The flow of blood through the aorta can be approximated with the volumetric flow rate equation. When the smooth muscle of the aorta contracts, the radius can decrease by as much as 10%. If the blood flow is to remain unchanged, what must happen to the velocity of blood flow in the aorta?

A) Decrease by approximately 10%

B) Increase by approximately 10%

C) Increase by approximately 25%

D) Remain unchanged

**Explanation**

The volumetric flow rate equation is Q = A x v where A is the cross-sectional area of the pipe, Q is the flow rate, and v is the velocity of the fluid.

The problem states that Q is to remain unchanged. That allows us to set up the following equation:

Q = A_{1}v_{1} = A_{2}v_{2}

Plugging in 1 as the original values, we get:

1 x 1 = 0.81 x v_{2}

Because the radius was decreased to 90% of its original value, the cross-sectional area will be reduced to 81% of its original value (A is proportional to radius squared and 0.92 = 0.81)

r_{2} = 0.9r_{1}

A_{2} = 0.81A_{1}

Solving for v_{2} we get 1.23. Thus the velocity increased by approximately 25%, making choice (C) correct.

Please is there a better explanation for this

Hi there, thanks for the question! Let me try to explain this one a little bit more thoroughly. First, we need to begin in the same place as the explanation: with the continuity equation, Q = A x v, which can also be written as A1v1 = A2v2. I’m not sure if you’ve used this equation before, but it’s very important to understand for MCAT physics. Q represents the volume flow rate of a fluid, A is the cross-sectional area, and v is the velocity of individual fluid particles. With this in mind, all that Q = A x v states is that (since Q is constant) an increase in A must lead to a decrease in v, and vice versa.

Knowing this, we can already eliminate some choices. The question stem says that the radius is decreasing. A decrease in radius must also imply a decrease in cross-sectional area, and (according to the reasoning above) a decrease in area will lead to an INcrease in velocity. We can eliminate choices A and D because they do not involve an increase.

Now, we need to find out how large this increase in velocity will be. Let’s return to the equation A1v1 = A2v2. (Again, this is just another way of writing the continuity equation. This equation holds true for any “ideal” fluid traveling through a system of vessels or pipes – since Q is equal at all points at the same time, the product of A and v at one position must be equal to the product of A and v at another.)

We know that radius decreased by 10%. This means that the new radius is 0.9 times the previous radius, as written here: r2 = 0.9 r1. But we care about area, not radius, and area is proportional to the SQUARE of the radius. Squaring what we have written above gives us A2 = 0.81 A1.

Finally, we plug back into our A1v1 = A2v2 equation, substituting in 0.81 A1 for A2.

A1v1 = A2v2

A1v1 = (0.81 A1)v2

v1 = 0.81 v2

But we’re solving for v2, not v1, so divide both sides by 0.81: v1 / 0.81 = v2. 1 divided by 0.81 is around 1.25, so 1.25 v1 = v2. The new velocity, then, is 25% larger than the old velocity, and C is correct.

How do we know to treat the area as if it was a square? instead of a circle ?

We are just squaring the radius rather than using pie.

Thanks