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A Plane is in Level Flight at Constant Speed and Each of Its Two Wings Has an Area of 25 M2. If the Speed of the Air is 180 Km/H Over the Lower Wing and 234 Km/H Over the Upper Wing Surface, Determine the Plane’S Mass. (Take Air Density to Be 1 Kg M–3). - Physics

A plane is in level flight at a constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).

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Solution 1

The area of the wings of the plane, A = 2 × 25 = 50 m2

Speed of air over the lower wing, V1 = 180 km/h = 50 m/s

Speed of air over the upper wing, V2 = 234 km/h = 65 m/s

Density of air, ρ = 1 kg m–3

Pressure of air over the lower wing = P1

Pressure of air over the upper wing= P2

The upward force on the plane can be obtained using Bernoulli’s equation as:

`P_1 + 1/2 rhoV_1^2 = P_2 + 1/2 rhoV_^2`

`P_1 - P_2= 1/2 rho(V_2^2 - V_1^2)` ...(i)

The upward force (F) on the plane can be calculated as:

`(P_1 - P_2)A`

`= 1/2 rho(V_2^2 - V_1^2)A` Using equation (i)

`= 1/2 xx 1 xxx((65)^2 - (50)^2)xx 50`

= 4400.51 Kg

~ 4400 kg

Hence, the mass of the plane is about 4400 kg.

Solution 2

Here speed  of air over lower wing `v_1 = 180 km/h = 180 xx 5/18 = 50 ms^(-1)`

Speed over the upper wing `v_2 = 234 "km/h" = 234 xx 5/18 = 65 ms^(-1)`

:. Pressure difference, `P_1 -P_2 = 1/2 rho (v_2^2 - v_1^2) = 1/2 xx 1(65^2 - 50^2) = 862.5 Pa`

:. Net upward force , `F = (P_1 -P_2)A`

This upward force balances the weight of the plane

:. `mg  = F =(P_1 - P_2)A`   [A = 25 x 2 = 50 `m^2`]

:. `m = ((P_1 - P_2)A)/g = (862.5xx50)/9.8 = 4400 N`

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 10 Mechanical Properties of Fluids
Q 27 | Page 271
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