Question

Water is conveyed through a uniform tube of $8cm$in diameter and$3140\,m$in length at the rate of$2 \times {10^{ - 3}}{m^3}$per second. The pressure required to maintain the flow is (viscosity of water$= {10^{ - 3}}$units):
(A) $6.25\,\,N{m^{ - 2}}$
(B) $0.625\,\,N{m^{ - 2}}$
(C) $6250\,\,N{m^{ - 2}}$
(D) $0.00625\,\,N{m^{ - 2}}$

Answer 1

According to the volume of liquid coming out of the tube per second is
(i) Directly proportional to the pressure difference$\left( P \right)$.
(ii) Directly proportional to the fourth power of radius $\left( r \right)$of the capillary tube.
(iii) Inversely proportional to the coefficient of viscosity$\left( \eta \right)$of the liquid.
(iv) Inversely proportional to the length$\left( l \right)$of the capillary tube.

Complete step by step answer:
Water is conveyed through a uniform tube of $8\,cm$in diameter and$3140\,m$in length at the rate of$2 \times {10^{ - 3}}{m^3}\,per\,\sec ond$.
Radius of cross section of tube is
$r\, = 4cm = 4 \times {10^{ - 2}}m$
Length of tube$\left( l \right) = 3140\,m$
Rate of flow$\left( v \right) = 2 \times {10^{ - 3}}{m^3}/s$
Coefficient of viscosity is also given
$\eta = {10^{ - 3}}$S.I. units.
According to poiseuille's formula rate of flow
$V = \dfrac{{\pi .{{\Pr }^4}}}{{8\eta l}}$ … (i)
$P = \dfrac{{v8\eta l}}{{\pi {r^4}}}$ … (ii)
In equation (i) $P$is pressure difference across end's of pipe to maintain flow of fluid, $v$is rate of flow, $\eta$is coefficient of viscosity, $l$ is length of tube and$r$radius of cross section of tube.
Put the above given value in equation (ii) to calculate pressure difference
$P = \dfrac{{\left( {2 \times {{10}^{ - 3}}} \right)\,8\left( {{{10}^{ - 3}}} \right)\left( {3140} \right)}}{{3.14{{\left( {4 \times {{10}^{ - 2}}} \right)}^4}}}$
$= 6.25 \times {10^3}\,N/{m^2}$

So, the correct answer is “Option C”.

Note:
The velocity of a layer in contact with the walls of the tube is negligible, i.e. almost zero. The velocity of the layers increases as we go towards the axis of the capillary tube.