Question

Air is blown through a pipe $AB$ at a rate of $15\text{litres/min.}$ The cross-sectional area of the broad portion of the pipe $AB$ is $2{\text{cm}}^2$ and that of the narrow portion is $0.5{\text{cm}}^2$. The difference in water level $h$ is (density of air $1.32{\text{kg/m}}^3$)

• A. $16\text{mm}$
• B. $1.5\text{mm}$
• C. $10\text{mm}$
• D. $3.2\text{mm}$

Answer 1

The correct option is B $1.5\text{mm}$

Volume flow rate of air is given as
$Q=15\text{litres/min.}=250{\text{cm}}^3/\text{s}$

Now, $Q=A_1{V}_1=A_2{V}_2$

$250=2V_1\Rightarrow V_1=125\text{cm/s}=1.25\text{m/s}$

And, $250=0.5\times V_2\Rightarrow V_2=500\text{cm/s}=5\text{m/s}$
For a fluid flowing through a horizontal pipe,
difference in pressure $=$ difference in kinetic energy

$\frac{\Delta P}{{\rho }_a}=\frac{1}{2}(V_2^2-V_1^2)$

$\frac{h{\rho }_{w}g}{{\rho }_a}=\frac{1}{2}(V_2^2-V_1^2)$

$\therefore h=\frac{{\rho }_a(V_2^2-V_1^2)}{2{\rho }_{w}g}$

$=\frac{1.32\times (5^2-{1.25}^2)}{2\times {10}^3\times 10}$

$=1.54\times {10}^{-3}\text{m}=1.5\text{mm}$

Hence, $\left(B\right)$ is the correct answer.

Why this Q? With the help of venturimeter we can measure flow speed in a pipe of non-uniform cross-section.