A soil having the average properties, bulk unit weight $=19\, \text{kN/m}^3$, angle of internal friction $=25^\circ$ and cohesion $=15\, \text{kPa}$, is being formed on a rock slope at an inclination of $35^\circ$ with the horizontal. The critical height (in m) of the soil formation up to which it would be stable without failure is \underline{\hspace{2cm}} (round off to one decimal place). [Assume the soil is formed parallel to the rock bedding plane and there is no ground water effect.]
A soil having the average properties, bulk unit weight $=19\\{kN/m}^3$, angle of internal friction =25 degree and cohesion $=15\\{kPa}$, is being formed on a rock slope at an inclination of 35 degree with the horizontal. The critical height (in m) of the soil formation up to which it would be stable without failure is (round off to one decimal place). [Assume the soil is formed parallel to the rock bedding plane and there is no ground water effect.]
An unsupported slope of height 15 m is shown in the figure (not to scale), in which the slope face makes an angle of 50° with the horizontal. The slope material comprises purely cohesive soil having undrained cohesion 75 kPa. A trial slip circle KLM, with a radius 25 m, passes through the crest and toe of the slope and it subtends an angle 60° at its center O. The weight of the active soil mass (W, bounded by KLMN) is 2500 kN/m, which is acting at a horizontal distance of 10 m from the toe of the slope. Consider the water table to be present at a very large depth from the ground surface.

Considering the trial slip circle KLM, the factor of safety against the failure of the slope under undrained condition (rounded off to two decimal places) is \(\underline{\hspace{1cm}}\).