Answer:
The correct option is A.
Step-by-step explanation:
Following the equation of continuum, AV remains constant.
Case a
(3A)(V0) = AV1 + AV1 + AV1
3AV0 = 3AV1
V1 = V0
Case b
(A)(V0) = (A/3)V2 + (A/3)V2 + (A/3)V2 + (A/3)V2
AV0 = 4V2/3
V2 = 3/4V0
Case c
(A/2)(V0) = AV3 + AV3 + AV3
AV0/2 = 3AV3
V3 = V0/6
Case d
(3A)(V0) = 2AV4 + 2AV4
3AV0 = 4AV4
V4 = 3V0/4
Comparing all the flow speeds, V1 is the largest.
Thus, the correct option is A.