The velocity of liquid (v) in steady flow at a location through cylindrical pipe is given by `v=v_(0)(1-(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 cm. volume rate of flow through the pipe is `pi//2 xx 10^(-2) m^3s^(-1)` and the coefficient of viscosity of the liquid is 0.75 N s`m^(-2)`, find the magnitude of the viscous force per unit area, in N`m^(-2)` at r = 4 cm.

Explanation

Magnitude of viscous force, `F=eta A (dv)/(dt)`
`rArr v=v_(0)(1-(r^(2))/(R^(2)))rArr (dv)/(dt)=-(2V_(0)r)/(R^(2))rArr sigma =eta.(2v_(0)r)/(R^(2))....(i)`
Volume rate of flow, Q
consider an annular element at r from axis, width dr.
`dA=2pir dr, dQ=v.dA =v_(0)(1-(r^(2))/(R^(2)))2pirdr`
`Q=intdQ=2piv_(0)[(r^(2))/2-(r^(4))/(4R^(2))]_(0)^(R)=(pi)/2R^(2) V_(0)rArr v_(0)=(2Q)/(piR^(2))`
`:. (i) rArr sigma =eta(4Q)/(piR^(4))r, R=0.1 m`
At r=0.04 m ,`sigma =(0.75) 4xx(pi)/2 xx10^(-2)xx0.04/(pixx10^(-4))=6 Nm^(-2)`