Question

Two pipes of same length and radii in the ratio $1:2$ are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination in $1\text{m}$ of water, the pressure difference across the first pipe is

• A. $9.4\text{m}$ of water
• B. $4.9\text{m}$ of water
• C. $0.49\text{m}$ of water
• D. $0.94\text{m}$ of water

Answer 1

The correct option is D $0.94\text{m}$ of water

Since pipes are of same length,
$l_1=l_2=l$
$\frac{r_1}{r_2}=\frac{1}{2}$
$\Rightarrow r_1=r$ and $r_2=2r$


In series combination, the flow rate through tubes will be same, pressure difference across ends will be equal to sum of pressure difference across individual pipe.
$\Rightarrow \Delta P=\Delta P_1+\Delta P_2........\left(i\right)$
$Q=\frac{\pi r^4\Delta P_1}{8\eta l}=\frac{\pi (2r{)}^4\Delta P_2}{8\eta l}........\left(ii\right)$
From Eq. (ii),
$\frac{\Delta P_1}{\Delta P_2}=\frac{(2r{)}^4}{r^4}=16$
$\Delta P_1=16\Delta P_2$
Given: Pressure difference across ends $\left(\Delta P\right)$ is equivalent to $1\text{m}$ water head
$\Rightarrow 1=\Delta P_1+\frac{\Delta P_1}{16}$
$\therefore \Delta P_1=\frac{16}{17}=0.94\text{m}$
Hence pressure difference across first tube is equivalent to $0.94\text{m}$ water head.