JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 14)
A fully loaded boeing aircraft has a mass of $$5.4\times10^5$$ 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. ($$\mathrm{g=10~m/s^2}$$)
16
8
6
10
Explanation
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%
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