A light rod of length of length `2 m` is suspended from a ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross - section `10^(-3) m^(2)` and the other is of brass of cross- section `2xx10^(-3) m^(2).x` is the distance from the steel wire end, at which a weight may be hung. ` Y_(steel) = 2xx10^(11) Pa and Y_(brass) = 10^(11) Pa`
Which of the following statement(s) is /are correct ?

Options

A. (A) `x= 1.2 m` , if the strains of both the wires are to be equal
B. (B) ` x= 1.42 m` , if the stresses of the wires are to be equal
C. (C) `x= 1m` , if the stains of both the wires are to be equal
D. (D) ` x = 1.33 m`, if the stresses of the wires are to be equal

Explanation

To solve the problem, we need to find the position \( x \) from the steel wire end where a weight can be hung such that the stress in both wires (steel and brass) is equal.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Length of the rod, \( L = 2 \, \text{m} \)
- Cross-sectional area of steel wire, \( A_s = 10^{-3} \, \text{m}^2 \)
- Cross-sectional area of brass wire, \( A_b = 2 \times 10^{-3} \, \text{m}^2 \)
- Young's modulus of steel, \( Y_s = 2 \times 10^{11} \, \text{Pa} \)
- Young's modulus of brass, \( Y_b = 10^{11} \, \text{Pa} \)

2. **Write the Stress Equations:**
- The stress in the steel wire can be expressed as:
\[ \sigma_s = \frac{T_s}{A_s} \]
- The stress in the brass wire can be expressed as:
\[ \sigma_b = \frac{T_b}{A_b} \]

3. **Set the Stresses Equal:**
- For the stresses to be equal:
\[ \frac{T_s}{A_s} = \frac{T_b}{A_b} \]
- Substituting the areas:
\[ \frac{T_s}{10^{-3}} = \frac{T_b}{2 \times 10^{-3}} \]
- Rearranging gives:
\[ \frac{T_s}{T_b} = \frac{1}{2} \]

4. **Torque Equilibrium:**
- Let \( x \) be the distance from the steel wire where the weight is hung. The distance from the brass wire will then be \( 2 - x \).
- The torque about the steel wire due to the weight is:
\[ \tau_s = T_s \cdot x \]
- The torque about the brass wire is:
\[ \tau_b = T_b \cdot (2 - x) \]
- For equilibrium:
\[ T_s \cdot x = T_b \cdot (2 - x) \]

5. **Substituting the Tension Ratio:**
- Using \( \frac{T_s}{T_b} = \frac{1}{2} \), we can express \( T_s \) in terms of \( T_b \):
\[ T_s = \frac{1}{2} T_b \]
- Substituting this into the torque equilibrium equation:
\[ \frac{1}{2} T_b \cdot x = T_b \cdot (2 - x) \]
- Cancel \( T_b \) (assuming \( T_b \neq 0 \)):
\[ \frac{1}{2} x = 2 - x \]

6. **Solving for \( x \):**
- Multiply through by 2 to eliminate the fraction:
\[ x = 4 - 2x \]
- Rearranging gives:
\[ 3x = 4 \implies x = \frac{4}{3} \, \text{m} \]

### Conclusion:
The position \( x \) from the steel wire where the weight can be hung is \( \frac{4}{3} \, \text{m} \).