The velocity of liquid (v) in steady flow at a location through cylindrical pipe is given by $v = v_{0} (1 - \frac{r^{2}}{R^{2}})$ , where r is the radial distance of that location from the axis of the pipe and R is the inner radius of pipe. If $R = 10 c m$ , volume rate of flow through the pipe is $\frac{π}{2} \times 10^{- 2} m^{3} s^{- 1}$ and the coefficient of viscosity of the liquid is $0.25 N s m^{- 2}$ , find the magnitude of the viscous force per unit area, in $N m^{- 2}$ , at $r = 8 c m$ .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Magnitude of viscous force, $F = η A = \frac{d v}{d r}$
$\Rightarrow$ Viscous force per unit area $σ = \frac{F}{A} = η \frac{d v}{d r}$
$v = v_{0} (1 - \frac{r^{2}}{R^{2}}) \Rightarrow \frac{d v}{d r} = - \frac{2 v_{0} r}{R^{2}} \Rightarrow σ = η \frac{2 v_{0} r}{R^{2}} \to (i)$

Volume rate of flow, Q
Consider an annular cylindrical element at r from axis and width dr.
$d A = 2 π r d r; d Q = v . d A = v_{0} (1 - \frac{r^{2}}{R^{2}}) 2 π r d r Q = \int d Q = 2 π v_{0} {[\frac{r^{2}}{2} - \frac{r^{4}}{4 R^{2}}]}_{0}^{R} = \frac{π}{2} R^{2} v_{0} \Rightarrow v_{0} = \frac{2 Q}{π R^{2}} σ = η \frac{4 Q}{π R^{4}} r$