Question

An inverted cone of height 12cm and base radius 6cm contains `20cm^3` of water . Calculate the depth of water in the cone ,measured from the vetex?

Answer 1

The depth of water in the cone measured from the vertex is
`4.243(3dp)` cm.

Explanation:

Let the radius and hight of water cone is `r_w and`d_w#

respectively. The ratio of radius and hight of cone is

`r/d=6/12 =1/2`. The ratio of radius and hight of water cone

is `r_w/d_w=1/2 or r_w=d_w/2` . The volume of water cone is

`20 cm^3`. We know Volume om cone is `1/3*pi*r^2*d`

`:.1/3*pi*r_w^2*d_w =20 or pi*r_w^2*d_w =60` or

`pi*(d_w/2)^2*d_w =60 or pi*d_w^3=60*4or d_w^3=240/pi`

or `dw^3~~76.394 :. d_w=root3 76.394~~4.243(3dp)` cm

The depth of water in the cone measured from the vertex is

`4.243(3dp)` cm [Ans]

Answer 2

Height = `4.243cm`

Explanation:

We are dealing with two similar figures:
There are two cones involved:

• The whole cone which has a base radius of `6 cm` and a depth (height) of `12cm`

The Volume of this cone is `V = 1/3 pi r^2 h`

`V = 1/3 pi (6^2) xx12`

`V = 144 pi`

• The smaller cone represented by the water which is filled with a volume of `20 cm^3`

The volumes of similar figures are in the same ratio as the ratio of the cube of their heights:

`(h_1)^3/ (H_2)^3 = 20/(144pi)" "(larr"smaller cone")/(larr"larger cone"`

`(h_1)^3/12^3 = 20/(144pi)`

`h_1^3 = (12^3 xx 20)/(144pi)`

`h_1^3 = 76.39437`

`h_1 = 4.243cm`