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

Hint: 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.