A brass of length `2 m` and cross-sectional area `2.0 cm^(2)` is attached end to end to a steel rod of length and cross-sectional area `1.0 cm^(2)`. The compound rod is subjected to equal and oppsite pulls of magnitude `5 xx 10^(4) N` at its ends. If the elongations of the two rods are equal, the length of the steel rod `(L)` is
{`Y_(Brass) = 1.0 xx 10^(11) N//m^(2) ` and `Y_(steel) = 2.0 xx 10^(11) N//m^(2)`

Options

A. `1.5 m`
B. `1.8 m`
C. `1 m`
D. `2 m`

Answer

To find the length of the steel rod \( L \) when attached end to end with a brass rod, we can use the concept of Young's modulus and the condition that the elongations of both rods are equal under the same load.

### Given Data:
- Length of brass rod, \( L_b = 2 \, \text{m} \)
- Cross-sectional area of brass rod, \( A_b = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2 \)
- Cross-sectional area of steel rod, \( A_s = 1.0 \, \text{cm}^2 = 1.0 \times 10^{-4} \, \text{m}^2 \)
- Force applied, \( F = 5 \times 10^4 \, \text{N} \)
- Young's modulus of brass, \( Y_b = 1.0 \times 10^{11} \, \text{N/m}^2 \)
- Young's modulus of steel, \( Y_s = 2.0 \times 10^{11} \, \text{N/m}^2 \)

### Step-by-Step Solution:

1. **Understanding the Elongation Formula**:
The elongation \( \Delta L \) of a material under stress can be expressed as:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where \( F \) is the force applied, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus.

2. **Set Up the Equations for Elongation**:
For the brass rod:
\[ \Delta L_b = \frac{F \cdot L_b}{A_b \cdot Y_b} \]
For the steel rod:
\[ \Delta L_s = \frac{F \cdot L}{A_s \cdot Y_s} \]

3. **Equate the Elongations**:
Since the elongations are equal, we have:
\[ \Delta L_b = \Delta L_s \]
Substituting the expressions we derived:
\[ \frac{F \cdot L_b}{A_b \cdot Y_b} = \frac{F \cdot L}{A_s \cdot Y_s} \]

4. **Cancel Out Common Terms**:
The force \( F \) cancels out from both sides:
\[ \frac{L_b}{A_b \cdot Y_b} = \frac{L}{A_s \cdot Y_s} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ L = \frac{L_b \cdot A_s \cdot Y_s}{A_b \cdot Y_b} \]

6. **Substituting the Known Values**:
Substitute \( L_b = 2 \, \text{m} \), \( A_b = 2.0 \times 10^{-4} \, \text{m}^2 \), \( A_s = 1.0 \times 10^{-4} \, \text{m}^2 \), \( Y_s = 2.0 \times 10^{11} \, \text{N/m}^2 \), and \( Y_b = 1.0 \times 10^{11} \, \text{N/m}^2 \):
\[ L = \frac{2 \cdot (1.0 \times 10^{-4}) \cdot (2.0 \times 10^{11})}{(2.0 \times 10^{-4}) \cdot (1.0 \times 10^{11})} \]

7. **Calculating the Length of the Steel Rod**:
Simplifying the equation:
\[ L = \frac{2 \cdot 2.0}{2.0} = 2 \, \text{m} \]

### Final Answer:
The length of the steel rod \( L \) is \( 2 \, \text{m} \).

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