The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere new sphere is half of that previous one. Calculate the ratio of the volumes of the portion cut off and remaining portion of the shere.

Explanation

Let the radius of the metalic sphere be R unit and that of the new sphere produced be r unit.
As per question, `4pir^(2)=(1)/(2)xx4piR^(2) rArrR^(2)=2r^(2)rArrR=sqrt(2r).`
`:.` the volume of the new sphere `=(4)/(3)pir^(3)` cubic.units.
Also, the volume of the portion cut off `=((4)/(3)piRr^(3)=(4)/(3)pir^(2))`
=`(4)/(3)pi(R^(3)-r^(3))` cubic.units.
`=(4)/(3)xx(22)/(7){(sqrt(2)r)^(3)-r^(3)}` cubic.units
`=(4)/(3)xx(22)/(7)(2sqrt(2)r^(3)-r^(3))` cubic.units
`=(4)/(3)xx(22)(2sqrt(2)-1)r^(3)` cubic.units
`:.` ratio of the volumes of the cut off the large sphere and the volume of the remaining part.
`=(4)/(3)xx(22)/(7)xx(2sqrt(2)-1)r^(3):(4)/(3)pir^(3)=(2sqrt(2)-1):1`
Hence the required ratio=`=(2sqrt(2)-1):1`