A liquid of density `800 kg//m^3` flowing steadily in a tube of varying cross-section. If area of cross-section at `A` is `4 cm^2` and at `B` is `2 cm^2`. If speed of liquid at `A` is `10 cm//s`, calculate
(i) the rate of flow
(ii) the difference in pressures at `A` and `B`.

Explanation

(i) Rate of flow `Q = a_A v_A = a_B v_B`
`Q = a_A v_A = 4 xx 10^-4 xx 10 xx 10^-2 = 4 xx 10^-5 m^3//s`
(ii) From continuity equation
`a_A v_A = a_B v_B`
`4 xx 10 = 2 xx v_B rArr v_2 = 20 cm//s`
Applying Bernoulli's equation
`P_A + rho gh_A + (1)/(2) rho v_A^2 = P_B + rho gh_B + (1)/(2) rho v_B^2`
Here `h_A = h_B`
`P_A - P_B = (1)/(2) rho (v_B^2 - v_A^2) = (1)/(2) xx 800(20^2 - 10^2) xx 10^-4`
=`400 xx 300 xx 10^-4`
=`12 N//m^2`.