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

• Jan 30th 2010, 08:55 AM
snigdha
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?
• Jan 30th 2010, 10:59 AM
Soroban
Hello, snigdha!

Quote:

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