Question

A conical vessel of radius 6cm 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

Hint: When the sphere just touches the inner surface of the cone, then the surface of the cone is nothing but a tangent to the sphere. Also, the centre of the sphere is R distance below the surface of the base of the cone (Where R is the radius of the sphere)
The volume of a sphere is
\[=\dfrac{4}{3}\pi {{r}^{3}}\]
(Where ‘r ‘ is the radius of the sphere)
The volume of a cone is
\[=\dfrac{1}{3}\pi {{r}^{2}}h\]
(Where ‘r’ is the radius of the cone and’ h’ is the height of the cone)

Complete step-by-step answer:
As mentioned in the question, the sphere is completely immersed in the cone.

Now, by referring to the figure, we can say that
\[\begin{align}
  & \angle CAB={{\tan }^{-1}}\dfrac{8}{6} \\
 & \angle CAB={{\tan }^{-1}}\dfrac{4}{3} \\
 & \angle CAB={{53}^{\circ }} \\
\end{align}\]

Now, if
\[\begin{align}
  & \angle CAB={{53}^{\circ }} \\
 & \therefore \angle BCA={{37}^{\circ }} \\
\end{align}\]

Now, on applying sin formula in \[\vartriangle PCO\] , we get
\[\begin{align}
  & \sin {{37}^{\circ }}=\dfrac{R}{\left( 8-R \right)} \\
 & \dfrac{3}{5}=\dfrac{R}{\left( 8-R \right)} \\
 & 24-3R=5R \\
 & R=3 \\
\end{align}\]
(Where R is the radius of the sphere)

Therefore, using the formula for finding the volume of a sphere as it is given in the hint, we get
\[\begin{align}
  & =\dfrac{4}{3}\pi {{(3)}^{3}} \\
 & =4\times 9\pi \\
 & =36\pi \\
\end{align}\]

Now, the fraction of water which overflows is given as follows
\[\begin{align}
  & =\dfrac{36\pi }{\dfrac{1}{3}\pi {{6}^{2}}\times 8} \\
 & =\dfrac{3}{8} \\
\end{align}\]
Hence, the fraction of water that overflows is \[\dfrac{3}{8}\] .

Note: The students can make an error if they don’t know the formulas for volume of sphere and cone and also the trigonometric ratio which are given in hint as
The volume of a sphere is
\[=\dfrac{4}{3}\pi {{r}^{3}}\]
(Where ‘r ‘ is the radius of the sphere)
The volume of a cone is
\[=\dfrac{1}{3}\pi {{r}^{2}}h\]
(where ‘r’ is the radius of the cone and’ h’ is the height of the cone)