Question

Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm, and the areas of cross section at A and B are 4 cm² and 2 cm² respectively, find the rate of flow of water across any section. Given $\sqrt{\frac{2}{15}}=0.365$
approx.

• A. 146 cc/s
• B. 146 m³/s
• C. 246 cc/s
• D. 246 m³/s

Answer 1

The correct option is A

146 cc/s

from equation of continuity,
$A_1{V}_1=A_2{V}_2$
$\Rightarrow 4V_A=2V_B$
$V_B=2V_A......\left(1\right)$
From Bernoulli's equation,
$P_1+\stackrel{,}{\rho }gH+\frac{1}{2}{V}^2=constant$
$P_A-P_B=\frac{1}{2}[V_B^2-V_A^2]\stackrel{,}{\rho }$
$But{P}_A-P_B=\stackrel{,}{\rho }$ $gh[h=differenceinheightofwatercolumn=2cm]$
$\Rightarrow \frac{1}{2}\stackrel{,}{\rho }{V}_A^2-V_A^2=1000\times \frac{2}{100}$
From(1)
$\frac{1}{2}x10004V_A^2=1000\times \frac{2}{10}$
$3V_A^2=\frac{4}{10}$
$V_A=0.365\frac{m}{s}$
Flow rate=$A_A{V}_A=\left(4c{m}^2\right)(0.365\times 100)\frac{cm}{s}$
$=146\frac{c{m}^3}{s}$