Pls solve the following question:

Solution

Let VAB be cone of height 30 cm and base radius r₁ cm.

Suppose it is cut off by a plane parallel to the base at a height h₂ from the base of the cone.

Clearly $∆VOD~∆VO\text{'}B$

Therefore,

$$\begin{gathered} \frac{OV}{O\text{'}V}=\frac{OD}{O\text{'}B} \\ \Rightarrow \frac{h_1}{30}=\frac{r_2}{r_1} \end{gathered}$$

But,

$$\begin{gathered} Volume of cone VCD=\frac{1}{27}Volume of cone VAB \\ \Rightarrow \frac{1}{3}\mathrm{\pi }{\left({\mathrm{r}}_2\right)}^2{\mathrm{h}}_1=\frac{1}{27}\left(\frac{1}{3}\mathrm{\pi }{\left({\mathrm{r}}_1\right)}^{2}30\right) \\ \Rightarrow {\left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)}^2{\mathrm{h}}_1=\frac{10}{9} \\ \Rightarrow {\left(\frac{{\mathrm{h}}_1}{30}\right)}^2{\mathrm{h}}_1=\frac{10}{9} \\ \Rightarrow {\mathrm{h}}_1=10 \end{gathered}$$

Hence,

Required height = 30 - 10 = 20 cm