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?

• A. 0.375
• B. 0.625
• C. 0.125
• D. 0.875

Answer 1

The correct option is

A

0.375

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

Height of the conical vessel, $h=OC=8cm$

Radius of the sphere, $PD=PC=r$

$\therefore PC=PD=r$
AC = AD = 6 $\text{cm}$

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

$△OCA\&△OPD$ are right triangle.
[$\because$ Tangent and radius are perpendicular to each other]

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

$OD=OA-AD=10-6=4\text{cm}$
$OP=OC-PC=8-r$
$(8-r{)}^2=4^2+r^2$
$64-16r+r^2=16+r^2$
$16r=48\Rightarrow r=3\text{cm}$

Volume of water overflown

= Volume of sphere
$=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times (3{)}^3=36\pi {\text{cm}}^3$

Original volume of water

= volume of cone

$=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 6^2\times 8=96\pi {\text{cm}}^3$

$\therefore$ Fraction of water overflown $=\frac{\text{Volume of water overflown}}{\text{Original volume of water}}=\frac{36\pi }{96\pi }=\frac{3}{8}=0.375$

$\therefore$ Fraction of water overflown is $0.375$