Question:medium

If the diameter of the well is doubled, the water yield will increase by about:

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The yield of a well is logarithmically dependent on its radius (\(Q \propto 1/\ln(R/r)\)). This means that doubling the radius gives only a small, incremental increase in yield (typically 10-15%), making it generally uneconomical to drill a much larger well just for a minor increase in discharge.
Updated On: Feb 19, 2026
  • 5%
  • 8%
  • 10%
  • 15%
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The Correct Option is C

Solution and Explanation

Step 1: The Thiem equation for steady-state discharge \(Q\) from a well in an unconfined aquifer is \( Q = \frac{\pi k (H^2 - h^2)}{\ln(R/r)} \), where \(k\) is permeability, \(H\) is initial saturated thickness, \(h\) is head in the well, \(R\) is radius of influence, and \(r\) is well radius.
Step 2: To analyze the effect of doubling the well radius, let initial yield be \(Q_1\) with radius \(r_1\) and new yield be \(Q_2\) with radius \(r_2 = 2r_1\). The ratio of yields is: \[ \frac{Q_2}{Q_1} = \frac{\ln(R/r_1)}{\ln(R/r_2)} = \frac{\ln(R/r_1)}{\ln(R/(2r_1))} = \frac{\ln(R) - \ln(r_1)}{\ln(R) - \ln(2r_1)} = \frac{\ln(R) - \ln(r_1)}{\ln(R) - \ln(2) - \ln(r_1)} \] Step 3: Evaluate the increase with typical values, assuming \(R = 200\) m and \(r_1 = 0.2\) m, where \(R\) is significantly larger than \(r\). \[ \frac{Q_2}{Q_1} = \frac{\ln(200/0.2)}{\ln(200/0.4)} = \frac{\ln(1000)}{\ln(500)} = \frac{6.907}{6.214} \approx 1.11 \] This indicates an approximate 11% increase in yield. Doubling the well diameter results in a modest yield increase, not a doubling, with 10% being a common approximate value.
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