Question

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

• 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 $1.5\text{mm}$

Applying equation of continuity for the flow section,
Flowrate: $Q=Av$
$\Rightarrow Q=A_1{v}_1=A_2{v}_2=15\text{L/min}$
$=\frac{15\times {10}^{-3}}{60}=2.5\times {10}^{-4}{\text{m}}^3\text{/s}$
$\Rightarrow v_1=\frac{Q}{A_1}=\frac{2.5\times {10}^{-4}}{2\times {10}^{-4}}=1.25\text{m/s}$
& $v_2=\frac{Q}{A_2}=\frac{2.5\times {10}^{-4}}{0.5\times {10}^{-4}}=5.0\text{m/s}$
Since the elevation is same for the horzontal section $AB$, applying Bernoulli's equation gives:
$\Delta P=\frac{1}{2}{\rho }_{\text{air}}(v_2^2-v_1^2)...\left(i\right)$
where $v_1=v_A\&v_2=v_B$
$\Delta P=P_A-P_B$
Because $A_A>A_B$, hence $v_A<v_B$ and in accordance with Bernoulli's theorem, $P_A>P_B$
The pressure difference can be related with height difference in water columns.
$\Delta P={\rho }_{w}gh...\left(ii\right)$

From Eq.$\left(i\right)$ and $\left(ii\right)$,
${\rho }_{w}gh=\frac{1}{2}{\rho }_{\text{air}}(v_2^2-v_1^2)$
$h=\frac{{\rho }_{\text{air}}}{2{\rho }_{w}g}(v_2^2-v_1^2)$
On substituting the values, we get,
$h=\frac{1.32}{2\times {10}^3\times 10}\left[\right(5{)}^2-\left(1.25{)}^2\right]$
$\Rightarrow h=1.54\times {10}^{-3}\text{m}$
$\therefore h\simeq 1.5\text{mm}$