$$\begin{gathered} \text{An infinite slope with slope angle}\beta =22°\text{ consists of soil with the following properties:} \\ \end{gathered}$$

\text{<br>}

\text{Unit weight }\gamma =15.72\text{ kN/}{m}^3

$$\begin{gathered} {\mathrm{}}^{} \\ \text{Cohesion }{c}^{\prime }=12\text{ kPa} \\ \end{gathered}$$

\text{Angle of internal}\text{ friction }{\varphi }^{\prime }=15°

\text{<br>}\text{The critical height of the slope (in m) is \_\_\_\_\_\_\_\_ (rounded off to two decimal places).}

The correct answer is 6.52.

Step-by-step Explanation:

For an infinite slope with a slope angle $\beta$ consisting of cohesive-frictional soil, the factor of safety ($FS$) against sliding at a depth $H$ is given by the following formula:
$$FS=\frac{c^{\prime }+\gamma H{\cos }^2\left(\beta \right)\tan \left({\varphi }^{\prime }\right)}{\gamma H\sin \left(\beta \right)\cos \left(\beta \right)}$$
where:
• $c^{\prime }$ is the cohesion of the soil = 12 kPa = 12 kN/m²
• $\gamma$ is the unit weight of the soil = 15.72 kN/m³
• $\beta$ is the slope angle = 22°
• ${\varphi }^{\prime }$ is the angle of internal friction = 15°

The critical height of the slope ($H_{\mathrm{cr}}$) occurs when the factor of safety is exactly equal to 1 ($FS=1$):
$$1=\frac{c^{\prime }+\gamma H_{\mathrm{cr}}{\cos }^2\left(\beta \right)\tan \left({\varphi }^{\prime }\right)}{\gamma H_{\mathrm{cr}}\sin \left(\beta \right)\cos \left(\beta \right)}$$

Rearranging the equation to solve for $H_{\mathrm{cr}}$:
$$\gamma H_{\mathrm{cr}}\sin \left(\beta \right)\cos \left(\beta \right)=c^{\prime }+\gamma H_{\mathrm{cr}}{\cos }^2\left(\beta \right)\tan \left({\varphi }^{\prime }\right)$$
$$\gamma H_{\mathrm{cr}}\cos \left(\beta \right)\left[\sin \right(\mathrm{��})-\cos (\beta \left)\tan \right({\varphi }^{\prime }\left)\right]=c^{\prime }$$
$$H_{\mathrm{cr}}=\frac{c^{\prime }}{\gamma \cos \left(\beta \right)\left[\sin \right(\beta )-\cos (\beta \left)\tan \right({\varphi }^{\prime }\left)\right]}$$

Now, let us calculate the trigonometric values:
• $\sin (22°)\approx 0.3746$
• $\cos (22°)\approx 0.9272$
• $\tan (15°)\approx 0.2679$

Substitute the calculated trigonometric values into the equation:
• Term inside the brackets:
$$\sin (22°)-\cos (22°)\tan (15°)\approx 0.3746-0.9272\times 0.2679$$
$$\approx 0.3746-0.2484=0.1262$$
• Denominator:
$$15.72\times 0.9272\times 0.1262\approx 1.8394$$
• Critical Height:
$$H_{\mathrm{cr}}\approx \frac{12}{1.8394}\approx 6.5238\text{ m}$$

Rounding to two decimal places, we obtain:
$$H_{\mathrm{cr}}=6.52\text{ m}$$