MCQ
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
  • A
    $4Q$
  • B
    $Q$ 
  • C
    $\frac{Q}{4}$
  • $\frac{Q}{8}$

Answer

Correct option: D.
$\frac{Q}{8}$
d
(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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