The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be `1/(27)` of the volume of the given cone, at what height above the base is the section mode?

Explanation

Let OA=h `rArr` AB=30-h
and let AC=r,BD=R
`triangle` OAC `triangle` OBD
`(h)/(30)=(r )/(R )rArr r=(hR)/(30)`
Now volume of smaller cone =`(1)/(27)xx`volume of larger cone
`(1)/(3)pir^(2)h=(1)/(27)xx(1)/(3)xx(piR^(2)xx30`
`(1)/(3)pi(hr)/(30)^(2).h=(1)/(27)xx(1)/(3)piR^(2)xx30`
`(h^(2))/(30^(2))=(30)/(2)rArrh^(3)=(30^(3))/(30^(3))rArrh=10`
Required height =30-10=20 cm
1kl=`1m^(3)`
1000=`1000xx100xx100cm^(3)`
1 litre =1000 `cm^(3)`
1cm `5^(3)=(1)/(1000)`litre