JEE Physics 2023 Question with Solution

Given: mass of aircraft $= 5.4 \times 10^5 \, \text{kg}$, wing area $= 500 \, \text{m}^2$, speed $= 1080 \, \text{km/h} = 300 \, \text{m/s}$, air density $= 1.2 \, \text{kg/m}^3$, and $g = 10 \, \text{m/s}^2$.

Find: the fractional increase in air speed on the upper surface relative to the lower surface, in percentage.

In level flight, the lift balances the weight, so

$$(P_2 - P_1)A = mg$$

Hence,

$$P_2 - P_1 = \frac{5.4 \times 10^5 \times 10}{500} = 10.8 \times 10^3 \, \text{N/m}^2$$

Using Bernoulli’s principle,

$$P_2 - P_1 = \frac{1}{2}\rho (V_1^2 - V_2^2) = \rho (V_1 - V_2)(V_1 + V_2)$$

Substituting the given values,

$$10.8 \times 10^3 = 1.2 \times (V_1 - V_2) \times 600$$

So,

$$V_1 - V_2 = 30 \, \text{m/s}$$

Therefore,

$$\frac{V_1 - V_2}{V_1} \times 100 = \frac{30}{300} \times 100 = 10\%$$

Therefore, the fractional increase in speed is $10\%$. The correct option is D. The solution marks B, but the extracted working gives $10\%$, which matches option D.