Question:

If water is flowing at the same depth in most hydraulically efficient triangular and rectangular channel sections then the ratio of hydraulic radius of triangular section to that of rectangular section is

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In hydraulically efficient channel sections, the hydraulic radius of a triangular section is \( \frac{1}{\sqrt{2}} \) times that of a rectangular section at the same depth.
Updated On: Jan 2, 2026
  • \( \frac{1}{\sqrt{2}} \)
  • \( \sqrt{2} \)
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The Correct Option is A

Solution and Explanation

In hydraulically efficient channel sections, the hydraulic radius \( R \) is defined as the ratio of the cross-sectional area \( A \) to the wetted perimeter \( P \): \[ R = \frac{A}{P}. \] For a triangular section, the hydraulic radius is \( R_{\text{tri}} = \frac{A_{\text{tri}}}{P_{\text{tri}}} \), and for a rectangular section, the hydraulic radius is \( R_{\text{rect}} = \frac{A_{\text{rect}}}{P_{\text{rect}}} \). When comparing the hydraulic radii of the two sections, the ratio is: \[ \frac{R_{\text{tri}}}{R_{\text{rect}}} = \frac{1}{\sqrt{2}}. \] Final Answer: \( \frac{1}{\sqrt{2}} \)
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