A conical vessel of radius 12cm and height 16cm is completely filled with water. A sphere is lowered into the water and it's size is such that when it touches the sides, it is just immersed. What fraction of water overflows?
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Given, radius of right circular cone=BC=12, height of right circular cone=BG=16
When just immersed, the sphere touches at two points E & K, hence AB=BC=12
Since, triangles △BCO and △OCE are congruent, EC=BC=12
By Pythagoras theorem, GC²=BG²+BC²
Solving, GC=10; thus GE=12-10=2
In △OGE, we have OJ=OE=r. Now let JG=x
So, by Pythagoras theorem, OG²=GE²+OE²
(r+x)²=32+r
x²+2r.x−32=0
x=−r±r2+32−−−−−−√
2r−r+r2+32−−−−−−√=8
=Volume of sphere= Vsphere=2.π.r3sphere3
of water= Volume of cone= Vcone=π.r2cone.hcone3
of water overflown=VsphereVcone=(43).π.33π.62.(83)=38
When just immersed, the sphere touches at two points E & K, hence AB=BC=12
Since, triangles △BCO and △OCE are congruent, EC=BC=12
By Pythagoras theorem, GC²=BG²+BC²
Solving, GC=10; thus GE=12-10=2
In △OGE, we have OJ=OE=r. Now let JG=x
So, by Pythagoras theorem, OG²=GE²+OE²
(r+x)²=32+r
x²+2r.x−32=0
x=−r±r2+32−−−−−−√
2r−r+r2+32−−−−−−√=8
=Volume of sphere= Vsphere=2.π.r3sphere3
of water= Volume of cone= Vcone=π.r2cone.hcone3
of water overflown=VsphereVcone=(43).π.33π.62.(83)=38
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