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Question 18
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
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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