Introduction:
In this problem, we are given a system consisting of two bars and a pin. A pull of 80 kN is transmitted from bar X to the bar through the pin. We need to determine the diameter of the bars and the pin based on the maximum permissible tensile stress in the bars and the permissible shear stress in the pin.

Assumptions:
- The bars and the pin are made of the same material.
- The applied load is uniformly distributed across the cross-section of the bars.
- The bars and the pin are in equilibrium under the applied load.

Analysis:
To determine the diameter of the bars and the pin, we need to consider the stress in both the bars and the pin separately.

1. Maximum Permissible Tensile Stress in the Bars:
The maximum permissible tensile stress in the bars is given as 120 N/mm^2.

The tensile stress in a bar can be calculated using the formula:
Tensile Stress = Force/Area

Since the applied load is uniformly distributed, the force acting on each bar can be calculated as:
Force = 80 kN / 2 = 40 kN

Let's assume the diameter of the bars as 'd' mm. The area of each bar can be calculated as:
Area = (π/4) * d^2

Substituting the values, we get:
Tensile Stress = 40 kN / [(π/4) * d^2]

According to the given condition, the maximum permissible tensile stress is 120 N/mm^2. Therefore, we can write the equation as:
120 = 40 kN / [(π/4) * d^2]

Simplifying the equation, we get:
d^2 = (40 kN * 4) / (120 * π)
d^2 = (40 * 4) / (120 * π)
d^2 = 4 / (3 * π)
d^2 = 1.273
d ≈ √1.273
d ≈ 1.13 mm

Therefore, the diameter of each bar is approximately 1.13 mm.

2. Permissible Shear Stress in the Pin:
The permissible shear stress in the pin is given as 90 N/mm^2.

The shear stress in a pin can be calculated using the formula:
Shear Stress = Force/Area

Since the applied load is uniformly distributed, the force acting on the pin can be calculated as:
Force = 80 kN

Let's assume the diameter of the pin as 'd_pin' mm. The area of the pin can be calculated as:
Area = (π/4) * d_pin^2

Substituting the values, we get:
Shear Stress = 80 kN / [(π/4) * d_pin^2]

According to the given condition, the permissible shear stress is 90 N/mm^2. Therefore, we can write the equation as:
90 = 80 kN / [(π/4) * d_pin^2]

Simplifying the equation, we get:
d_pin^2 = (80 kN * 4) / (90 * π)
d_pin^2 = (80 * 4) / (90 * π)
d_pin^2 = 3.56