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 . What fraction of water overflows ?

Given,

Radius of conical vessel (R) = AC = 6 cm

Height of conical vessel (H) = OC = 8 cm

Radius of sphere (r)

∴ PC = PD = r cm.

We know that,

Since, lengths of two tangents from an external point to a circle are equal.

∴ AC = AD = 6 cm

As radius from center to the point of tangent are perpendicular to each other.

△OCA and △OPD are right angle triangles.

In △OCA,

By pythagoras theorem,

$\Rightarrow OA^2 = OC^2 + AC^2 \\[1em] \Rightarrow OA = \sqrt{OC^2 + AC^2} \\[1em] \Rightarrow OA = \sqrt{8^2 + 6^2} \\[1em] \Rightarrow OA = \sqrt{64+ 36} \\[1em] \Rightarrow OA = \sqrt{100} = 10 \text{ cm}.$

In △OPD,

By pythagoras theorem,

$\Rightarrow OP^2 = OD^2 + PD^2 \text{ ……..(1)}$

From figure,

OD = OA - AD = 10 - 6 = 4 cm.

OP = OC - PC = 8 - r

Substituting values of OP, OD and PD in equation (1), we get :

$\Rightarrow (8 - r)^2 = 4^2 + r^2 \\[1em] \Rightarrow 64 + r^2 - 16r = 16 + r^2 \\[1em] \Rightarrow r^2 - r^2 + 64 - 16 - 16r = 0 \\[1em] \Rightarrow 16r = 48 \\[1em] \Rightarrow r = \dfrac{48}{16} = 3 \text{ cm}.$

Volume of water overflown = Volume of sphere

= $\dfrac{4}{3}πr^3 = \dfrac{4}{3}π(3)^3$

= $\dfrac{4}{3}π \times 27= 36π$ cm³.

Original volume of water = Volume of cone

= $\dfrac{1}{3}πR^2H = \dfrac{1}{3}π \times 6^2 \times 8$ = 96π cm³.

Fraction of water overflown = $\dfrac{\text{Volume of water overflown}}{\text{Original volume of water}} = \dfrac{36π}{96π} = \dfrac{3}{8}$.

Hence, fraction of water overflown = $\dfrac{3}{8}$.