Problem Analysis:
We are given a rod of length 4m suspended from two wires. One wire is made of steel and the other is made of copper. We need to find the distance x such that both wires experience equal stress. The cross-sectional areas of the wires are given as 10 m^2 and 20 m^2 respectively, and the Young's modulus for steel and copper are given as 200 and 400 respectively.

Solution:
To solve this problem, we can use the formula for stress:

Stress = Force/Area

For the steel wire:
Let the force acting on the steel wire be F1.
The cross-sectional area of the steel wire is given as 10 m^2.
The stress experienced by the steel wire is equal to the stress experienced by the copper wire, so we can write:

Stress1 = Stress2

F1/10 = F2/20 ...(1)

For the copper wire:
Let the force acting on the copper wire be F2.
The cross-sectional area of the copper wire is given as 20 m^2.

Since the rod is suspended from the two wires, the total force acting on the rod is equal to the sum of the forces acting on the two wires:

F1 + F2 = Weight of the rod

The weight of the rod can be calculated using the formula:

Weight = mass × acceleration due to gravity

Since the mass of the rod is not given, we can assume a mass of 1 kg for simplicity. Therefore, the weight of the rod is:

Weight = 1 kg × 9.8 m/s^2 = 9.8 N

Substituting this value into the equation:

F1 + F2 = 9.8 N ...(2)

Now we have two equations (equations 1 and 2) with two unknowns (F1 and F2). We can solve these equations to find the values of F1 and F2.

Solving equation 1 for F1, we get:

F1 = (10/20) × F2

Substituting this value into equation 2, we get:

(10/20) × F2 + F2 = 9.8 N

Simplifying the equation, we get:

1.5 × F2 = 9.8 N

Solving for F2, we find:

F2 = 9.8 N / 1.5 = 6.53 N

Substituting this value back into equation 1, we can solve for F1:

F1 = (10/20) × 6.53 N = 3.265 N

Now, we can find the distance x by considering the moments acting on the rod about the point where the steel wire is attached.

The moment due to the force F1 about this point is:

M1 = F1 × x

The moment due to the weight of the rod about this point is:

M2 = Weight × (4 - x)

Since the rod is in equilibrium, the moments M1 and