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 the sides, it just immersed. what fraction of water overflows.

Explanation

By Pythagoras theorem,
`AB^(2)=AC^(2)+CB^(2)`
`=(6)^(2)+(8)^(2)=(10)^(2)`
`rArr AB = sqrt(100)=10 cm`
Area of `Delta ABE = ar(Delta AOE)+ ar(Delta EOB)+ ar(Delta BOA)`
`=(1)/(2)xx12xx r+(1)/(2)xx10xx r+(1)/(2)xx10xx r`
= 16 r ......(1)
But `ar(Delta ABC)=sqrt(s(s-a)(s-b)(s-c)` (where, `s=(a+b+c)/(2)=(12+10+10)/(2)=16`)
`sqrt(16(16-12)(16-10)(16-10))`
`sqrt(16xx4xx6xx6)`
`= 4xx2xx6`
`therefore` From (1) and (2), we have
`16 r=4xx2xx6 rArr r = 3 cm` .....(2)
Now, volume of water in the tank `=(1)/(3)pi(6)^(2)xx8`
Water overflows due to the sphere.
`therefore` Volume of water overflows = Volume of sphere `= (4)/(3)pi(3)^(3)`
`therefore` Required fraction of water overflows `=((4)/(3)pi(3)^(3))/((1)/(3)pi(6)^(2)xx8)=(3)/(8)`