QJEEM23Ph6f27: JEE Main 2023 Physics - Class 11 JEEM23Ph6f27 Solution: A fully loaded boeing aircraft has a mass of kg. Its total wing …

Velocity of aircraft = 1050 km/h = 300 m/s

Now, weight of aircraft = $\Delta PA$

$\Delta P = {{5.4 \times {{10}^5} \times g} \over {500}} = 10800$ $\mathrm{Pa}$

From Bernoulli's principle

$\Delta P = {1 \over 2}\rho \left[ {V_{upper}^2 - V_{lower}^2} \right]$

$10800 = {1 \over 2} \times 1.2 \times V_{lower}^2\left[ {{{\left( {{{{V_{upper}}} \over {{V_{lower}}}}} \right)}^2} - 1} \right]$

${\left( {{{{V_{upper}}} \over {{V_{lower}}}}} \right)^2} = 1 + {{10800 \times 2} \over {1.2 \times {{(300)}^2}}} = 1.2$

${{{V_{upper}}} \over {{V_{lower}}}} = 1.096$

$\Rightarrow$ Fractional increases = 9.6%