Consider a circulation distribution over a finite wing given by the equation below:
$$\Gamma(y) = \begin{cases} \Gamma_0 \left(1 - \frac{2y}{b}\right) & \text{if } 0 \le y \le \frac{b}{2} \\[6pt] \Gamma_0 \left(1 + \frac{2y}{b}\right) & \text{if } -\frac{b}{2} \le y \le 0 \end{cases}$$
The wingspan $b$ is $10\text{ m}$, the maximum circulation $\Gamma_0$ is $20\text{ m}^2/\text{s}$, density of air is $1.2\text{ kg/m}^3$ and the free stream speed is $80\text{ m/s}$. The lift over the wing is ______ $\text{N}$ (rounded off to the nearest integer).
Scientific schematic diagram reproduced from official IIT paper
Official Answer Key:
9650.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2022 Q.51)**
**1. Kutta-Joukowski Lift Integration**:
$$L = \rho_\infty V_\infty \int_{-b/2}^{b/2} \Gamma(y) \, dy$$
Since $\Gamma(y)$ is a symmetric triangular distribution of base $b$ and height $\Gamma_0$:
$$\int_{-b/2}^{b/2} \Gamma(y) \, dy = \frac{1}{2} \times b \times \Gamma_0 = \frac{1}{2} \times 10 \times 20 = 100\text{ m}^3/\text{s}$$
**2. Total Lift**:
$$L = 1.2 \times 80 \times 100 = 9600\text{ N}$$
- **Official Answer Key**: **9600** (Tolerance Range: 9550 to 9650)
