Question

Cups A and B are cone shaped and have heights of `32 cm` and `24 cm` and openings with radii of `15 cm` and `8 cm`, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

Answer 1

The cup does not overflow and it will be filled to a height of 19.12 cm.

Explanation:

The volume of a cone is given by

`V=pir^2h/3`

For our two cups we have volumes of

`V_A=pi*15^2*32/3=2400pi" "cm^3`

`V_B=pi*8^2*24/3=512pi" "cm^3`

Because the volume of B is less than A, we know that if we pour B into A it won't overflow.

So we need to solve for the height of fluid in cone A. We know the volume taken up but we don't know the new height or radius. But we do know that the new shape that the fluid takes up will be another cone.

The volume the fluid takes up is the volume of cone B

`V=512pi=pir^2h/3" "cm^3`

We must solve for height, but there are two unknowns. So now we must find an expression for radius in terms of height, so that we can substitute it into the volume expression and get rid of the radius term. This means there is only one unknown and allows us to solve for height.

To do this, we use the concept of similar triangles.

Because the angle `theta` in the picture below of Cone A is constant for any cone drawn within the cone, we can draw similar triangles.

This means that the ratio of radius to height is the same for each triangle. We can use this to get an expression for our unknown `r` in terms of `h`

`15/32=r/hrArrr=15h/32`

Now we can substitute this back into our volume expression

`V=pir^2h/3=pi(15h/32)^2h/3=(225pih^3)/3072`

We know that

`V=512pi" "cm^3`

So now we can solve for `h`

`512pi=(225pih^3)/3072rArrh=((512pi*3072)/(225pi))^(1/3)=19.12" "cm`