Question

A liquid flows through a horizontal tube as shown in figure. The velocities of the liquid in the two sections, which have areas of cross-section $A_1$ and $A_2$, are $v_1$ and $v_2$, respectively. The difference in the levels of the liquid in the two vertical tubes is $h$. Then

• A. $v_2^2-v_1^2=2gh$
• B. $v_2^2+v_1^2=2gh$
• C. $v_2^2-v_1^2=gh$
• D. $v_2^2+v_1^2=gh$

Answer 1

Answer: (A)

Let $\rho$ be density of liquid flowing in the tube
$A_1{v}_1=A_2{v}_2$
$\frac{v_1}{v_2}=\frac{A_2}{A_1}$
According to Bernoullis equation for horizontal flow of liquid,
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
$P_1-P_2=\frac{1}{2}\left(\rho v_2^2-\rho v_1^2\right)$
$h\rho g=\frac{1}{2}\rho \left(v_2^2-\rho v_1^2\right)$
$h\rho g=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)\left(\because P_1-P_2=h\rho g\right)$
$v_2^2-v_1^2=2hg$