Question:

The total observed runoff volume during a $6 \, \text{hr}$ storm with a uniform intensity of $1.5 \, \text{cm/hr}$ is $21.6 \times 10^6 \, \text{m}^3$. What will be the average infiltration rate of the basin if the area of the basin is $300 \, \text{km}^2$?

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For infiltration problems, always ensure rainfall, runoff, and basin area are expressed in consistent units before calculating f = I − R.
Updated On: Jan 7, 2025
  • 2.4 mm/hr
  • 3 mm/hr
  • 1.2 mm/hr
  • 4 mm/hr
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The Correct Option is B

Solution and Explanation

The infiltration rate ($f$) is calculated using:
\[f = I - R\]
where:
$I =$ Total rainfall intensity,
$R = \frac{\text{Runoff volume}}{\text{Area}} = \frac{21.6 \times 10^6}{300 \times 10^6} = 0.072 \, \text{m/hr}$ (converted to mm/hr: 72 mm/hr).
Rainfall intensity over 6 hours:
\[I = 1.5 \, \text{cm/hr} \times 6 = 9 \, \text{cm} = 90 \, \text{mm}.\]
Runoff converted to mm/hr:
\[R = 72 \, \text{mm/hr}.\]
Infiltration rate:
\[f = 90 - 72 = 18 \, \text{mm/hr over 6 hours}.\]
Average infiltration rate:
\[f_{\text{avg}} = \frac{18}{6} = 3 \, \text{mm/hr}.\]

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