Two exactly similar wires of steel and copper are stretched by equal force. If the total elongation is 2 cm, then how much is the elongation in steel and copper wire respectively? Given, `Y_("steel")=20xx10^(11)"dyne/cm"^(2),Y_("copper")=12xx10^(11)"Dyne/cm"^(2)`

Options

A. `1.25cm, 0.75cm`
B. `0.75cm, 1.25cm`
C. `1.15cm, 0.85cm`
D. `0.5cm, 1.15cm`

Answer

To solve the problem, we will use the relationship between Young's modulus, stress, and strain. Let's break down the solution step by step.

### Step 1: Understand the relationship of Young's Modulus
Young's modulus (Y) is defined as the ratio of stress to strain. Mathematically, it can be expressed as:

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

Where:
- \( F \) = Force applied
- \( A \) = Cross-sectional area
- \( \Delta L \) = Change in length (elongation)
- \( L \) = Original length

### Step 2: Write the equations for both materials
For the steel wire:
\[ Y_{steel} = \frac{F}{A} \cdot \frac{L}{\Delta L_{steel}} \]

For the copper wire:
\[ Y_{copper} = \frac{F}{A} \cdot \frac{L}{\Delta L_{copper}} \]

### Step 3: Set up the ratio of Young's Modulus
Since the wires are similar, we can set up the ratio of Young's moduli:

\[ \frac{Y_{steel}}{Y_{copper}} = \frac{\Delta L_{copper}}{\Delta L_{steel}} \]

### Step 4: Substitute the values of Young's Modulus
Given:
- \( Y_{steel} = 20 \times 10^{11} \, \text{dyne/cm}^2 \)
- \( Y_{copper} = 12 \times 10^{11} \, \text{dyne/cm}^2 \)

Substituting these values into the ratio gives:

\[ \frac{20 \times 10^{11}}{12 \times 10^{11}} = \frac{\Delta L_{copper}}{\Delta L_{steel}} \]

This simplifies to:

\[ \frac{20}{12} = \frac{\Delta L_{copper}}{\Delta L_{steel}} \quad \Rightarrow \quad \frac{5}{3} = \frac{\Delta L_{copper}}{\Delta L_{steel}} \]

### Step 5: Express elongation in terms of one variable
From the ratio, we can express the elongation of copper in terms of steel:

\[ \Delta L_{copper} = \frac{5}{3} \Delta L_{steel} \]

### Step 6: Use the total elongation
We know that the total elongation is given as:

\[ \Delta L_{copper} + \Delta L_{steel} = 2 \, \text{cm} \]

Substituting the expression for \(\Delta L_{copper}\):

\[ \frac{5}{3} \Delta L_{steel} + \Delta L_{steel} = 2 \]

### Step 7: Combine and solve for \(\Delta L_{steel}\)
Combining the terms gives:

\[ \frac{5}{3} \Delta L_{steel} + \frac{3}{3} \Delta L_{steel} = 2 \]

\[ \frac{8}{3} \Delta L_{steel} = 2 \]

Multiplying both sides by \(\frac{3}{8}\):

\[ \Delta L_{steel} = \frac{2 \times 3}{8} = \frac{6}{8} = 0.75 \, \text{cm} \]

### Step 8: Find \(\Delta L_{copper}\)
Now, substituting back to find \(\Delta L_{copper}\):

\[ \Delta L_{copper} = 2 - \Delta L_{steel} = 2 - 0.75 = 1.25 \, \text{cm} \]

### Final Answer
The elongation in steel is \(0.75 \, \text{cm}\) and in copper is \(1.25 \, \text{cm}\).

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