The correct answer is A. 0.5 N/mm2 and 30°.
The cohesion of a soil is the maximum shear stress that can be developed in the soil when there is no relative movement between the soil particles. The angle of internal friction is the angle between the major principal stress and the failure plane.
In this case, the soil failed under an axial vertical stress of 100 kN/m2. The failure plane was inclined to the horizontal plane at an angle of 45°. This means that the major principal stress was acting at an angle of 45° to the horizontal plane.
The cohesion of the soil can be calculated using the following equation:
$c = \frac{1}{2} \sigma_1 \tan \phi$
where $c$ is the cohesion, $\sigma_1$ is the major principal stress, and $\phi$ is the angle of internal friction.
Substituting the values given in the question, we get:
$c = \frac{1}{2} (100 \text{ kN/m}^2) \tan 45° = 0.5 \text{ N/mm}^2$
The angle of internal friction can be calculated using the following equation:
$\phi = \arctan \frac{c}{\sigma_1}$
Substituting the values given in the question, we get:
$\phi = \arctan \frac{0.5 \text{ N/mm}^2}{100 \text{ kN/m}^2} = 30°$
Therefore, the values of cohesion and angle of internal friction for the soil are respectively 0.5 N/mm2 and 30°.