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Q9.14: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.14 Solution: In a test experiment on a model aeroplane in a wind tunnel, the …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often subtract the velocities in the wrong order, using $\frac12\rho(v_{upper}^2-v_{lower}^2)$, which yields a negative pressure difference and wrong lift direction. Remember that the slower flow (lower surface) has higher pressure, so compute $\Delta P=\frac12\rho\big(v_{lower}^2- v_{upper}^2\big)$ before multiplying by the wing area.

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s__ and 63 m s__ respectively. What is the lift on the wing if its area is 2.5 m_? Take the density of air to be 1.3 kg m__.

Step-by-Step Verified Solution

**Answer:** 1.51×1031.51 \times 10^3 N **Step-by-Step Explanation:** Using Bernoulli's principle:
P1−P2=12ρ(v22−v12)P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)
Plower−Pupper=12×1.3×(702−632)=0.65×(4900−3969)=605 PaP_{lower} - P_{upper} = \frac{1}{2} \times 1.3 \times (70^2 - 63^2) = 0.65 \times (4900 - 3969) = 605 \text{ Pa}
Lift force: F=(Plower−Pupper)×A=605×2.5=1.51×103F = (P_{lower} - P_{upper}) \times A = 605 \times 2.5 = 1.51 \times 10^3 N.

All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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