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
Medium
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(d)$V = \frac{{\pi p{r^4}}}{{8\eta l}}$
$V \propto P\,{r^4}$($\eta \;{\rm{and }}l$ are constants)
$\frac{{{V_2}}}{{{V_1}}} = \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\;{\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4}$= $2 \times {\left( {\frac{1}{2}} \right)^4}$= $\frac{1}{8}$
${V_2} = \frac{Q}{8}$
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