Question

A liquid of density ${\rho }_1$ is flowing through a tube of varying cross-section. A manometer containing a liquid of density ${\rho }_2$ is connected to the tube as shown. The area of cross section of the tube at points A and B are $a_1$ and $a_2$ , respectively. The rate of flow of the liquid through the tube is

• A. $a_1{a}_2\sqrt{\frac{2{\rho }_{2}gh}{{\rho }_1(a_2^2-a_1^2)}}$
• B. $a_1{a}_2\sqrt{\frac{2gh}{{\rho }_1(a_2^2-a_1^2)}}$
• C. $\frac{a_1}{a_2}\sqrt{\frac{2{\rho }_{2}gh}{{\rho }_1(a_2^2-a_1^2)}}$
• D. $a_1{a}_2\sqrt{\frac{2{\rho }_{1}gh}{{\rho }_2(a_2^2-a_1^2)}}$

Answer 1

The correct option is A $a_1{a}_2\sqrt{\frac{2{\rho }_{2}gh}{{\rho }_1(a_2^2-a_1^2)}}$

Using Bernoulli equation at A and B.

$\frac{{\rho }_1{v}_1^2}{2}+P_A=\frac{{\rho }_1{v}_2^2}{2}+P_B$

Also $P_A-P_2={\rho }_{2}gh$ , putting this in above equation we get

$v_2^2-v_1^2=\frac{2{\rho }_{2}gh}{{\rho }_1}$....(1)

also using continuity equation $a_1{v}_1=a_2{v}_2$..(2)

putting the value of $v_2$ in (1)

we get $v_1=a_2\sqrt{\frac{2{\rho }_{2}gh}{{\rho }_1(a_1^2-a_2^2)}}$

flow rate Q=$v_1{a}_1$

$q=a_1{a}_2\sqrt{\frac{2{\rho }_{2}gh}{{\rho }_1(a_1^2-a_2^2)}}$