Question

A conical vessel of radius 6 centimeter and height 8 centimeter 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 inside the cone. Find the radius of the sphere and also find out the fraction of the water that flowed out.

Answer 1

Radius (R) of conical vessel = 6 cm
Height (H) of conical vessel = 8 cm
volume of conical vessel $\left(V_c\right)=\frac{1}{3}\pi R^{2}H$
$=\frac{1}{3}\times \pi \times 6^2\times 8$
$=96\pi c{m}^3$
Let the radius of the sphere be r cm
In right $\Delta P{O}^{\prime }R$, by pythagoras theorem:
$l^2=6^2+8^2$
$\Rightarrow l=\sqrt{36+64}=10cm$
Hence, si$n\theta =\frac{O^{\prime }P}{PR}=\frac{6}{10}=\frac{3}{5}$ ....(1)
In right$\Delta MRO$,
$sin\theta =\frac{OM}{OR}=\frac{r}{OR}\Rightarrow \frac{3}{5}=\frac{r}{8-r}$
(Using (1) and $O^{\prime }R=O^{\prime }O+OR\Rightarrow OR=O^{\prime }R-O^{\prime }O=8-r$)
$\Rightarrow 24-3r=5r\Rightarrow 8r=24\Rightarrow r=3cm\therefore Volumeofsphere\left(V_s\right)=\frac{4}{3}\pi r^3\frac{4}{3}\pi (3{)}^{3}c{m}^3=36\pi c{m}^3$
Now,
Volume of the water=Volume of cone $\left(V_c\right)=96\pi c{m}^3$
Clearly, Volume of the water that flows out of cone is same as the volume of the sphere i.e. $V_2$.
$\therefore$ Fraction of the water that flows out $\frac{V_s}{V_c}=\frac{36\pi }{96\pi }=3:8$