Question

Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are 5 cm² and 2 cm² respectively. The rate of flow of water through the tube is 500 cm³ s^-1. Find the difference of mercury levels in the U-tube. Assume this tube lying horizontally on a table.

• A. 1.93 cm
• B. 1.93 m
• C. 2.58 cm
• D. 2.58 m

Answer 1

The correct option is A

1.93 cm

As clear from the sitution,
$P_1-P_2=\stackrel{,}{\rho }gh$
h= difference in height of the mercury column.
Using Bernoulli's equation,
$P_1+\stackrel{}{\rho }gH+\frac{1}{2}{V}^2=constant$
As both the points are at the same horizontal level.
$P_1+\frac{1}{2}{\stackrel{}{V}}_1^2=P_2+\frac{1}{2}{V}_2^2\stackrel{}{\rho }$
$P_1-P_2=\frac{1}{2}[V_2^2-V_2^2]\stackrel{}{\rho }$
$\Rightarrow \frac{1}{2}\stackrel{}{\rho }(V_2^2-V_1^2)=\stackrel{,}{\rho }gh.....\left(1\right)$

From equation of continuity,
$A_1{V}_1=A_2{V}_2$
$5V_1=2V_2$
$V_2=\frac{5}{2}{V}_1......\left(2\right)$
$also{A}_1{V}_1=A_2{V}_2=500\frac{c{m}^3}{s}$
$\Rightarrow V_1=100\frac{cm}{s}=1\frac{m}{s}$......(3)
from (3) &(1)
$\frac{1}{2}\stackrel{}{\rho }\left\{\frac{21}{4}\right\}=\stackrel{,}{\rho }gh$
$\frac{1}{2}\times 1000\left\{\frac{21}{4}\right\}=13600\times 10\times h$
$h=0.0193m$
$h=1.93cm$