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Q. Water flows in a streamlined manner through a capillary tube of radius a the pressure difference being p and the rate of flow $Q$. If the radius is reduced to $a/2$ and the pressure increased to $2p$. The rate of flow becomes

AMUAMU 2012

Solution:

$V=\frac{\pi P ^{4}}{8 \eta l}$
$\therefore V \propto p r^{4}$ ( $\eta$ and I are contants)
$\therefore \frac{V_{2}}{V_{1}}=\left(\frac{p_{2}}{p_{1}}\right)\left(\frac{r_{2}}{r_{1}}\right)^{4}$
$=2 \times\left(\frac{1}{2}\right)^{4}=\frac{1}{8}$
$V_{2}=\frac{Q}{8}$