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 ?

Explanation

To solve the problem, we need to analyze the forces acting on the rod and the conditions for equilibrium. We will follow these steps:

### Step 1: Understand the Setup
We have a horizontal rod of length 2 m suspended by two vertical wires: one made of steel and the other made of brass. The cross-sectional areas are given, and we need to find the point \( x \) from the steel wire end where a weight can be hung such that the stresses in both wires are equal.

### Step 2: Define the Variables
- Let \( T_S \) be the tension in the steel wire.
- Let \( T_B \) be the tension in the brass wire.
- The cross-sectional area of steel wire \( A_S = 10^{-3} \, m^2 \).
- The cross-sectional area of brass wire \( A_B = 2 \times 10^{-3} \, m^2 \).
- Young's modulus for steel \( Y_S = 2 \times 10^{11} \, Pa \).
- Young's modulus for brass \( Y_B = 10^{11} \, Pa \).

### Step 3: Apply the Condition of Equal Stress
Since the stresses in both wires are equal, we have:
\[ \frac{T_S}{A_S} = \frac{T_B}{A_B} \]
Substituting the areas:
\[ \frac{T_S}{10^{-3}} = \frac{T_B}{2 \times 10^{-3}} \]
This simplifies to:
\[ T_B = 2 T_S \tag{1} \]

### Step 4: Apply the Condition of Equilibrium
For the rod to be in equilibrium, the sum of moments about any point must be zero. Taking moments about the point where the brass wire is attached (let's call this point B):
\[ T_S \cdot x = T_B \cdot (2 - x) \]
Substituting \( T_B \) from equation (1):
\[ T_S \cdot x = 2 T_S \cdot (2 - x) \]
Dividing both sides by \( T_S \) (assuming \( T_S \neq 0 \)):
\[ x = 2(2 - x) \]
Expanding this gives:
\[ x = 4 - 2x \]
Rearranging gives:
\[ 3x = 4 \implies x = \frac{4}{3} \, m \approx 1.33 \, m \]

### Step 5: Conclusion
The distance \( x \) from the steel wire end at which a weight may be hung such that the stresses in both wires are equal is approximately \( 1.33 \, m \).