The height of a cone is 20 cm. A small cone is cut off at the top by a plane parallel to the base. If the volume of smaller cone is \(\frac{1}{64}\)th of volume of the bigger cone, at what height above the base was the cut made? 

15 cm

Assume QB is the point of cut for smaller cone.

In ΔAQB, ΔAOC; ∠A is common and QB ll OC

⇒ ΔAQB    ∾     ΔAOC

⇒ \(\frac{AQ}{AO}\) = \(\frac{QB}{OC}\)

⇒ \(\frac{h}{H}\) = \(\frac{r}{R}\) ..... (i)

volume of smaller cone = \(\frac{1}{3}\) \(\pi \)r²h

volume of bigger cone = \(\frac{1}{3}\) \(\pi \)R²H

We know; \(\frac{1}{3}\) \(\pi \)r²h = \(\frac{1}{64}\) (\(\frac{1}{3}\) \(\pi \)R²H)

⇒ \(\frac{r^2}{R^2}\) = \(\frac{1}{64}\) (\(\frac{H}{h}\)) .... (ii)

From eqn. (i);   

\(\frac{r}{R}\)² =  \(\frac{h^2}{H^2}\)

⇒ Putting eqn. (ii) \(\frac{h^2}{H^2}\) = \(\frac{1}{64}\) × \(\frac{H}{h}\)

⇒ h³ = \(\frac{H^3}{64}\) 

⇒ h = \(\frac{H}{4}\) = \(\frac{20}{4}\) = 5 cm

So, required height= (20 - 5) = 15 cm