**Problem Statement**

We are given the following information:
- Length of the shaft (L) = 2 m
- Power transmitted (P) = 100 N/mm^2
- Maximum permissible shear stress (τ_max) = 60 kW
- Speed of rotation (N) = 150 rpm

We need to determine the required minimum diameter for the shaft and the corresponding angle of twist produced in the shaft.

**Solution**

To solve this problem, we will use the formula for power transmitted by a circular shaft:

\(P = \frac{2π}{60}NT\)

where P is the power transmitted, N is the speed of rotation in rpm, and T is the torque transmitted by the shaft. The torque can be calculated using the formula:

\(T = \frac{π}{16}d^3τ\)

where d is the diameter of the shaft and τ is the shear stress.

**Step 1: Calculate the Torque Transmitted**

Given that the power transmitted (P) is 100 N/mm^2 and the speed of rotation (N) is 150 rpm, we can calculate the torque transmitted (T) using the formula:

\(P = \frac{2π}{60}NT\)

Substituting the given values, we get:

\(100 = \frac{2π}{60} \times 150 \times T\)

Simplifying, we find:

\(T = \frac{100 \times 60}{2π \times 150}\)

\(T ≈ 0.636 \text{ kNm}\)

**Step 2: Calculate the Minimum Diameter**

Given that the maximum permissible shear stress (τ_max) is 60 kW and the torque transmitted (T) is 0.636 kNm, we can calculate the minimum diameter (d) using the formula:

\(T = \frac{π}{16}d^3τ\)

Substituting the given values, we get:

\(0.636 = \frac{π}{16}d^3 \times 60 \times 10^3\)

Simplifying, we find:

\(d^3 = \frac{0.636 \times 16}{π \times 60 \times 10^3}\)

\(d ≈ \sqrt[3]{\frac{0.636 \times 16}{π \times 60 \times 10^3}}\)

\(d ≈ 0.042 \text{ m}\)

**Step 3: Calculate the Angle of Twist**

Given that the length of the shaft (L) is 2 m and the minimum diameter (d) is 0.042 m, we can calculate the angle of twist (θ) using the formula:

\(θ = \frac{TL}{Gd^4}\)

Substituting the given values, we get:

\(θ = \frac{0.636 \times 10^3 \times 2}{90 \times 10^9 \times 0.042^4}\)

Simplifying, we find:

\(θ ≈ 0.001 \text{ rad}\)

Therefore, the required minimum diameter for the shaft is approximately 0.042 m and the corresponding angle of twist produced in the shaft is approximately 0.001 rad.