The rate of steady volume flow of water through a capillary tube of length $l$ and radius $r$, under a…
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 $p$. )
$\frac{V}{16}$
$\frac{V}{17}$
$\frac{16 \mathrm{~V}}{17}$
$\frac{17 \mathrm{~V}}{16}$
Solution
The rate of flow of water inside a capillary, $V=\frac{\pi p r^4}{8 n l^{\prime}}$
Pressure difference, $p=\frac{V(8 n l)}{\pi r^4}$
In series combination,
$p=p_1+p_2$
where, $p_1$ and $p_2$ are the pressure difference in the two tubes.
$\therefore \quad \frac{V(8 n l)}{\pi r^4}=\frac{V^{\prime}(8 n l)}{\pi r^4}+\frac{V^{\prime}(8 n l)}{\pi(r / 2)^4}$
In series combination, rate of flow of water $\left(V^{\prime}\right)$ will be same in both the tubes.
$\therefore \quad \frac{V}{r^4}=\frac{V^{\prime}}{r^4}+\frac{V^{\prime} \times 16}{r^4}$
$V=V^{\prime}+16 V^{\prime}$
$V^{\prime}=\frac{V}{17}$