Question:
An experimental train uses 1 g of nuclear material to run. 90% of the produced energy is wasted to overcome the frictional force between the wheels and the track. If the weight of each coach is 18 ton (18000 kg) and it runs at a speed of 100 m s$^{-1}$, the number of coaches the engine can drag at a time is:
An experimental train uses 1 g of nuclear material to run. 90% of the produced energy is wasted to overcome the frictional force between the wheels and the track. If the weight of each coach is 18 ton (18000 kg) and it runs at a speed of 100 m s$^{-1}$, the number of coaches the engine can drag at a time is:
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When only a fraction of energy is useful, divide total useful energy by energy per unit to find maximum number of units moved.
Updated On: Jul 18, 2026
- 100
- 1000
- 10000
- 100000
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The Correct Option is D
Solution and Explanation
Step 1: Energy produced by nuclear material.
Mass of nuclear fuel \(m_f = 1 \, \text{g} = 10^{-3} \, \text{kg}\). Energy produced by complete conversion (E = mc$^2$) is: \[ E = m_f c^2 = 10^{-3} \times (3 \times 10^8)^2 = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13} \, \text{J} \]
Step 2: Useful energy fraction.
90 \[ E_{\text{useful}} = 0.1 \times 9 \times 10^{13} = 9 \times 10^{12} \, \text{J} \]
Step 3: Kinetic energy of one coach.
Each coach has mass \(M = 18000 \, \text{kg}\), velocity \(v = 100 \, \text{m/s}\). Kinetic energy: \[ KE = \frac{1}{2} M v^2 = \frac{1}{2} \times 18000 \times (100)^2 = 9000 \times 10000 = 9 \times 10^7 \, \text{J} \]
Step 4: Number of coaches that can be dragged.
\[ n = \frac{E_{\text{useful}}}{KE_{\text{per coach}}} = \frac{9 \times 10^{12}}{9 \times 10^7} = 10^5 \]
Step 5: Final conclusion.
\[ \boxed{100000 \, \text{coaches}} \]
Mass of nuclear fuel \(m_f = 1 \, \text{g} = 10^{-3} \, \text{kg}\). Energy produced by complete conversion (E = mc$^2$) is: \[ E = m_f c^2 = 10^{-3} \times (3 \times 10^8)^2 = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13} \, \text{J} \]
Step 2: Useful energy fraction.
90 \[ E_{\text{useful}} = 0.1 \times 9 \times 10^{13} = 9 \times 10^{12} \, \text{J} \]
Step 3: Kinetic energy of one coach.
Each coach has mass \(M = 18000 \, \text{kg}\), velocity \(v = 100 \, \text{m/s}\). Kinetic energy: \[ KE = \frac{1}{2} M v^2 = \frac{1}{2} \times 18000 \times (100)^2 = 9000 \times 10000 = 9 \times 10^7 \, \text{J} \]
Step 4: Number of coaches that can be dragged.
\[ n = \frac{E_{\text{useful}}}{KE_{\text{per coach}}} = \frac{9 \times 10^{12}}{9 \times 10^7} = 10^5 \]
Step 5: Final conclusion.
\[ \boxed{100000 \, \text{coaches}} \]
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