The volumetric flow rate of fluid is found using the equation:

Where  is the velocity of the fluid and  is the cross-sectional area of the space through which the fluid is flowing. Use the continuity equation, we see that , therefore

In this problem, the cross-section of the pipe is a circle, which is

The area of the exit cross-section is:

Plug in these variables into the continuity equation and solve:

The volumetric flow rate of fluid is found using the equation:

Where  is the velocity of the fluid and  is the cross-sectional area of the space through which the fluid is flowing.

Use the continuity equation, we see that , therefore:

In this problem,

The cross-section of the pipe is a circle, which is:

Plugging our variables into the continuity equation gives us

Based on the continuity equation, both parts of the flow must have the same volumetric flow rate.

The volumetric flow rate of fluid can be defined using the equation:

Where  is the volume of the fluid and  is the time the fluid is flowing. To solve this problem, time must be converted from minutes to seconds:

Using the volumetric flow rate equation we find that the volume of molasses is:

The volumetric flow rate of fluid is found using the equation:

Where  is the velocity of the fluid and  is the cross-sectional area of the space through which the fluid is flowing. In this problem the cross-section of the water pipe is a circle. The area of the cross-section is:

The volumetric flow rate is:

The cross-section of the oil pipe is a square. The area of the cross-section is:

The volumetric flow rate is:

The water pipe has the larger volumetric flow rate.

For this question, we need to consider the flow of a fluid through a cylindrical pipe. We then need to determine which parameter, when changed, will decrease the flow rate of fluid through the pipe.

First, let's consider a few variables that are implicated in fluid flow. In fact, the flow of a fluid through a pipe (or a connection of pipes) is analogous to a circuit and Ohm's law. For example, the expression for Ohm's law is as follows:

From this expression, we can see that the voltage difference (driving force for movement of charge) is equal to the current (flow of charge) multiplied by the resistance (which impedes the flow of charge). Similarly, the flow of a fluid through a pipe can be expression as follows:

Where  is equal to the pressure difference,  is equal to flow rate, and  is equal to resistance. This expression takes the same form as Ohm's law. The pressure (driving force for movement of the fluid) is equal to the flow rate (movement of fluid) multiplied by the resistance (which, again, impedes the flow of the fluid).

Making this comparison between Ohm's law and flow rate should help make it easier to remember. But to answer the question, let's rearrange the expression slightly.

In this form, we can see that as the pressure difference is increased, so too does the flow rate increase. Moreover, flow rate is higher when resistance is lower. We need to keep these things in mind when determining what will reduce flow rate. Since increasing the pressure difference increases the flow rate, we can rule this answer choice out.

Next, let's see what things contribute to resistance. While the equation for resistance is quite complicated, we can remember a few generalities. First, the resistance of flow is inversely related to the radius of the pipe. In other words, as the radius increases, there is more room for the fluid to flow through, and thus the flow rate increases.

When the length of the tube increases, the resistance also increases. Think of it this way: it's easier to move a given amount of water through a straw than it is a garden hose. A big reason for this is that the increased length of the tube provides more area to come into contact with the moving fluid, which thus allows more opportunity for friction between the fluid and the walls of the pipe.

Fluids flow fastest when their flow is laminar. Turbulent flow is characterized as disordered and involves "wasting" energy through the movement of liquid molecules in directions other than the main direction of flow, which also impedes other fluid molecules and causes them to bump into the walls of the container/other fluid molecules irregularly, ultimately disrupting flow.

Finally, increasing the viscosity of the fluid also increases its resistance, and thus decreases its flow rate. This is because viscosity is a measure of the frictional interactions between molecules of the fluid itself. Think of it this way: it's harder to get something like syrup (very viscous) to flow as opposed to water (much less viscous). Thus, when we decrease the viscosity of a fluid, the resistance to flow decreases, and flow rate increases.

We will solve this problem in two steps. First, will use the law of continuity:

Rearranging for velocity 2:

Now we can use the expression for conservation of energy to adjust this velocity for the change in height:

Assuming an initial height of 0, we get:

Plugging in expressions, we get:

Rearranging for final velocity:

Plugging in our values, we get:

We will begin this problem with the expression for conservation of energy:

If we assume a final height of 0, we get:

Plugging in expressions, we get:

Rearranging for final velocity, we get:

This would be the velocity if the cross-sectional area of the tube stayed constant. However, that is not the case, so we can use the following expression to adjust for the difference of the cross-sectional area:

Rearranging for the ratio we are looking for:

Where:

Plugging these in, we get:

Initial volume rate must equal final volume rate:

Solving for

Plugging in values:

We can model the volumetric flow through the tube as the following expression:

Where:

so:

Applying this to the first and second scenario, we get:

According to the law of continuity, we know that the volumetric flow through the tube (when neglecting friction and assuming that it is horizontal) is constant. Therefore, we can say:

Rearranging for the ratio :

From the problem statement, we are told that the circumference is doubled. Thus, we know that the radius of the tube doubles as well:

Plugging this into the expression, we get: