Question

At $20^\circ C$ temperature in horizontal pipe pressure falls 60 KPa per 100 m and water flow in pipe with rate 3 litre per minute, what is approximate radius of pipe $($coefficient of viscosity, $20^\circ C = {10^{ - 3}}Pa)$
(A) $2.52 cm$
(B) $0.38 cm$
(C) $1.23 cm$
(D) $4.63 cm$

Answer 1

In given numerical rate of water flow, coefficient of viscosity length of pipe and pressure difference is given at fixed temperature. Using poiseuille’s law and on substituting the values of all these physical quantities we get a radius of pipe.

Formula used:
According to poiseuille’s law, the rate of flow of water is given as
$Q = \dfrac{{\Delta \rho \pi {r^4}}}{{8\eta \ell }}$
Where
$\Delta \rho =$ Pressure difference
r $=$ Radius of pipe
$\eta =$ Coefficient of viscosity
$\ell =$ Length of pipe

Complete step by step answer:
Given that
Pressure difference $\Delta \rho = 60kPa$
$\Delta \rho = 60 \times {10^{ - 3}}Pa$
Coefficient of viscosity $\eta = {10^{ - 3}}$
Length of pipe $\ell = 100m$
Rate of water flow $Q = \dfrac{{3 \times {{10}^{ - 3}}}}{{60}}$
$Q = 5 \times {10^{ - 5}}{m^3}{s^{ - 1}}$
According to poiseuille’s law, the rate of flow of water is given by
$\Rightarrow Q = \dfrac{{\Delta \rho \pi {r^4}}}{{8\eta \ell }}$
$\Rightarrow {r^4} = \dfrac{{8Q\eta \ell }}{{\Delta \rho \pi }}$ $[\because \pi = 3.14]$
On putting all the values in above expression
$\Rightarrow {r^4} = \dfrac{{8 \times 5 \times {{10}^{ - 5}} \times {{10}^{ - 3}} \times 100}}{{60 \times {{10}^3} \times 3.14}}$
$\Rightarrow {r^4} = \dfrac{{40 \times 100 \times {{10}^{ - 8}}}}{{188.4 \times {{10}^3}}}$
$\Rightarrow {r^4} = \dfrac{{4000}}{{1884}} \times {10^{ - 8}} \times {10^{ - 2}}$
$\Rightarrow {r^4} = 2.12 \times {10^{ - 10}}$
$\Rightarrow {r^2} = \sqrt {2.12 \times {{10}^{ - 10}}}$
$\Rightarrow {r^2} = 1.457 \times {10^{ - 5}}$
$\Rightarrow r = \sqrt {1.457 \times {{10}^{ - 5}}}$
$\Rightarrow r = \sqrt {14.57 \times {{10}^{ - 6}}}$
$\Rightarrow r = 3.187 \times {10^{ - 3}}m$
$\Rightarrow r \approx 3.82 \times {10^{ - 3}}m$
$\Rightarrow r \approx 3.82 \times {10^{ - 1}}cm$
$\therefore r \approx 0.38cm$

Hence, the radius of pipe is $0.38 cm$. So, option B is the correct answer.

Note: Many times students may get confused between dynamic viscosity and kinematic viscosity.Dynamic viscosity is the measurement of the fluid’s internal resistance to flow.Kinematic viscosity if the ratio of dynamic viscosity to density.