Thread: a sum dealing with cone, sphere...help!

1. a sum dealing with cone, sphere...help!

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 will overflow?

2. Hello, snigdha!

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 will overflow?
Code:
      A      6        P       6       B
- o - - - - - - * o * - - - - - - o
:\        *     |     *        /
: \     *       |r      *     /
:  \   *        |        *   /
:   \           |           /
:    \*         oO        */ 6
:     \         | *  r    /
:      *        |   *    *
8 :       *       |r    * *
:        *      |      o Q
:         \   * o *   /
:          \    |    /
:           \   |   /
:            \  |  / 4
:             \ | /
:              \|/
-               o
C

$ABC$ is the cone, with altitude $H \,=\,PC \,=\, 8$
Its radius is: . $R \,=\,AP\,=\,PB\,=\,6$

In right triangle $BPC$, Pythagorus tell us: . $BC \,=\,10$

Note that: . $BQ \,=\,BP \,=\,6$. . Hence: . $QC \,=\,4$

The center of the sphere is $O$ with radius $r\!:\;\;OP \,=\, OQ \,=\, r$

In right triangle $OQC\!:\;\;OQ^2 + QC^2 \:=\:OC^2$

We have: . $OQ \,=\,r,\;QC\,=\,4,\;OC \,=\,8-r$

Then: . $r^2 + 4^2 \:=\:(8-r)^2 \quad\Rightarrow\quad 16r \:=\:48 \quad\Rightarrow\quad\boxed{ r \:=\:3}$

The volume of the sphere is: . $V_s \:=\:\tfrac{4}{3}\pi r^3 \:=\:\tfrac{4}{3}\pi (3^3) \:=\:36\pi$
. . This is the volume of water that will overflow.

The volume of the cone is: . $V_c\:=\:\tfrac{1}{3}R^2H \:=\:\tfrac{1}{3}\pi(6^2)(8) \:=\:96\pi$

The fraction is: . $\frac{36\pi}{96\pi} \:=\:\frac{3}{8}$