Question

The rate of steady volume flow of water through a capillary tube of length $l$ and radius $r$ , under a pressure difference of $p \, is \, V$ . This tube is connected with another tube of the same length but half the radius, in series. Then the rate of steady volume flow through them is (the pressure difference across the combination is p)

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

Answer 1

Answer: (B)

The rate of flow of water inside a capillary
$V=\frac{\pi \, p \, r^{4}}{8 \, n \, l},$
Path difference $ \, p=\frac{V \left(8 \, n \, l\right)}{\pi \, r^{4}}$
In series combination $ \, \, p=p_{1}+p_{2}$
where $p_{1}$ and $p_{2}$ are the pressure difference in the two tubes.
In series combination, the rate of flow of water $\left(V '\right)$ will be the same in both the tubes.
$\frac{V}{r^{4}}=\frac{V^{′}}{r^{4}}+\frac{V^{′} \times 16}{r^{4}}$
$V=V^{′}+16V^{′}$
$ \, \, V^{′}=\frac{V}{17}$