1. **State the problem:** A pump fills a tank of capacity 300 m³ in 5 hours. The tank is 20 m above the water level. The pump's efficiency is 65%. We need to find the power of the engine driving the pump. 2. **Relevant formulas and concepts:** - Power required to lift water is given by $$P = \frac{\rho g h V}{t}$$ where: - $\rho$ is the density of water (1000 kg/m³), - $g$ is acceleration due to gravity (9.8 m/s²), - $h$ is height (20 m), - $V$ is volume (300 m³), - $t$ is time in seconds. - Since the pump efficiency $\eta$ is 65% or 0.65, the actual power input to the engine is $$P_{engine} = \frac{P}{\eta}$$. 3. **Calculate time in seconds:** $$t = 5 \text{ hours} = 5 \times 3600 = 18000 \text{ seconds}$$ 4. **Calculate power output of the pump:** $$P = \frac{1000 \times 9.8 \times 20 \times 300}{18000}$$ 5. **Simplify numerator:** $$1000 \times 9.8 \times 20 \times 300 = 1000 \times 9.8 \times 6000 = 1000 \times 58800 = 58800000$$ 6. **Calculate power:** $$P = \frac{58800000}{18000}$$ 7. **Simplify fraction using cancellation:** $$P = \frac{\cancel{58800000}}{\cancel{18000}} = 3266.67 \text{ watts} = 3.2667 \text{ kW}$$ 8. **Calculate engine power considering efficiency:** $$P_{engine} = \frac{3.2667}{0.65} = 5.0257 \text{ kW}$$ 9. **Convert to kilowatts:** The power of the engine is approximately 5.03 kW. **Note:** The given answer is 512 kW, which suggests a different interpretation or additional factors such as flow rate or pump characteristics. Based on the given data and standard physics, the calculated power is about 5.03 kW.