Question

A capillary tube of length $L$ and radius $r$ is connected with another capillary tube of the same length but half the radius in series. The rate of steady volume flow of water through first capillary tube under a pressure difference of $p$ is $V$. The rate of steady volume flow through the combination will be (the pressure difference across the combination is $p$)

• A. $17V$
• B. $\frac{16}{17}V$
• C. $\frac{V}{17}$
• D. $\frac{17}{16}V$

Answer 1

Answer: (C)

$\frac{V}{17}$

Problems of series and parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is potential difference is replaced by $\Delta p$ and electrical resistance as, $R_1=\frac{8\eta L}{\pi r^4}=R$
and $R_2=\frac{8\eta L}{\pi {\left(\frac{r}{2}\right)}^4}=16R$

Electric current is replaced by rate of volume flow $V^{\prime }$.
$\Rightarrow p=V{R}_1=VR$ ....(i)
$p^{\prime }=p=V^{\prime }{R}_{eq}=V^{\prime }(R_1+R_2)=V^{\prime }(R+16R)$
$=17V^{\prime }R$ ....(ii)
From equations (i) and (ii), we get

$VR=17V^{\prime }R$
$\Rightarrow V^{\prime }=\frac{V}{17}$