Water flows in a streamlined manner through a capillary 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 $'4p'$, the rate of flow becomes :-
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Flow rate $Q=\frac{\pi}{8} \frac{p r^{4}}{\eta \ell}$
$\frac{Q^{\prime}}{Q}=\frac{p^{\prime} r^{\prime 4}}{p r^{4}} \Rightarrow \frac{Q^{\prime}}{Q}=\left(\frac{4 p}{p}\right)\left(\frac{r^{4}}{2^{4}}\right) \frac{1}{r^{4}}$
$\Rightarrow \frac{Q^{\prime}}{Q}=\frac{4}{2^{4}}=\frac{1}{4} \Rightarrow Q^{\prime}=\frac{Q}{4}$
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