Final answer:
The rate of flow of water in the wide section of the pipe is the same as in the narrow section, which is 9.0 L/min, based on the conservation of mass and the incompressibility of water.
Step-by-step explanation:
The rate of flow of water in the wide section of a pipe, when the diameter at the wide section is 4 times the diameter at the narrow section, can be found using the principle of conservation of mass. This principle requires that the mass of water entering a section of the pipe must be equal to the mass leaving that section. Since the rate of flow or volume flow rate is the product of the cross-sectional area and the velocity of the fluid, we can express this principle with the equation A1V1 = A2V2, where A represents the area and V represents the velocity.
Using the information given, we have the diameter of the wide section (D1) is 4 times the diameter of the narrow section (D2), so the area of the wide section (A1) will be 16 times larger than the area of the narrow section (A2), since area depends on the square of the diameter (A = π(d/2)^2). The rate of flow at the narrow section is given as 9.0 L/min. Because the wide section has a larger cross-sectional area, the velocity of the water will decrease to maintain the same rate of flow. To find the rate of flow in the wide section, we can use the relationship between the velocity and area while the rate of flow in terms of volume remains constant (Volume flow rate = Area × Velocity).
Therefore, the rate of flow in the wide section is also 9.0 L/min, due to the incompressibility of water and the conservation of mass.