Question

Water is conveyed through a uniform tube of $ 8 \,cm $ in diameter and $ 3140\, m $ in length at the rate $ 2 \times 10^{-3 } \,m^{3} $ per second. The pressure required to maintain the flow is (Viscosity of water $ = 10^{-3} $ )

• A. $ 6.25 \times 10^{3} \, N\, m^{-2} $
• B. $ 0.625 \,N \, m^{-2} $
• C. $ 0.0625 \,N \, m^{-2} $
• D. $ 0.00625 \,N \, m^{-2} $

Answer 1

Answer: (A)

According to Poiseuille formula,
Rate of flow, $ V=\frac{\pi Pr^{4}}{8\eta l} $
where the symbols have their usual meanings.
$ \therefore P=\frac{V 8\eta l}{\pi r^{4}} $
Here, $ V= 2 \times10^{-3}\, m^{3}\, s^{-1} $
$ r=\frac{8}{2} cm =4\, cm=4\times10^{-2}\,m $
$ l=3140 \,m, \eta=10^{-3} \, N \, s\,m^{-2} $
Substituting the given values, we get
$ P=\frac{\left(2\times10^{-3}\, m^{3}\,s^{-1}\right)\left(8\right)\left(10^{-3}\,N\,sm^{-2}\right)\left(3140\,m\right)}{\left(3.14\right)\left(4\times10^{-2}\,m\right)^{4}} $
$ = 6.25 \times 10^{3} N\, m^{-2} $