Question

Air streams flows horizontally past an air plane. The speed over the top surface is $50\text{m/s}$ and that under the bottom surface is $40\text{m/s}$. If the density of air is $1.2{\text{kg/m}}^3$, then the difference in pressure across the plane is

• A. $1160{\text{N/m}}^2$
• B. $540{\text{N/m}}^2$
• C. $280{\text{N/m}}^2$
• D. $12.84{\text{N/m}}^2$

Answer 1

The correct option is B $540{\text{N/m}}^2$

Let $P_1$ and $P_2$ be the pressure at the top and bottom surface respectively.
given, $v_1=50\text{m/s}$
$v_2=40\text{m/s}$
Applying Bernoulli's equation, we get
$P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_1+\rho g{h}_2+\frac{1}{2}\rho v_2^2$
$[\because h_1=h_2]$
$\Rightarrow P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
$\Rightarrow \frac{1}{2}\rho (v_1^2-v_2^2)=P_2-P_1$
$\Rightarrow \frac{1}{2}\times 1.2({50}^2-{40}^2)=\Delta P$
$\Rightarrow \Delta P=540{\text{N/m}}^2$