For a catchment area of 120 km2, the equilibrium discharge in m3/hour of an S-curve obtained by the summation of 6 hour unit hydrograph is
A. 0.2 x 106
B. 0.6 x 106
C. 2.4 x 106
D. 7.2 x 106
Answer: Option A
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A=120 km²=120 *10^6 m²
t=6 hr
h=1 cm= 0.01 m (unit hydrograph)
Q=V/t
=(120*10^6*0.01)/6
=0.2*10^6 m³/hr
Jsk
Q= A/D *10 to the power 4 bcz m3/hr
Where D =duration of time
A= Area of catchment
Q= 120/6 *10 4= 0.2*10 to the power 6
Explain please