The height of a cone is 30 cm. A small cone is cut off at the top by…

ATQ,

Vol. of bigger cone : Vol. of smaller cone = H³ : h³ = 27 : 1

⇒ H : h = 3 : 1

⇒ H = 3R = 30 cm (given)

Hence, h = 1R = 10 cm

Therefore, Small cone cut above = 30 - 10 = 20 cm from the base.

 

Alternate Method:

In ΔAQC & ΔAOB

∠A is common and QC ∥ OB

⇒ ΔAQC  ∼ ΔAOB ..... (i)

⇒ \(\frac{AQ}{AO}\) = \(\frac{QC}{OB}\)

⇒ \(\frac{h}{30}\) = \(\frac{r}{R}\) ..... (ii)

Volume of small cone = \(\frac{1}{3}\) \(\pi \) r²h

Volume of bigger cone = \(\frac{1}{3}\) \(\pi \) R²30

Now;  \(\frac{1}{3}\) \(\pi \) r²h = \(\frac{1}{27}\) (\(\frac{1}{3}\) \(\pi \) R²× 30)

⇒ r²h = \(\frac{10R^2}{9}\)

⇒ (\(\frac{r}{R}\))² = \(\frac{10}{9h}\)

putting in eqn. (ii)

⇒ \(\frac{10}{9h}\) = (\(\frac{h}{30}\))²

⇒ h³ = \(\frac{900 × 10}{9}\)

h = 10 cm

Required height = 30 - 10 = 20 cm