An artery in a certain person has been widened $1 \frac{1}{2}$ times the original diameter. If the pressure difference across the artery is maintained constant, the blood flow through the artery will be increased:

\frac{3}{2}

times

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\frac{9}{4}

times

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No change

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\frac{81}{16}

times

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Solution

The correct option is C $\frac{81}{16}$ times
Q flow initially $= \frac{π Δ P r^{4}}{8 η L}$
New radius $= \frac{3}{2} r$
$Q_{n e w} = \frac{π Δ P r^{4}}{8 η L} {(\frac{3}{2})}^{4}$
So, increase in blood flow = ${(\frac{3}{2})}^{4} = \frac{81}{16}$

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An artery in a certain person has been widened 112 times the original diameter. If the pressure difference across the artery is maintained constant, the blood flow through the artery will be increased:

The correct option is C 8116 timesQ flow initially =πΔPr48ηLNew radius =32rQnew=πΔPr48ηL(32)4So, increase in blood flow =(32)4=8116

An artery in a certain person has been widened $1 \frac{1}{2}$ times the original diameter. If the pressure difference across the artery is maintained constant, the blood flow through the artery will be increased:

The correct option is C $\frac{81}{16}$ times
Q flow initially $= \frac{π Δ P r^{4}}{8 η L}$
New radius $= \frac{3}{2} r$
$Q_{n e w} = \frac{π Δ P r^{4}}{8 η L} {(\frac{3}{2})}^{4}$
So, increase in blood flow = ${(\frac{3}{2})}^{4} = \frac{81}{16}$