The discharge over a Cipolletti weir of length 2.185 m when the head over the weir is 1 m, is

ANS:D - 4.0 m³.

I apologize for the confusion, but it seems there might have been a miscalculation or misunderstanding in the explanation provided. Let's recalculate the discharge over the Cipolletti weir. The discharge over a Cipolletti weir can be calculated using the Francis formula: Q=C×L×H^3/2 Where:

• Q is the discharge,
• C is the weir coefficient (dimensionless),
• L is the length of the weir,
• H is the head of water over the weir.

Given:

• Length of the weir, L=2.185 m
• Head over the weir, H=1 m

We need the weir coefficient C to calculate the discharge accurately. However, without the specific value of C, we cannot calculate the exact discharge. Let's assume a typical value of C for a Cipolletti weir, which is often around 3.0. Q=3.0×2.185×1^3/2 Q=3.0×2.185×1 Q=6.555m3/s Therefore, the discharge over a Cipolletti weir of length 2.185 m when the head over the weir is 1 m is approximately 6.555 m3/s6.555m3/s. It seems there might have been a mistake in the calculation. The correct discharge is approximately 6.555 m3/s6.555m3/s, not 4.0 m³.