Question

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?

Answer 1

Radius of the conical vessel, R = AC = 6 cm.

Height of the conical vessel, H = OC = 8 cm.

Let the radius of the sphere be r.

Then, PC = PD = r.

Now, AC = AD = 6 cm.

[Since lengths of two tangents from an external point to a circle are equal]

$OA=\sqrt{O{C}^2+A{C}^2}=\sqrt{8^2+6^2}=\sqrt{100}=10cm$

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

OP = (OC - PC) = (8 - r).

In right angled $\Delta ODP$, we have:

$O{P}^2=O{D}^2+P{D}^2$

$(8-r{)}^2=4^2+r^2$

$64-16r+r^2=16+r^2$

$16r=48\Rightarrow r=3$.

Volume of water overflown = Volume of sphere
$=\frac{4}{3}\pi r^3=[\frac{4}{3}\pi \times (3{)}^3]c{m}^3=\left(36\pi \right)c{m}^3$ .

Volume of water in the cone before immersing the sphere = volume of cone
$=\frac{1}{3}\pi r^{2}h=(\frac{1}{3}\pi \times (6{)}^2\times 8)c{m}^3=\left(96\pi \right)c{m}^3$

$\therefore$ Fraction of water overflown = $\frac{Volumeofwateroverflown}{Originalvolumeofwater}$

$\Rightarrow$ Fraction of water overflown $=\frac{36\pi }{96\pi }=\frac{3}{8}$