Question

How do you find the rate at which water is pumped into an inverted conical tank that has a height of 6m and a diameter of 4m if water is leaking out at the rate of `10,000(cm)^3/min` and the water level is rising `20 (cm)/min`?

Answer 1

This question has already been answered although you seem to be missing the height of the water in the cone at the time the water level is rising `20 (cm)/(min)`.
Assuming this question came from the same source, the specified height of water was `2 m` or `200 cm`.

The cone has a radius of 2 m (half the diameter) and a height of 6 m
for a ratio of `(radius)/(height) = 1/3`

This ratio is constant for volumes of water contained in the cone,

Therefore the volume of the cone (or water in the cone), normally written as
`V(r,h) = (pi r^2h)/3`

can be re-written as
`V(h) = ( pi (h/3)^2 h)/3`
`= (pi h^3)/(27)`

and therefore
`(d V(h))/(dh) = pi/9 h^2` `(cm^3)/(cm)`

We are told
`(d h)/(dt) = 20 (cm)/(min)`

The increase in volume contained in the cone is given by
`(d V)/(dh) xx (d h)/(dt)` at water level height of `200 cm`

`= pi/9 (200 cm)^2 xx 20 (cm)/(min)`
`= 2,792,527 (cm^3)/(min)` (approx. assuming I haven't slipped up somewhere)

The inflow of water must be the total of
the outflow (leakage)
plus
the amount needed to raise the water level:
`10,000 ((cm)^3)/(min)`
`+ 2,792,527 ((cm)^3)/(min)`

`= 2,802,527 ((cm)^3/min)`