Water flows through a horizontal tube as shown in figure. If the difference of heights of water colun in the vertical tubes is 2 cm and the area of cross section at A and B are `4cm^2 and 2cm^2` respectively, find the rate of flow of water across any section.

Answer

A, C, D

Explanation

v_A=a_A=v_Bxxa_B`
`rarr v_Axx4=v_Bxx2`
`rarr v_B=2v_A`…i
`-Again
`1/2rhov_A^2+rhogh_A+p_A=1/2rhov+B^2+rhogh_B+p_B`
`rarr 1/2rhov_A^2+p_A=1/2rho(v_B^2-v_A^2)`
`rarr p_A-p_B=1/2rho(v_b^2-v_A^2)`
`=1/2xx1xx(4v_A^2-v_A^2)`
`rarr 2xx1xx1000=1/2xx1x3v_A^2`
=[p_A-p_B=2cm` of water column
`=2xx1xx1000dyne/cm^2]`
`rarrv_A^2=sqrt(4000/3)=36.51cm/sec`
`So, rate of flow =V_aa_A=36.50xx4`
`=146cm^3/sec`