Question

At two points on a horizontal tube of varying cross section carrying water, the radii are $1 cm$ and $0.4 cm$. The pressure difference between these points is $4.9 cm$ 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
$v_{2}^{2}-v_{1}^{2}=\frac{2\left(P_{1}-P_{2}\right)}{\rho}=\frac{2 \rho g \times 4.9}{\rho}=(2 g) \times 4.9=2 \times 980 \times 4.9$
or $ v_{2}^{2}-v_{1}^{2}=98^{2} cm ^{2} / sec ^{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}}=50 c.c $ per sec