Question

A conical vessel of radius $6cm$ and height $8cm$ is completely filled with water. A sphere is lowered into water and its size is such that when it touches the sides, it is just immersed as shown in figure. What fraction of water over flows

Answer 1

Let the radius of the sphere be $r$ cm
In $△V{O}^{\prime }A$

we have
$\tan \theta =\frac{6}{8}=\frac{3}{4}\Rightarrow \sin \theta =\frac{3}{5}$

In $△VPO$, we have
$\sin \theta =\frac{r}{VO}\Rightarrow \frac{3}{5}=\frac{r}{8-r}\Rightarrow 24-3r=5r\Rightarrow r=3cm$

Volume of the sphere $V_1=\frac{4}{3}\pi {\times 3}^3{cm}^3=36\pi {cm}^3$

Volume of the water $V_2=\frac{1}{3}\pi \times 6^2\times 8{cm}^3=96\pi {cm}^3$

Clearly the volume of the water that flows out of the cone is same as the volume of the sphere $V_1$

Fraction of the water that flows out $V_1:V_2=36\pi :96\pi =36:96$