RESULTED from 1 cm of rainfall excess depthWe know that∴ \(rainfall\;excess = \frac{{Volume\;of\;DIRECT\;runoff}}{{Catchment\;area}}\)The area under triangle = Volume of direct runoff\(\frac{1}{2} × {q_p} × Base\;period = Area\;of\;catchment × 1\;cm\;runoff\;\;\)Where qp is peak discharge of unit hydrographGiven,Base period = 20 hours, Area of catchment = 500 HA = 50 × 104 m2\(\frac{1}{2} × {q_p} × Base\;period = Area\;of\;catchment × 1\;cm\;runoff\;\;\)\(\frac{1}{2} × {q_p} × 20 = 50 \times 10^4m^2 × 1\;cm\;runoff\;\;\)qp = 5000 m3/hr∴ Peak Discharge = 5000 m3/hr