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A water tank with a cross-sectional area of 750 cm² is placed on a rooftop. The water surface is at a height h metres above a tap whose cross-sectional area is 500 mm². At the instant when water flows out of the tap at 30 cm/s, the rate of fall of the water level in the tank is x * 10⁻³ m/s. Find x.
- x = 2
- x = 4
- x = 20
- x = 0.2
Correct answer: x = 2
Solution
By continuity, the volume leaving the tap per second equals the volume lost by the tank per second: A_tap * v = A_tank * |dh/dt|. With A_tap = 500 mm² = 5 * 10⁻⁴ m² * (1/10000)... Careful unit conversion gives |dh/dt| = 2 * 10⁻³ m/s, so x = 2.
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