Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm. if the amount of glycerine collected per second at one end is `4.0 xx 10^(-3) kg s^(-1)`, what is the pressuer difference between the two ends of the tube? (density of glycerine = `1.3xx10^(3)kgm^(-3)` and viscosity of glycerine =`0.83Ns m^(-2)`).

Explanation

`M=4 xx 10^(-3) kg s^(-1)`
l=1.5 m
r=1 cm `=10^(-2) m`
P=Pressure difference between two ends of a tube.
According to Poiseuille.s equation volume of glycerine flowing per second is given by :
`Q=(pi Pr^(4))/(8 eta l)`
`rArr P=(Q(8 etal))/(pi r^(4))`
Here, `Q=(4 xx 10^(-3))/(1.3 xx 10^(3)) m^(3)//s`
`=((4 xx 10^(-3))/(1.3 xx 10^(3))) xx (8 xx 0.83 xx 1.5)/(3.14 xx (10^(-2))^(4))`
`=9.75 xx10^(2)Pa`