Water leaks out of a tank so that the depth h (in metres) of the water after t hours satisfies \(\displaystyle\frac{dh}{dt} = -k\sqrt{h}\), where k is a positive constant. The depth is 16 m at the start and 9 m after 1 hour. The tank becomes empty after
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4 hours
The variables separate:
\( \displaystyle \frac{dh}{\sqrt{h}} = -k\,dt \)
Integrating both sides:
\(2\sqrt{h} = -kt + c\)
At t = 0, h = 16:
\(c = 2 \times 4 = 8\)
At t = 1, h = 9:
\(2 \times 3 = -k + 8\)
\(k = 2\)
So
\(2\sqrt{h} = 8 - 2t\)
\(\sqrt{h} = 4 - t\)
The tank is empty when h = 0:
\(4 - t = 0\)
\(t = 4\)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{16}{7}\) hours — Incorrect. This assumes the depth falls by 7 m every hour; it is \(\sqrt{h}\) that falls by 1 every hour.
Option (3) 3 hours — Incorrect. At t = 3, \(\sqrt{h} = 1\), so 1 m of water is still left.
Option (4) 2 hours — Incorrect. At t = 2, \(\sqrt{h} = 2\), so 4 m of water is still left.