Question Bank • QB_SWCE_SM_002 NAT • 2 Marks

Section 4: Soil and Water Conservation Engineering

Soil Mechanics

Subtopic: Permeability and seepage analysis

Question Statement:

A falling-head permeability test is conducted on a soil sample of length $L = 12\text{ cm}$ and cross-sectional area $A = 30\text{ cm}^2$. The standpipe cross-sectional area is $a = 1.5\text{ cm}^2$. The head in the standpipe drops from $h_1 = 60\text{ cm}$ to $h_2 = 30\text{ cm}$ in $t = 180\text{ seconds}$. The hydraulic conductivity of the soil $K$ in $\text{cm/s}$ expressed as $K \times 10^3$ is ________ (round off to 2 decimal places).

Answer Key:

2.31

Step-by-Step Solution & Explanation

In a falling head permeameter: $K = \frac{a L}{A t} \ln\left( \frac{h_1}{h_2} \right)$ Given: • $a = 1.5\text{ cm}^2$ • $L = 12\text{ cm}$ • $A = 30\text{ cm}^2$ • $t = 180\text{ s}$ • $\ln(60/30) = \ln(2) \approx 0.69315$ $K = \frac{1.5 \times 12}{30 \times 180} \times 0.69315 = \frac{18}{5400} \times 0.69315 = 0.003333 \times 0.69315 = 2.3105 \times 10^{-3}\text{ cm/s}$ Therefore, $K \times 10^3 = 2.31$.
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