10 m cut in cohesive soil, bedrock 15 m

Note: the rest of the question and the cut slope are missing. The usual form asks for the factor of safety (or safe height) of the cut. The solution below assumes a vertical cut (slope angle $\beta = 90^\circ$) and uses Taylor's stability number method; the same steps apply for any given slope angle.

Data: $H = 10$ m, $c = c_u = 35$ kN/m², $\gamma = 18.5$ kN/m³, $\phi = 0$. Bedrock at 15 m below the original ground, so the depth factor is

$$n_d = \frac{D}{H} = \frac{15}{10} = 1.5$$

Taylor's stability number (for $\phi = 0$)

$$N_s = \frac{c_m}{\gamma H}$$

For steep slopes ($\beta > 53^\circ$) the critical circle is a toe circle regardless of the depth factor, and Taylor's chart gives, for $\beta = 90^\circ$, $N_s \approx 0.261$.

Critical height

$$H_c = \frac{c}{\gamma N_s} = \frac{35}{18.5\times0.261} = 7.25\ \text{m}$$

Factor of safety

$$F = \frac{H_c}{H} = \frac{7.25}{10} = 0.72 \quad \left(\text{or } F = \frac{c}{N_s\gamma H} = \frac{35}{0.261\times185} = 0.72\right)$$

Result: $F < 1$, so a 10 m vertical cut in this soil is unsafe and would fail. The maximum safe height of a vertical cut is about 7.25 m (with $F = 1$). For $F = 1.5$, the cut must be flattened (read $N_s$ for a smaller $\beta$ from Taylor's chart) or supported by a retaining system.

Answer: $F \approx 0.72$ for a vertical cut (unsafe). The depth factor $n_d = 1.5$ matters only for flatter slopes ($\beta<53^\circ$).