Step 1: Recall the duty-delta-base period relation.
For an irrigation outlet, duty \(D\) (in hectares per cumec, hectares irrigated per \(m^3/s\) of continuous flow) relates to the base period \(B\) (in days) and the total depth of watering \(\Delta\) (in metres) by \[ \Delta = \frac{8.64\,B}{D} \] The constant 8.64 converts a discharge of \(1\ m^3/s\) running for 1 day into a depth over 1 hectare (\(1\ m^3/s \times 86400\ s = 86400\ m^3\), and \(86400\ m^3\) over 1 hectare \((10000\ m^2)\) gives a depth of 8.64 m).
Step 2: Substitute the given base period and depth to get duty.
Here \(B=140\) days and \(\Delta=40\ cm=0.4\ m\). Rearranging for \(D\): \[ D=\frac{8.64B}{\Delta}=\frac{8.64\times140}{0.4}=\frac{1209.6}{0.4}=3024\ \text{ha/cumec} \]
Step 3: Convert duty and CCA into discharge. \[ Q=\frac{\text{CCA}}{D}=\frac{10000}{3024}=3.307\ m^3/s \]
Step 4: Match against the options. \(3.307\ m^3/s\) lies strictly between 3 and 4, ruling out the other bands (13-14, 23-24, 33-34), which would need a duty roughly ten, twenty, or thirty times smaller than 3024 ha/cumec.
Final Answer:
The outlet discharge is about 3.31 \(m^3/s\), between 3 and 4. \[ \boxed{Q \approx 3.31\ m^3/s} \]