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?

2 Answers
Dec 27, 2017

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]

Dec 27, 2017

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