Question

A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered into the water such that when it touches the sides it is just immersed. What fraction of water overflows?

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 $\Delta$ PO’R, by Pythagoras theorem
$l^2=6^2+8^2$
$\Rightarrow l=\sqrt{(36+64)}=10cm$
Hence, $sin\theta =\frac{O^{\prime }P}{OR}=\frac{6}{10}=\frac{3}{5}$ ....(1)
in right $\Delta MRO$,
$sin\theta =\frac{OM}{OR}=\frac{r}{OR}$
$\Rightarrow \frac{3}{5}=\frac{r}{8-r}$ (Using (1) and $O^{\prime }R=O^{\prime }O+OR\Rightarrow OR=O^{\prime }R–O^{\prime }O=8-r)$
$\Rightarrow 24–3r=5r$
$\Rightarrow 8r=24$
$\Rightarrow r=3cm$
$\therefore$ Volume of sphere $\left(V_s\right)=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (3{)}^{3}c{m}^3=36\pi c{m}^3$
Now,
Volume of the water = volume of cone $\left(V_c\right)=96nc{m}^3$
Clearly, volume of the water that flows out of cone is same as the volume of the sphere i.e. $V_s$
$\therefore$ Fraction of the water that flows out $\frac{V_s}{V_c}=\frac{36\pi }{96\pi }=3:8$