A light rod of length `2 m` is suspended from the 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 `0.1 cm^(2)`. The other wire is a brass of cross section `0.2 cm^(2)`. A weight is suspended from a certain point of the rod such that equal stress are produced in both the wires. Which of the following are correct?

Options

A. The ratio of tension in the steel and brass wires is `0.5`
B. The load is suspended at a distance of `400//3` cm from the steel wire.
C. Both (a) and (b) are correct
D. Neither (a) nor b) is correct.

Explanation

To solve the problem step by step, we will analyze the situation involving the light rod suspended by two wires made of different materials (steel and brass) and determine the conditions for equal stress and the position of the suspended weight.

### Step 1: Understand the Setup
We have a light rod of length \( L = 2 \, \text{m} \) suspended horizontally from the ceiling. It is held by two vertical wires: one made of steel with a cross-sectional area \( A_s = 0.1 \, \text{cm}^2 \) and the other made of brass with a cross-sectional area \( A_b = 0.2 \, \text{cm}^2 \).

### Step 2: Equal Stress Condition
According to the problem, the stresses in both wires are equal. Stress (\( \sigma \)) is defined as:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force in the wire and \( A \) is the cross-sectional area.

For the steel wire:
\[ \sigma_s = \frac{F_s}{A_s} \]
For the brass wire:
\[ \sigma_b = \frac{F_b}{A_b} \]
Setting these equal gives:
\[ \frac{F_s}{A_s} = \frac{F_b}{A_b} \]

### Step 3: Calculate the Ratio of Forces
From the above equation, we can express the ratio of the forces:
\[ \frac{F_s}{F_b} = \frac{A_s}{A_b} \]
Substituting the given areas:
\[ \frac{F_s}{F_b} = \frac{0.1 \, \text{cm}^2}{0.2 \, \text{cm}^2} = \frac{1}{2} \]
This means:
\[ F_s = \frac{1}{2} F_b \]

### Step 4: Rotational Equilibrium
To find the position of the weight suspended from the rod, we need to consider rotational equilibrium about the point of application of the load. Let the distance from the steel wire to the load be \( x \) and from the brass wire to the load be \( L - x \).

The condition for rotational equilibrium is:
\[ F_s \cdot x = F_b \cdot (L - x) \]

### Step 5: Substitute the Force Ratio
Substituting \( F_s = \frac{1}{2} F_b \) into the equilibrium equation:
\[ \left(\frac{1}{2} F_b\right) \cdot x = F_b \cdot (L - x) \]
Dividing both sides by \( F_b \) (assuming \( F_b \neq 0 \)):
\[ \frac{1}{2} x = L - x \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ \frac{1}{2} x + x = L \]
\[ \frac{3}{2} x = L \]
\[ x = \frac{2}{3} L \]
Substituting \( L = 2 \, \text{m} \):
\[ x = \frac{2}{3} \cdot 2 = \frac{4}{3} \, \text{m} = 400 \, \text{cm} \]

### Conclusion
The weight is suspended at a distance of \( \frac{400}{3} \, \text{cm} \) from the steel wire.

### Summary of Correct Options
1. The ratio of the forces in the wires is \( F_s : F_b = 1 : 2 \).
2. The distance from the steel wire to the load is \( \frac{400}{3} \, \text{cm} \).
3. Therefore, both statements A and B are correct.

### Final Answer
The correct options are:
- Option 1: Correct
- Option 2: Correct
- Option 3: Both A and B are correct
- Option 4: Incorrect