+4 A bowl is a hemisphere with 6cm. Water fills two-fifths of the volume in a bowl. The water is poured into a hollow cone. The depth of the water in the cone is 12cm. Work out the radius of the surface of the water in the cone
+9665 Volume of water: \(\dfrac{1}{2}\left(\dfrac{4\pi(6)^3}{3}\right)\times\dfrac{2}{5}=144\pi \times\dfrac{2}{5}=\dfrac{288\pi}{5}\)cm3
If the cone is inverted, the shape that the water formed in the cone is still a cone:
\(\dfrac{1}{3}\pi r^2(12)=\dfrac{288\pi}{5}\\ 60\pi r^2=864\pi\\ r^2=\dfrac{864}{60}=\dfrac{72}{5}\\ r=\dfrac{6\sqrt{10}}{5}\)
ANS: r = 6/5 x sqrt(10)
If the cone is not inverted, the shape that the water formed in the cone is a fructum(or something like that?) and the surface area can't be worked out without the exact height of the hollow cone.
Therefore ANS = 6sqrt(10)/5
