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JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2023
- Hard
- Numerical
A fully loaded boeing aircraft has a mass of \(5.4 \times 10^5\,kg\). Its total wing area is \(500\,m ^2\). It is in level flight with a speed of \(1080\,km / h\). If the density of air \(\rho\) is \(1.2\,kg\,m ^{-3}\), the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be \(\left( g =10\,m / s ^2\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\(P _2 A - P _1 A =5.4 \times 10^5 \times g\)
\(P _2- P _1=\frac{5.4 \times 10^6}{500}=5.4 \times 2 \times 10^2 \times 10\)
\(=10.8 \times 10^3\)
\(P _2+0+\frac{1}{2} \rho V _2^2= P _1+0+\frac{1}{2} \rho V _1^2\)
\(P_2-P_1=\frac{1}{2} \rho\left(V_1^2-V_2^2\right)=\frac{1}{2} \rho\left(V_1-V_2\right)\left(V_1+V_2\right)\)
\(10.8 \times 10^3=\frac{1}{2} \times 1.2\left( V _1- V _2\right) \times 2 \times 3 \times 10^2\)
\(10.8 \times 10=3.6\left( V _1- V _2\right)\)
\(V _1- V _2=30\)
\(\left(\frac{ V _1- V _2}{ V }\right) \times 100=\frac{30}{300} \times 100=10 \%\)
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