Water is conveyed through a uniform tube of 8 $\mathrm{cm}$ in diameter and $3140 \mathrm{~m}$ in length at…
Water is conveyed through a uniform tube of 8 $\mathrm{cm}$ in diameter and $3140 \mathrm{~m}$ in length at the rate $2 \times 10^{-3} \mathrm{~m}^3$ per second. The pressure required in maintain the flow is (Viscosity of water $=10^{-3}$ SI units)
$6.25 \times 10^3 \mathrm{Nm}^{-3}$
$0.625 \mathrm{Nm}^{-2}$
$0.0625 \mathrm{Nm}^{-2}$
$0.00625 \mathrm{Nm}^{-2}$
Solution
Radius $r=\frac{8}{2} \mathrm{~cm}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$,
Rate of flow of water
$\begin{aligned} & Q=2 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s} \\ & Q=\frac{\pi P r^2}{8 \eta l}\end{aligned}$
$\Rightarrow \quad P=\frac{Q(8 \eta l)}{\pi r^4}$
$\eta=$ coefficient of viscosity of water $=10^{-3}$ SI units
$\begin{aligned} \therefore \quad P & =\frac{2 \times 10^{-3} \times 8 \times 10^{-3} \times 3140}{3.14 \times\left(4 \times 10^{-2}\right)^4} \\ & =\frac{2 \times 8 \times 3140 \times 10^{-6}}{3.14 \times 256 \times 10^{-8}} \\ & =\frac{3140 \times 10^2}{3.14 \times 16} \\ & \left.=\frac{100 \times 10^3}{16}\right] \\ & =6.25 \times 10^3 \mathrm{~N} / \mathrm{m}^2\end{aligned}$