Explanation

For a rectangular channel to have maximum discharge for a given cross-sectional area, the hydraulic radius must be maximized. This occurs when the width is twice the depth (or depth is half the width). Wait, let me double check. Hydraulic radius R = A/P. For rectangle, A = b*d, P = b + 2d. To maximize A for fixed P, or maximize Q for fixed A? The question says 'discharge will be maximum if its depth is...'. Usually this implies for a given cross-sectional area and slope/roughness. Max discharge occurs when wetted perimeter is minimum for a given area. For rectangle, min perimeter for fixed area means b = 2d (width is twice depth), so d = b/2. Thus depth is half the width. Wait, option B says 'Equal to its width'. Option C says 'Half of its width'. So C is correct. Let me verify standard fluid mechanics MCQs. 'The most economical rectangular channel section has width equal to twice the depth.' So depth = half width. Therefore C is correct. Current answer is C. I will confirm it's correct.