Q40 / 60 · KVPY 2011 (SA) · Mathematics
The diameter of one of the bases of a truncated cone is $100 \mathrm{~mm}$. If the diameter of this base is increased by $21 \%$ such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by $21 \%$. The radius of the other base (in $\mathrm{mm}$ ) is-
- A65
- Correct answerB55
- C45
- D35
Solution
Let initially 2 bases have radii $5 \mathrm{~cm}$ and $\mathrm{r} \mathrm{cm}$. Finally base have radii $(1.21 \times 5)$ and $\mathrm{r}$.
Ratios of volumes $=\frac{V_{2}}{V_{1}}=1.21$
$\begin{array}{l}
\mathrm{V}_{2}=\frac{\pi \mathrm{h}}{3}\left[(6.05)^{2}+(6.05) \mathrm{r}+\mathrm{r}^{2}\right] \\
\mathrm{V}_{1}=\frac{\pi \mathrm{h}}{3}\left[5^{2}+5 \mathrm{r}+\mathrm{r}^{2}\right] \\
\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=1.21 \Rightarrow \frac{(6.05)^{2}+(6.05) \mathrm{r}+\mathrm{r}^{2}}{5^{2}+5 \mathrm{r}+\mathrm{r}^{2}}=1.21 \\
\Rightarrow \mathrm{r}^{2}=\frac{6.3525}{21} \\
\Rightarrow \mathrm{r}=\frac{11}{2} \mathrm{~cm}=55 \mathrm{~mm}
\end{array}$
Hence, the correct answer is option: 'B'.