Question

A plane is in a level flight at a constant speed and each of its two wings has an area of $50{\text{m}}^2$. If the speed of air is $180\text{km/h}$ over the lower wing surface and $252\text{km/h}$ over the upper wing surface, the plane's mass is [Lift force on the wings = weight of plane]
(Take density of air = $1{\text{kg/m}}^3$ and $g=10{\text{m/s}}^2$)

• A. $2.4\times {10}^4\text{kg}$
• B. $4.4\times {10}^4\text{kg}$
• C. $1.2\times {10}^4\text{kg}$
• D. $3.6\times {10}^4\text{kg}$

Answer 1

The correct option is C $1.2\times {10}^4\text{kg}$

Applying the Bernoulli's theorem above and below the wings, we get
$P_1+\frac{1}{2}\rho V_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho V_2^2+\rho g{h}_2$
$[\because h_1=h_2]$
$P_1-P_2=\frac{1}{2}\rho [V_2^2-V_1^2].....\left(1\right)$
Given, $v_1=180\text{km/h}=50\text{m/s}$ and
$v_2=252\text{km/h}=70\text{m/s}$
$\rho =1{\text{kg/m}}^3$
$P_1-P_2=\frac{1}{2}({70}^2-{50}^2)$
$P_1-P_2=1200\text{Pa}.....\left(2\right)$
Area of both wings $=2\times 50=100{\text{m}}^2$

Now,
Upward force on the plane $=(P_1-P_2)A=1200\times 100=1.2\times {10}^5\text{N}$

As the plane is in level flight, therefore upward force balances the weight of the plane.
$\therefore mg=1.2\times {10}^5\text{N}$

So, mass of the plane $=1.2\times {10}^4\text{kg}$