Figure shows a liquid of densith `1200kgm^-3` flowing steadily in a tube of varying cross section. The cross section at a point A is `1.0cm^2` and that at B is `20mm^2`, the points A and B are in the same horizontal plane.The speed of the liquid at A is 10 `cms^-1`. Calculate the difference in pressure at A and B.

Explanation

From equation of continuity the speed `v_2` at B is given by
`A_1v_1=A_2v_2` ltbrrarr `(1.0cm^2)(10cms^-1)=(20mm^2)v_2`
`or v_2=(1.0cm^2)/(20mm^2)xx10cms^-1=50cms^-1`
By Bernoullil equation
`P_1+rhogh_1+1/2 rhov_1^2=P_2+rhogh_2+1/2rhov_2^2`.
here `h_1=h_2` Thus,
`P_1-P_2=1/2rhov_2^2-1/2rhov_1^2`
`=1/2xx91200kgm^-3)(2500cm^2s^-2-100cm^2^-2)`
`=600kgm^-3xx2400cm^2s^-2=144Pa`