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

Explanation

To solve the problem, we need to find the length of the steel rod (L) when a brass rod and a steel rod are connected end to end and subjected to equal and opposite pulls. The elongations of both rods are equal.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Length of brass rod, \( L_{brass} = 2 \, \text{m} \)
- Cross-sectional area of brass rod, \( A_{brass} = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2 \)
- Cross-sectional area of steel rod, \( A_{steel} = 1.0 \, \text{cm}^2 = 1.0 \times 10^{-4} \, \text{m}^2 \)
- Young's modulus of brass, \( Y_{brass} = 1.0 \times 10^{11} \, \text{N/m}^2 \)
- Young's modulus of steel, \( Y_{steel} = 2.0 \times 10^{11} \, \text{N/m}^2 \)
- Force applied, \( F = 5 \times 10^4 \, \text{N} \)

2. **Use the Formula for Young's Modulus:**
The formula for Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]
Rearranging gives:
\[ \Delta L = \frac{F L_0}{A Y} \]

3. **Calculate Elongation of Brass Rod:**
For the brass rod:
\[ \Delta L_{brass} = \frac{F L_{brass}}{A_{brass} Y_{brass}} = \frac{(5 \times 10^4) \times 2}{(2.0 \times 10^{-4}) \times (1.0 \times 10^{11})} \]
Simplifying this:
\[ \Delta L_{brass} = \frac{10^5}{2.0 \times 10^{-4} \times 1.0 \times 10^{11}} = \frac{10^5}{2.0 \times 10^7} = 0.005 \, \text{m} = 5 \, \text{mm} \]

4. **Set Up Equation for Steel Rod:**
Since the elongations are equal:
\[ \Delta L_{steel} = \Delta L_{brass} \]
Thus,
\[ \Delta L_{steel} = \frac{F L_{steel}}{A_{steel} Y_{steel}} = 5 \, \text{mm} \]

5. **Calculate Elongation of Steel Rod:**
\[ 5 \times 10^{-3} = \frac{(5 \times 10^4) L_{steel}}{(1.0 \times 10^{-4}) \times (2.0 \times 10^{11})} \]
Rearranging gives:
\[ L_{steel} = \frac{5 \times 10^{-3} \times (1.0 \times 10^{-4}) \times (2.0 \times 10^{11})}{5 \times 10^4} \]
Simplifying:
\[ L_{steel} = \frac{(1.0 \times 10^{-4}) \times (2.0 \times 10^{11})}{10^4} = \frac{2.0 \times 10^{7}}{10^4} = 2000 \, \text{m} = 2 \, \text{m} \]

6. **Final Answer:**
The length of the steel rod \( L_{steel} = 2 \, \text{m} \).