GATE AE 2021 • Q18 MCQ • 1 Mark Propulsion

Propulsion

Rocket Propulsion

Subtopic: Solid, Liquid & Hybrid Rocket Engines

Question Statement:

A solid propellant rocket producing 25 MN thrust is fired for 150 seconds. If the propellant mass consumed during this duration is $1.25 \times 10^6$ kg, the effective exhaust velocity (in m/s) is:

Options:

  • (A) 2000
  • (B) 2500
  • (C) 3000
  • (D) 3500

Official Answer Key:

C

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2021 Q.18 / AE Q.8)** **1. Intuitive Physical Setup & Given Parameters**: - Thrust: $F = 25\text{ MN} = 25 \times 10^6\text{ N}$. - Burn time: $t_b = 150\text{ s}$. - Total propellant mass consumed: $m_p = 1.25 \times 10^6\text{ kg}$. **2. Step-by-Step Propulsion Derivation**: 1. **Propellant Mass Flow Rate ($\dot{m}_p$)**: \dot{m}_p = \frac{m_p}{t_b} = \frac{1.25 \times 10^6\text{ kg}}{150\text{ s}} = \frac{1250000}{150} = \frac{25000}{3}\text{ kg/s} \approx 8333.33\text{ kg/s} 2. **Effective Exhaust Velocity ($c$)**: From the rocket thrust equation $F = \dot{m}_p c$: c = \frac{F}{\dot{m}_p} = \frac{F \times t_b}{m_p} = \frac{I_{total}}{m_p} 3. **Evaluating $c$**: I_{total} = F \times t_b = (25 \times 10^6\text{ N}) \times 150\text{ s} = 3.75 \times 10^9\text{ N s} c = \frac{3.75 \times 10^9\text{ N s}}{1.25 \times 10^6\text{ kg}} = 3000\text{ m/s} The effective exhaust velocity is exactly **3000 m/s**. **3. Comprehensive Options & Distractor Analysis**: - **(A)** 2000: ❌ Incorrect. - **(B)** 2500: ❌ Incorrect. - **(C)** 3000: ✅ **CORRECT**. Exactly matches $c = I_{total}/m_p = 3.75 \times 10^9 / 1.25 \times 10^6 = 3000\text{ m/s}$. - **(D)** 3500: ❌ Incorrect. **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **C** - **Key Formula**: $c = \frac{F}{\dot{m}} = \frac{F \cdot t_b}{m_p}$.
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