Question:

In a laboratory experiment, water discharge through a porous rock sample in 2 hours was 10 cm3. The cylindrical rock sample is 10 cm long and has a diameter of 50 mm. If the discharge occurred at a constant head of 300 cm, the coefficient of permeability of the rock sample is _________ ×10–6 cm/s. (Round off to two decimal places.)

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Use Darcy’s Law: \( K = \frac{Q L}{A h} \). Always convert discharge time to seconds for consistent units.
Updated On: Dec 6, 2025
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Correct Answer: 2.3

Solution and Explanation

Step 1: Given data.
Discharge \( Q = 10 \, \text{cm}^3 \) in 2 hours = \( 10 / (2 × 3600) = 1.39 × 10^{-3} \, \text{cm}^3/s \).
Length \( L = 10 \, \text{cm}, \) Head \( h = 300 \, \text{cm}, \) Diameter \( D = 5 \, \text{cm}. \)
Area \( A = \pi (D/2)^2 = 3.14 × (2.5)^2 = 19.63 \, \text{cm}^2. \)
Step 2: Using Darcy’s Law.
\[ Q = K \, A \, \frac{h}{L} \] \[ K = \frac{Q L}{A h} \] Step 3: Substitution.
\[ K = \frac{1.39 × 10^{-3} × 10}{19.63 × 300} = 2.36 × 10^{-6} \, \text{cm/s} \] Step 4: Conversion to ×10–6.
\( K = 2.36 × 10^{-6} \, \text{cm/s} = 2.36 ×10^{-6} \). Step 5: Conclusion.
Coefficient of permeability = 2.36 ×10–6 cm/s.
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