(II) If cholesterol buildup reduces the diameter of an artery ...

Question

(II) If cholesterol buildup reduces the diameter of an artery by 25%, by what % will the blood flow rate be reduced, assuming the same pressure difference?

Accepted Answer

Everyone in this problem, we're told that a physics student is experimenting with a faucet. At first she turns on the faucet and the water flows out afterwards, she reduces the diameter of the nozzle by 35%. If the pressure difference is kept constant or has to determine by what percent the flow rate decreases. We're given four answer choices. Option A 18%. Option B 42% option C 58% and option D 82%. Now, if we want to think about the flow rate, we can think about Po's equation. OK. So recall that that flow rate QE is equal to pi multiplied by R to the exponent four. They were R is the radius of the faucet or the pipe or the hose that the, the liquid is moving through multiplied by P one minus P two. OK. So the pressure difference divided by eight ADA L where A A is the coefficient of a scarcity and L is the length of the P. And so this is the flow equation. And in this case, we are going to have to compare the flow rate originally to this new flow rate. When we change the diameter of the nozzle. So let's start by thinking about what that flow rate looks like initially. OK. So the flow rate initially is going to be equal to pi multiplied by the initial radius R not to the exponent four multiplied by the change in pressure P one minus P two or the pressure difference throughout that pipe divided by eight. The L. Now, what you'll notice is that I did not put a subscript on Ada, I did not put a subscript on L. And that's because we are dealing with the same pipe. We are dealing with water in both cases, the same fluid flowing through. And so, a, a, the coefficient of scarcity and the length of the pipe are not going to change when we consider the initial case to the new case, all that's changing is that diameter of the No. OK. So we have Q not. Now, let's think about the flow rate once that nozzle is changed. So we're gonna call this Q new, our new flow rate that's gonna be equal to pi multiplied by our new radius to the exponent four, multiplied by P one minus P two. And we're told that the pressure difference is kept constant and then we're dividing by eight A L again, that denominator is the same because we have the same pipe or the same Os made the same length, same viscosity, same fluid. Now, what we're told is that the diameter is reduced by 35%. If the diameter is re reduced by 35% then the radius is reduced by 35% as well. And so our new is going to be 65% of whatever the initial radius was are not. So we can write our new flow rate. Q new as pi multiplied by 0.65 are not to the exponent four multiplied by the pressure difference divided by eight A L. OK. Now this is starting to look a lot like QN. OK. So we can break our 0.65 R, not all to the exponent four up, we can write this as 0.65 to the exponent four multiply by pi are not to the exponent four multiplied by that pressure difference P one minus P two. And this is all still divided by eight beta L. So what we have is 0.65 to the exponent four and everything else is just QN. OK. So we can write the new low rate Q new as 0.65 to the exponent four multiplied by original flow rate. Q not OK. So we have this relationship between the two flow rates which is gonna allow us to find the percent decrease that we wanna find. So if we think about finding a percent decrease and the percent decrease is going to be equal to the original. All right, we had tuna minus the new flow rate Q now, so that's gonna show how much we decreased. We divide by the initial original flow rate Q knot and multiply by 100 to get that percent decrease. OK. So let's substitute in our values you not, we're just gonna leave us Q not OK. Q new, we know we can write a 0.65 to the exponent four multiplied by Q not, this is all divided by Q not and multiplied by 100. So all of these Q not values will divide out. Now, we can divide every single thing by Q not numerator and denominator. And when we do that we're left with is that this percent decrease in the flow rate is going to be one minus 0.65 to the exponent four multiplied by 100 which gives us a value of about 82.15%. And so in this experiment, when the diameter of the faucet decreases by 35% the flow rate is going to decrease by 82%. This corresponds with answer choice. D. Thanks everyone for watching. I hope this video helped see you in the next one.

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(II) If cholesterol buildup reduces the diameter of an artery by 25%, by what % will the blood flow rate be reduced, assuming the same pressure difference?

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