A right circular cone has height 9 cm and radius of the base 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of `(pi/"A")`cm/sec, where A is the area of the water surface A at that instant, show that the vessel will be full in 75 seconds.

#### Solution

Let r be the radius of the water surface and h be the height of the water at time t.

∴ area of the water surface A = πr^{2} sq cm.

Since the height of the right circular cone is 9 cm and radius of the base is 5 cm.

`"r"/"h" = 5/9` ∴ r = `5/9"h"`

∴ area of water surface, i.e. A = `pi(5/9 "h")^2`

∴ A = `(25 pi"h"^2)/81` .....(1)

The water level, i.e. the rate of change of h is `"dh"/"dt"` rises at the rate of `(pi/"A")` cm/sec.

∴ `"dh"/"dt" = pi/"A" = (pi xx 81)/(25 pi "h"^2)` ....[By (1)]

∴ `"dh"/"dt" = 81/(25 "h"^2)`

∴ `"h"^2 "dh" = 81/25 "dt"`

On integrating, we get

`int "h"^2 "dh" = 81/25 int "dt" + "c"`

∴ `"h"^3/3 = 81/25 * "t + c"`

Initially, i.e. when t = 0, h = 0

∴ 0 = 0 + c ∴ c = 0

∴ `"h"^3/3 = 81/25 "t"`

When the vessel will be full, h = 9

∴ `(9)^3/3 = 81/25 xx "t"`

∴ t = `(81 xx 9 xx 25)/(3 xx 81) = 75`

Hence, the vessel will be full in 75 seconds.