Question

a conical vessel of radius 12cm and height 16cm 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 the water overflows?

Answer 1

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

Answer 2

hi here is your answer..
hope it helps