MCAT Physics Question — Blood Flow
 by
 Sep 24, 2017
 MCAT Question of the Day
 Reviewed By: Liz Flagge
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
Click for Explanation
The volumetric flow rate equation is Q = A x v where A is the crosssectional 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 crosssectional 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.
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