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Question: In a wind tunnel experiment the pressures on the upper and lower surfaces of the wings are \(0.90 \t...

In a wind tunnel experiment the pressures on the upper and lower surfaces of the wings are 0.90×105 Pa0.90 \times 10 ^ { 5 } \mathrm {~Pa}and 0.91×105 Pa0.91 \times 10 ^ { 5 } \mathrm {~Pa} respectively. If the area of the wing is 40 m240 \mathrm {~m} ^ { 2 }the net lifting force on the wing is

A

2×104 N2 \times 10 ^ { 4 } \mathrm {~N}

B

4×104 N4 \times 10 ^ { 4 } \mathrm {~N}

C

6×104 N6 \times 10 ^ { 4 } \mathrm {~N}

D

8×104 N8 \times 10 ^ { 4 } \mathrm {~N}

Answer

4×104 N4 \times 10 ^ { 4 } \mathrm {~N}

Explanation

Solution

Pressure difference which provides the lift

= Pressure difference × Area of the wing

=(P2−P1)×A= \left( \mathrm { P } _ { 2 } - \mathrm { P } _ { 1 } \right) \times \mathrm { A }

=(0.91×105−0.90×105)Pa×40 m2= \left( 0.91 \times 10 ^ { 5 } - 0.90 \times 10 ^ { 5 } \right) \mathrm { Pa } \times 40 \mathrm {~m} ^ { 2 }

=0.01×105 Pa×40 m2= 0.01 \times 10 ^ { 5 } \mathrm {~Pa} \times 40 \mathrm {~m} ^ { 2 }

=4×104 N= 4 \times 10 ^ { 4 } \mathrm {~N}