Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 N m^-2 between A and B where the areas of cross section are `30cm^2 and 15cm^2 respectively. Find the rate of flow of water through the tube.
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Explanation

Let the velocity at `A=v_A and that at B=v_B` By the equation of continuity `v_B/V_A=(30cm^2)/(15cm^2)=2`
By Bernoulli equation
`P_A+1/2rhov_A^2=P_B+1/2rhov_B^2`
or `P_A-PB=1/2rho(2v_A)^2-1/2rho v_A^2=3/2rhov_A^2`
or v_A=sqrt(0.4ms^2s^-2)=0.63ms^-1`
the rate of flow =`(30cm^2)(0.63ms^-1)=1890cm^3s^-1`