A student has two open-topped containers: the larger one is a right
circular cylinder of height 21 cm and radius 12 cm, the smaller one is a right
circular cone of height 20 cm and radius 10 cm (at the open end). (The walls
of the two containers are thin enough so that their widths can be ignored.) The
larger container initially has water in it to a depth of 18 cm. The student slowly
lowers the cone into the larger container, keeping the top (open side) of the cone
horizontal. As the cone is lowered, water may first overflow out of the larger
container, but eventually water pours into the cone. When the tip of the cone is
resting on the bottom of the cylinder, what is the depth of the water in the cone?
The larger container has a volume of 144*21*pi ~= 9500.17618445553 cm^3. Its initial content has a volume of 144*18*pi ~= 8143.00815810474 cm^3.
The smaller container has a volume of 20*100*pi/3 ~= 2094.39510239319 cm^3.
The volume of the cone plus the initial water is pi*(144*18+2000/3) ~= 10237.4032604979, which exceeds the capacity of the larger container, so indeed some water has spilled outside the larger container by the time the top of the cone is aligned with the top of the larger container as well as the surface of the water. The amount that has spilled is pi*(2000/3 - 144*3) ~= 737.227076042405 cm^3, leaving pi*(144*21 - 2000/3) ~= 7405.78108206232. Indeed we could have started with this step, knowing that there'd be spillage outside the larger container.
When the tip of the cone reaches the bottom of the cylinder, either the cone is completely filled, or the cone is partially filled and the surface of the water outside the cone is at the same level as the top of the cone. We'll assume the latter case to see if it accounts for enough water; in fact it seems the point of the puzzle.
The portion of the cylinder that's below the level of the top of the cone has volume 20*144*pi, of which (20*144 - 2000/3)*pi ~= 6953.39173994539 cm^3 is outside the cone, leaving pi*(144*21 - 2000/3 + 2000/3 - 20*144) = 144*pi ~= 452.38934211693 cm^3 to have gone inside the cone.
From there, pi*100 * h / 3 = 144*pi allows us to solve for h, the depth of the water in the cone.
h = 144*3/100 = 4.32 cm.
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Posted by Charlie
on 2019-06-06 15:54:26 |
solution
