The volume of the hemisphere is given by:
V_hemisphere = (2/3)πr^3
The volume of water in the bowl is two-fifths of the volume of the hemisphere:
V_water = (2/5)(2/3)πr^3 = (4/15)πr^3
So the remaining volume in the bowl is:
V_bowl = V_hemisphere - V_water = (1/3)πr^3
We are told that this remaining volume is poured into a hollow cone with depth 12 cm. Let's call the radius of the surface of the water in the cone "R". We can set up an equation for the volume of the cone in terms of R and solve for R:
V_cone = (1/3)πR^2(12)
Setting V_cone equal to V_bowl, we have:
(1/3)πR^2(12) = (1/3)πr^3
Simplifying:
R^2 = (r^3)/(4312) = r^3/144
Taking the cube root of both sides:
R = (r^3/144)^(1/3) = r/3
Substituting the given radius of the hemisphere, r = 6 cm, we get:
R = (6 cm)/3 = 2 cm
Therefore, the radius of the surface of the water in the cone is 2 cm.