A jet-powered airplane is steadily climbing at a rate of 10 m/s. The air density is
0.8 kg/m3, and the thrust force is aligned with the flight path. Using the
information provided in the table below, the airplane's thrust to weight ratio is
______ (rounded off to one decimal place).
Airplane speed 80 m/s
Lift coefficient, CL 0.8
Aspect ratio 6
Oswald efficiency factor 0.8
Zero-lift drag coefficient 0.02
Official Answer Key:
0.2
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.52)**
**1. Intuitive Physical Setup & Given Parameters**:
Steady climbing flight:
- Rate of climb: $R/C = V_c = 10\text{ m/s}$
- Airspeed: $V = 80\text{ m/s}$
- Air density: $\rho = 0.8\text{ kg/m}^3$
- Thrust aligned with flight path ($\alpha_T = 0$)
- Lift coefficient: $C_L = 0.8$
- Wing aspect ratio: $AR = 6$
- Oswald efficiency: $e = 0.8$
- Zero-lift drag coefficient: $C_{D0} = 0.02$
We need to find the airplane's thrust-to-weight ratio $T/W$ (rounded off to one decimal place).
**2. Step-by-Step Flight Mechanics Derivation**:
1. **Climb Angle ($\gamma$)**:
$\sin\gamma = \frac{R/C}{V} = \frac{10}{80} = 0.125$
$\cos\gamma = \sqrt{1 - \sin^2\gamma} = \sqrt{1 - 0.015625} \approx 0.99215 \approx 1$
2. **Total Drag Coefficient ($C_D$)**:
Induced drag coefficient:
$C_{Di} = \frac{C_L^2}{\pi AR e} = \frac{0.8^2}{\pi \times 6 \times 0.8} = \frac{0.64}{4.8\pi} = \frac{0.64}{15.0796} \approx 0.04244$
Total drag coefficient:
$C_D = C_{D0} + C_{Di} = 0.02 + 0.04244 = 0.06244$
3. **Lift-to-Drag Ratio**:
$\frac{L}{D} = \frac{C_L}{C_D} = \frac{0.8}{0.06244} \approx 12.812$
$\frac{D}{L} = \frac{1}{12.812} \approx 0.07805$
4. **Equation of Motion in Climb**:
Along the flight path:
$T = D + W\sin\gamma$
Perpendicular to the flight path:
$L = W\cos\gamma \implies W = \frac{L}{\cos\gamma}$
Dividing by $W$:
$\frac{T}{W} = \frac{D}{W} + \sin\gamma = \left(\frac{D}{L}\right)\cos\gamma + \sin\gamma$
$\frac{T}{W} \approx 0.07805(0.992) + 0.125 \approx 0.0774 + 0.125 = 0.2024 \approx 0.2$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $T/W \approx 0.202$.
- **Rounding Requirement**: One decimal place $\implies \mathbf{0.2}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **0.2** (Acceptable Range: **0.18 to 0.22**)
- **Exam Takeaway**: Thrust-to-weight ratio in steady climb is $T/W = \frac{D}{L} + \sin\gamma$. Here $\frac{C_D}{C_L} \approx 0.078$ and $\sin\gamma = 0.125$, summing to $0.20$.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.