Question

At two points on a horizontal tube of varying cross section carrying water, the radii are $1cm$ and $0.4cm$. The pressure difference between these points is $4.9cm$ of water. How much liquid flows through the tube per second?

• A. 100 c.c. per sec
• B. 80 c.c. per sec
• C. 50 c.c. per sec
• D. 70 c.c. per sec

Answer 1

Answer: (C)

Using Bernoulli's equation,
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2\dots (i)$
where $\rho$ is the density of liquid, $v$ its velocity, $P$ its pressure and subscripts 1 and 2 refer to two points.
Also $A_1{v}_1=A_2{v}_2$ by equation of continuity $\dots (ii)$
$P_1-P_2=\rho g\times 4.9\dots (iii)$
From (i) and (iii), we get
$v_2^2-v_1^2=\frac{2\left(P_1-P_2\right)}{\rho }=\frac{2\rho g\times 4.9}{\rho }=(2g)\times 4.9=2\times 980\times 4.9$
or $v_2^2-v_1^2=98^{2}c{m}^2/se{c}^2$
Using (ii), $\frac{v_1}{v_2}=\frac{A_2}{A_1}=\frac{\pi \times 0.4^2}{\pi \times 1^2}=0.16$
Substituting, $v_1=0.16\times v_2$ in (iv), we get
$v_2^2\left[1-(0.16{)}^2\right]=98^2$ or $v_2=\sqrt{\frac{98^2}{0.9744}}$
Quantity of water flowing
$=A_1{v}_1=A_2{v}_2=\pi \times 0.4^2\times \sqrt{\frac{98^2}{0.9744}}=50c.c$ per sec