Under a constant pressure head, the rate of flow of liquid through a capillary tube is $V$. If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become
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(d) Rate of flow under a constant pressure head,
$V = \frac{{\pi p{r^4}}}{{8\eta l}}$==> $V \propto \frac{{{r^4}}}{l}$==> $\frac{{{V_2}}}{{{V_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^4} \times \frac{{{l_1}}}{{{l_2}}} = {\left( {\frac{1}{2}} \right)^4} \times \frac{1}{2}$
==> ${V_2} = \frac{{{V_1}}}{{32}} = \frac{V}{{32}}$
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Figure: $Image$
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