The height of a right circular cone is 30 cm . A small part of the cone is cut off from the above part of the cone with the help of a plane parallel to the base of the cone. If the volume of the small part be `(1)/(27)` part of the volume of the original cone , then at what distance above the base of the cone , it has been cut off ?

Explanation

Let OAB is the original cone of radius of base R cm and its height be H cm.
`therefore` the volume of the cone `=(1)/(3)piR^2H c c =(1)/(3)pi R^2xx30 c c =10pi R^2c c`
The cut off small part of the cone is OCD, the radius of base of which is r cm and height is h cm. Then the volume of this part `=(1)/(3) pi r^2h c c`.
As per question , `(1)/(3)pi r^2h=(1)/(27)(10pi R^2)rArr h=(10pi R^2)/(27)xx(3)/(pi r^2)rArr h =(10)/(9).(R^2)/(r^2)`..................(1)
Now , from `Delta OQB and Delta OPD` we get , `(QB)/(PD)=(OQ)/(OP)=(30)/(h)rArr (R)/(r)=(30)/(h)`.
`therefore` from (1) we get , ` h=(10)/(9)((R)/(2))^2=(10)/(9)((30)/(h))^2=(10)/(9)xx(900)/(h^2)or , h^3=1000 or h^3=(10)^3 or , h=10` .
`therefore ` the small part is cut off at a distance of (30-10) cm=20 cm high above the base of the cone.
Hence the required height =20cm.