A conical vessel of radius $6 \; cm$ and height $8 \; cm$ is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. Find the fraction of water that overflows.
Figure shows the cross-sectional view of the conical vessel (triangle ABC) and the sphere (circle at center $C_1$).
Since the sphere touches the sides of the conical vessel when it is just immersed, the sides of the cone are tangents to the circle.
In the figure,
radius of the conical vessel $= CM = MB = r = 6 \; cm$
height of the conical vessel $= AM = h = 8 \; cm$
slant height of the conical vessel $= AC = \ell = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \; cm$
radius of the sphere $= C_1 K = R \; cm$
In triangles $AKC_1$ and $AMC$,
$\angle AKC_1 = \angle AMC = 90^\circ$
[angle between the tangent to a circle and the radius of the circle at the point of contact]
$\angle A = \angle A$ $\;\;\;$ [common angle]
$\therefore \;$ By AA postulate, $\;$ $\triangle AKC_1 \sim \triangle AMC$
$\therefore \;$ $\dfrac{MC}{KC_1} = \dfrac{AC}{AC_1}$ $\;\;\;$ [corresponding parts of similar triangles are in proportion]
i.e. $\;$ $\dfrac{r}{R} = \dfrac{\ell}{8 - R}$
i.e. $\;$ $\dfrac{6}{R} = \dfrac{10}{8 - R}$
i.e. $\;$ $48 - 6R = 10R$
i.e. $\;$ $16 R = 48$ $\implies$ $R = 3 \; cm$
Now,
Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3^3 = 36 \pi \; cm^3$
Volume of conical vessel $= V_C = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \pi \times 6^2 \times 8 = 96 \pi \; cm^3$
$\therefore \;$ Fraction of water that overflows
$= \dfrac{V_C - V_S}{V_C} = \dfrac{96 \pi - 36 \pi}{96 \pi} = \dfrac{60}{96} = \dfrac{5}{8}$ of the total volume of water