10.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.

Speed of the wind on the upper surface of the wing, V1 = 70 m/sSpeed of the wind on the lower surface of the wing, V2 = 63 m/sArea of the wing, A = 2.5 m2Density of air, ρ = 1.3 kg/ m3According to Bernoulli’s theorem, we have the relation:P1 + 12ρV12 = P2 + 12ρV22( P2 - P1 )= 12ρ(V12 - V22) , where P1 = pressure on the upper surface of the wing and P2 - pressure on the lower surface of the wingThe pressure difference provides lift to the aeroplaneLift on the wing = ( P2 - P1 )A = 12ρ(V12 - V22) A = 12×1.3×(702 - 632)×2.5 N = 1512.8 N = 1.5 ×103 N

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physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Fluids

10.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^–1and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^–3.

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Vishal Baghel | Contributor-Level 10

5 months ago

Speed of the wind on the upper surface of the wing, $V_{1}$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_{2}$ = 63 m/s

Area of the wing, A = 2.5 $m^{2}$

Density of air, $ρ$ = 1.3 kg/ $m^{3}$

According to Bernoulli’s theorem, we have the relation:

$P_{1}$ + $\frac{1}{2} ρ {V_{1}}^{2}$ = $P_{2}$ + $\frac{1}{2} ρ {V_{2}}^{2}$

( $P_{2}$ - $P_{1}$ )= $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ , where $P_{1}$ = pressure on the upper surface of the wing and $P_{2}$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_{2}$ - $P_{1}$ )A = $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ A = $\frac{1}{2} \times 1.3 \times (70^{2}$ - $63^{2}) \times 2.5$ &nbs

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Speed of the wind on the upper surface of the wing, $V_{1}$ = 70 m/s

Speed of the wind on the lower surface of the wing, $V_{2}$ = 63 m/s

Area of the wing, A = 2.5 $m^{2}$

Density of air, $ρ$ = 1.3 kg/ $m^{3}$

According to Bernoulli’s theorem, we have the relation:

$P_{1}$ + $\frac{1}{2} ρ {V_{1}}^{2}$ = $P_{2}$ + $\frac{1}{2} ρ {V_{2}}^{2}$

( $P_{2}$ - $P_{1}$ )= $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ , where $P_{1}$ = pressure on the upper surface of the wing and $P_{2}$ - pressure on the lower surface of the wing

The pressure difference provides lift to the aeroplane

Lift on the wing = ( $P_{2}$ - $P_{1}$ )A = $\frac{1}{2} ρ ({V_{1}}^{2}$ - ${V_{2}}^{2})$ A = $\frac{1}{2} \times 1.3 \times (70^{2}$ - $63^{2}) \times 2.5$ N = 1512.8 N = 1.5 $\times 10^{3}$ N

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10.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^–1and 63 m s^-1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^–3.

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