Question

A larger tank with a nozzle attached contains three immiscible fluids as shown in figure. Assuming that the changes in $h_1,h_2$ and $h_3$ are negligible, the discharge velocity at this instant is

• A. $\sqrt{2g{h}_3}(1+\frac{{\rho }_1}{{\rho }_3}\frac{h_1}{h_3}+\frac{{\rho }_1}{{\rho }_3}\frac{h_2}{h_3})$
• B. $\sqrt{2g(h_1+h_2+h_3)}$
• C. $\sqrt{2g\left(\frac{{\rho }_1{h}_1+{\rho }_2{h}_2+{\rho }_3{h}_3}{{\rho }_1+{\rho }_2+{\rho }_3}\right)}$
• D. $\sqrt{2g\left(\frac{{\rho }_3{h}_2{h}_3+{\rho }_2{h}_3{h}_1+{\rho }_1{h}_2{h}_3}{{\int }_1{h}_1+{\int }_2{h}_2+{\int }_3{h}_3}\right)}$

Answer 1

The correct option is A $\sqrt{2g{h}_3}(1+\frac{{\rho }_1}{{\rho }_3}\frac{h_1}{h_3}+\frac{{\rho }_1}{{\rho }_3}\frac{h_2}{h_3})$

Gauge pressure at point A is
$P_A={\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2+{\rho }_{3}g{h}_3\dots \left(i\right)$
Gauge Pressure at point B is
$P_B=0$
Applying Bernoulli's equation between point A and B.
$P_A+(\rho gh{)}_A+\frac{1}{2}\rho v_A^2=P_B+(\rho gh{)}_B+\frac{1}{2}\rho v_B^2$
$[\because P_B=0\&(\rho gh{)}_A=(\rho gh{)}_B$ at same height; $v_A=0]$
$\Rightarrow P_A=\frac{1}{2}{\rho }_3{v}_B^2[\rho ={\rho }_3$ at point B]
$\Rightarrow {\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2+{\rho }_{3}g{h}_3=\frac{1}{2}{\rho }_3{v}_B^2$
[using (i)]
$\Rightarrow v_B=\sqrt{\frac{2g}{{\rho }_3}({\rho }_1{h}_1+{\rho }_2{h}_2+{\rho }_3{h}_3)}$
$=\sqrt{2g{h}_3(\frac{{\rho }_1{h}_1}{{\rho }_3{h}_3}+\frac{{\rho }_2{h}_2}{{\rho }_3{h}_3}+1)}$
$\Rightarrow v_B=\sqrt{2g{h}_3(1+\frac{{\rho }_1{h}_1}{{\rho }_3{h}_3}+\frac{{\rho }_2{h}_2}{{\rho }_3{h}_3})}$