The velocities of cylindrical layers of liquid flowing through a tube , situated at distances `0.8cm and 0.82cm` from the axis of the tube are `3cms^(-1)and2.5cms^(-1)` respectively . Find the viscous force acting between these layers , if the length of the tube is 10 cm and the coefficient of viscosity of the liquid is 8 poise .

Explanation

`r_(1)=0.8cm,r_(2)=0.82cm`
`Deltav=3-2.5=0.5cms^(-1)`
`Deltax=` distance between the layers `=0.02cm`
`L=10cm`
A = Area of contact of two layers
`=2pi((r_(1)+r_(2))/(2))L`
`eta=8` poise
Now `F_(v)=etaA(Deltav)/(Deltax)`
`=eta[2pi(r_(1)+r_(2))/(2)L](Deltav)/(Deltax)`
`=8[2xx3.14((0.8+0.82)/(2))10](0.5)/(0.02)`
`=16xx3.14xx0.81xx10xx25`
`F_(v)=10173.6` dyne