AS Daily A Level Physics question
A garden fountain pump draws a steady electrical power of 200 W and raises water to a spout 10 m above the reservoir. When running, the flow reaches a steady value. If only 50% of the input power goes into increasing the water’s gravitational energy (the rest is lost as heating and turbulence), which value is closest to the maximum steady volume flow rate at the spout? Assume 1 litre of water has a mass of 1.0 kg.
Answer
The correct answer is C.
Correct: C — 1.0 L/s. Only half the 200 W raises water, so 100 W goes into gravitational energy; each litre (≈1 kg) needs about 9.8 × 10 ≈ 98 J to reach 10 m, so flow ≈ 100/98 ≈ 1.0 L/s. A uses the full 200 W, ignoring the stated 50% losses, so it overestimates the flow by a factor of about 2. B omits the factor g, effectively assuming each kilogram needs only h joules rather than g h, giving an answer about 10× too large. C is the correct application: output power divided by energy per litre (g h) gives ≈1.0 L/s. D mis-scales by roughly a factor of 10 (e.g. double-counting a 10), yielding a value far below what 100 W can sustain at 10 m.