Question

The velocity of a liquid ($v$) in steady flow at a location through cylindrical pipe is given by $v=v_0\left(1-\frac{r^2}{R^2}\right)$, where $r$ is the radial distance of that location from the axis of the pipe and $R$ is the inner radius of the pipe, if $R=10cm$, volume flow rate through the pipe is $\pi /2\times {10}^{-2}{m}^2{s}^{-1}$ and the coefficient of viscosity of the liquid is $0.75Ns{m}^{-2}$

• A. Magnitude of the viscous force per unit area at $r=4cm$ is $6N/m^2$
• B. Maximum velocity of flow of liquid is $1m/s$
• C. Viscous force per unit area increases with radial distance
• D. Viscous force per unit area decreases with radial distance

Answer 1

Answer: (A), (B), (C)

The correct options are
A Magnitude of the viscous force per unit area at $r=4cm$ is $6N/m^2$
B Maximum velocity of flow of liquid is $1m/s$
C Viscous force per unit area increases with radial distance

Magnitude of viscous force, $F=\eta A\frac{dv}{dr}$
$\Rightarrow$ Viscous force per unit area $\tau =\frac{F}{A}=-\eta \frac{dv}{dr}$
$v=v_0\left(1-\frac{r^2}{R^2}\right)\frac{dv}{dr}=-\frac{2v_{0}r}{R^2}---\left(i\right)\Rightarrow \tau =\eta .\frac{2v_{0}r}{R^2}$
From above equation, we can conclude that viscous force per unit area increases with radial distance.

Consider an annular element at distance $r$ from axis having width $dr$
$dA=2\pi rdrdQ=V.dA=v_0\left(1-\frac{r^2}{R^2}\right)2\pi rdrQ=\int dQ=2\pi v_0{\left(\frac{r^2}{2}-\frac{r^4}{4R^2}\right]}_0^R=\frac{\pi }{2}{R}^2{v}_0\Rightarrow v_0=\frac{2Q}{\pi R^2}\therefore \tau =\eta .\frac{4Q}{\pi R^4}.r,\left(fromequation\right(i\left)\right)whereR=0.1mAtr=0.04m,\tau =0.75\times 4\times \frac{\pi }{2}\times {10}^{-2}\times \frac{0.04}{\pi \times {10}^{-4}}=6N/m^2$
Also, maximum velocity will be at $r=0$ which is equal to $v_0=\frac{2Q}{\pi R^2}=1m/s$