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
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$\displaystyle ABC$ is the cone, with altitude $\displaystyle H \,=\,PC \,=\, 8$
Its radius is: .$\displaystyle R \,=\,AP\,=\,PB\,=\,6$
In right triangle $\displaystyle BPC$, Pythagorus tell us: .$\displaystyle BC \,=\,10$
Note that: .$\displaystyle BQ \,=\,BP \,=\,6$. . Hence: .$\displaystyle QC \,=\,4$
The center of the sphere is $\displaystyle O$ with radius $\displaystyle r\!:\;\;OP \,=\, OQ \,=\, r$
In right triangle $\displaystyle OQC\!:\;\;OQ^2 + QC^2 \:=\:OC^2$
We have: .$\displaystyle OQ \,=\,r,\;QC\,=\,4,\;OC \,=\,8-r$
Then: .$\displaystyle r^2 + 4^2 \:=\:(8-r)^2 \quad\Rightarrow\quad 16r \:=\:48 \quad\Rightarrow\quad\boxed{ r \:=\:3}$
The volume of the sphere is: .$\displaystyle V_s \:=\:\tfrac{4}{3}\pi r^3 \:=\:\tfrac{4}{3}\pi (3^3) \:=\:36\pi$
. . This is the volume of water that will overflow.
The volume of the cone is: .$\displaystyle V_c\:=\:\tfrac{1}{3}R^2H \:=\:\tfrac{1}{3}\pi(6^2)(8) \:=\:96\pi$
The fraction is: .$\displaystyle \frac{36\pi}{96\pi} \:=\:\frac{3}{8}$