A molasses can is made up with a cylindrical base having a radius of 6 cm that is topped by a smaller cylindrical pouring spout with a radius 2 cm. When upright the top of the molasses is 14 cm above the base of the can but when turned over the top of the molasses is 19 1/3 cm above the base. Find the total height of the can in cm.

I know someone already answered this question however this is all the information I have for this question. Can someone please solve this with the information given. Please and Thank You.

Guest Dec 6, 2017

#1**0 **

Pretty sure I cannot answer this unless you give me the height of the spigot or the height of the body of the can. Sorry!

Guest Dec 6, 2017

#2**+2 **

I'm making one assumption in this problem...I'm assuming that when the larger cylinder is on the bottom [the natural position ]....the molasses fills all of that cylinder and none of the top cylinder...this would seem to be logical

Let us call the height of the smaller cylinder, h

First....when inverted so that the smaller cylinder is on the bottom, the molasses completely fills the smaller cylinder and the height of molasses that fills the larger cylinder is just (19 + 1/3 - h) = (58/3 - h)

So.....the total volume, V, of the molasses in this position is just

pi (2^2)h + pi (6^2) (58/3 - h) = V

4h * pi + 36pi (58/3 - h) = V

[4h + 36 ( 58/3 - h)] pi = V

[ 4h + 696 - 36h ] pi = V

[696 - 32h] pi = V (1)

And when the can is in its natural position, it fills all of the larger cylinder...so....the volume can be expressed as :

14* 36 * pi = V

504 pi = V (2)

Equate (1) and (2)

504 = 696 - 32h

32h = 192

h = 6 cm = the height of the smaller cylinder

So.... the total height of the can [ under my assumption] = 6cm + 14cm = 20 cm

CPhill Dec 6, 2017