( F \) = Force applied
- \( A \) = Cross-sectional area
- \( \Delta L \) = Elongation
- \( L \) = Original length

### Step 2: Set up the equations for both wires
Let \( \Delta L_S \) be the elongation of the steel wire and \( \Delta L_C \) be the elongation of the copper wire.

From the definition of Young's modulus, we can write:

\[ Y_S = \frac{F \cdot L}{A \cdot \Delta L_S} \quad \text{(for steel)} \]
\[ Y_C = \frac{F \cdot L}{A \cdot \Delta L_C} \quad \text{(for copper)} \]

### Step 3: Relate the elongations using Young's moduli
Since the wires are similar (same dimensions), we can relate their elongations:

\[ \frac{Y_S}{Y_C} = \frac{\Delta L_C}{\Delta L_S} \]

Substituting the values of Young's moduli:

\[ \frac{2.0 \times 10^{11}}{1.2 \times 10^{11}} = \frac{\Delta L_C}{\Delta L_S} \]

Calculating the ratio:

\[ \frac{2.0}{1.2} = \frac{\Delta L_C}{\Delta L_S} \implies \Delta L_C = \frac{2.0}{1.2} \Delta L_S \approx 1.67 \Delta L_S \]

### Step 4: Use the difference in elongations
We know that the difference in elongations is given as:

\[ \Delta L_C - \Delta L_S = 0.5 \text{ cm} \]

Substituting \( \Delta L_C \) in terms of \( \Delta L_S \):

\[ 1.67 \Delta L_S - \Delta L_S = 0.5 \]

This simplifies to:

\[ 0.67 \Delta L_S = 0.5 \]

### Step 5: Solve for \( \Delta L_S \)
Now, we can solve for \( \Delta L_S \):

\[ \Delta L_S = \frac{0.5}{0.67} \approx 0.746 \text{ cm} \approx 0.75 \text{ cm} \]

### Step 6: Find \( \Delta L_C \)
Now, substitute \( \Delta L_S \) back to find \( \Delta L_C \):

\[ \Delta L_C = 1.67 \Delta L_S = 1.67 \times 0.75 \approx 1.25 \text{ cm} \]

### Final Results
- Elongation of steel wire \( \Delta L_S \approx 0.75 \text{ cm} \)
- Elongation of copper wire \( \Delta L_C \approx 1.25 \text{ cm} \)