MCQ
A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is:
- A$1 : 3$
- ✓$1:\sqrt{3}$
- C$1 : 1$
- D$\sqrt{3}:1$
✓
Answer
Correct option: B.
$1:\sqrt{3}$
Surface area of hemispherical part $=$ surface area of conical part
$\Rightarrow2\pi\text{r}^2=\pi\text{rl}$
$\Rightarrow2\text{r}\text{l}$
$\Rightarrow2\text{r}=\sqrt{\text{r}^2+\text{h}^2}$
$\Rightarrow4\text{r}^2=\text{r}^2+\text{h}^2$
$\Rightarrow3\text{r}^2=\text{h}^2$
$\Rightarrow\frac{\text{r}^2}{\text{h}^2}=\frac{1}{3}$
$\Rightarrow\frac{\text{r}}{\text{h}}=\frac{1}{\sqrt{3}}$
$\therefore\text{Roots}=1:\sqrt{3}$
$\Rightarrow2\pi\text{r}^2=\pi\text{rl}$
$\Rightarrow2\text{r}\text{l}$
$\Rightarrow2\text{r}=\sqrt{\text{r}^2+\text{h}^2}$
$\Rightarrow4\text{r}^2=\text{r}^2+\text{h}^2$
$\Rightarrow3\text{r}^2=\text{h}^2$
$\Rightarrow\frac{\text{r}^2}{\text{h}^2}=\frac{1}{3}$
$\Rightarrow\frac{\text{r}}{\text{h}}=\frac{1}{\sqrt{3}}$
$\therefore\text{Roots}=1:\sqrt{3}$
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