Cups A and B are cone shaped and have heights of #25 cm# and #27 cm# and openings with radii of #12 cm# and #6 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?

1 Answer
Jul 17, 2016

Cup A will be filled to a height of 16.2 cm.

Explanation:

The volume of a cone can be found using the formula;

#V_"cone" = 1/3 pi r^2 h#

Here, #r# is the radius of the opening, and #h# is the height of the cup. For a proof of this formula, check here. Using the values for the two cups given we get volumes of;

#V_A = 1/3 pi (12"cm")^2 (25"cm") = 3.77 xx 10^3"cm"^3#

#V_B = 1/3 pi (6"cm")^2 (27"cm") = 1.02 xx 10^3 "cm"^3#

It seems that the volume of cup A is larger than the volume for cup B, so cup A will not overflow. Instead, the water will settle into a cone of water with some height, #y#, and some radius, #x#. The volume of this cone can be expressed as;

#V = 1/3 pi x^2 y = V_B#

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We do not yet know the values of #x# or #y#, but we do know that they make up a similar cone to cup A, therefore they should have similar ratios to the total volume of cup A.

#x/y = r/h = 12/25#

Now we can solve for #x# in terms of #y# and we get;

#x = 12/25 y#

Going back to our volume expression above, we can plug in this #x# value, as well as our expression for the volume of cup B from before.

#color(red)(cancel(color(black)(1/3 pi))) (12/25 y)^2 y = color(red)(cancel(color(black)(1/3 pi))) (6"cm")^2 (27"cm")#

The volume formula constants will cancel out, and the #y# terms can be combined on the right side, leaving;

#(12/25)^2 y^3 = 972 "cm"^3#

The rest is algebra.

#y^3 = 972 (25/12)^2 "cm"^3#

#y = root(3)(972 (25/12)^2 "cm"^3)#

#y = 16.2 "cm"#