Question 268047
a sphere of diameter 25 cm is half full with acid all of which is drained to a tall cyliderical beaker 16 cm in diameter. what is the depth of the acid in the beaker

volume of a sphere is 4/3 * pi * r^3 (diameter is 2 times the radius r)
volume = 4/3 * pi * (12.5)^3
volume = 4/3 * 1953.125 * pi
volume = 2604 1/6 * pi
volume of sphere is 1/2 full --> 1302 1/12 * pi cm^3 acid
volume of beaker = area of base * (height H)
volume = r^2 * H * pi
volume = 8^2 * H * pi
volume = 64 * H * pi cm^3 is volume of beaker
set the volumes equal and solve for H which will be the depth of the acid in the beaker
1302 1/12 * pi = 64 * H * pi (divide out pi from both sides)
1302 1/12 = 64 * H
20.345052 or 20.345 cm is approximately H