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

`l=1.5m,r=1xx10^(-2)m`
volumn/s `V=("mass/s")/("Density")=(4xx10^(-3))/(1.3xx10^(3))m^(3)s^(-1)`
`=(4)/(1.3)xx10^(-6)m^(3)s^(-1)`
`eta=0.83Pa` s
Now, `V=(pi pr^(4))/(8etal)'`
where p is the pressure difference across the capillary.
or `p=(V etal)/(pir^(4))`
substituting values
`p=8xx(4)/(1.3)xx10^(-6)xx0.83xx1.5xx(7)/(22)xx(1)/(10^(-8))Pa`
the reynolds number is 0.3 So, the flow is laminar.