2025 TKSS PRELIMS P1 Q3

Water is drained from a conical tank at a constant rate of \(10\pi\ \mathrm{cm}^3/\mathrm{s}\).

The volume of water, \(V\ \mathrm{cm}^3\), when the depth of water is \(h\) cm, is given by \(V=\frac{1}{12}\pi h^3\) and

the surface area of the water, \(A\ \mathrm{cm}^2\), is \(A=\frac{1}{4}\pi h^2\).

1. Find \(\frac{dh}{dt}\) in terms of \(h\).[3]
2. Find the rate at which the surface area of the water is decreasing when the depth of water is 10 cm.[2]