Air is streaming past a horizontal airplane's wing such that its speed is 120 ms -1 over the upper surface and 90 ms -1 at the lower surface. If the density of air is 1.3 kgm -3 and the wing is 10 m long and has an average width of 2 m, then the difference of the pressure on the two sides of the wing is

Question

Air is streaming past a horizontal airplane's wing such that its speed is 120 ms -1 over the upper surface and 90 ms -1 at the lower surface. If the density of air is 1.3 kgm -3 and the wing is 10 m long and has an average width of 2 m, then the difference of the pressure on the two sides of the wing is (A) 4095.0 Pa (B) 409.50 Pa (C) 40.950 Pa (D) 4.0950 Pa

Tardigrade Admin · Accepted Answer

From the Bernoulli's theorem, p1-p2 =(1/2) ρ(v22-v12) Given, ρ =1.3 kgm -3 v2 =120 ms -1 v1 =90 ms -1 ⇒ p1-p2 =(1/2) × 1.3 ×[(120)2-(902)] =4095 Nm -2 or Pa

$4095.0 P a$

$409.50 P a$

$40.950 P a$

$4.0950 P a$

From the Bernoulli's theorem,
$p_{1} - p_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
Given, $ρ = 1.3 k g m^{- 3}$
$v_{2} = 120 m s^{- 1}$
$v_{1} = 90 m s^{- 1}$
$\Rightarrow p_{1} - p_{2} = \frac{1}{2} \times 1.3 \times [(120)^{2} - (9 0^{2})]$
$= 4095 N m^{- 2}$ or $P a$

4095.0Pa

409.50Pa

40.950Pa

4.0950Pa

From the Bernoulli's theorem, p1​−p2​=21​ρ(v22​−v12​) Given, ρ=1.3kgm−3 v2​=120ms−1 v1​=90ms−1 ⇒p1​−p2​=21​×1.3×[(120)2−(902)] =4095Nm−2 or Pa

$4095.0 P a$

$409.50 P a$

$40.950 P a$

$4.0950 P a$

From the Bernoulli's theorem,
$p_{1} - p_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$
Given, $ρ = 1.3 k g m^{- 3}$
$v_{2} = 120 m s^{- 1}$
$v_{1} = 90 m s^{- 1}$
$\Rightarrow p_{1} - p_{2} = \frac{1}{2} \times 1.3 \times [(120)^{2} - (9 0^{2})]$
$= 4095 N m^{- 2}$ or $P a$