Question
A solid sphere and a solid hemi $-$ sphere have the same total surface area. Find the ratio between their volumes.
✓
Answer
Let the radius of the sphere be $I r_1 I$.
Let the radius of the hemisphere be $I r_2 I$
$\text{TSA }$ of sphere $=4 \pi r_1^2$
$\text{TSA }$ of hemisphere $=3 \pi r_2^2 e$
$\text{TSA }$ of sphere $=\text{TSA }$ of hemi $-$ sphre
$4 \pi r_1^2=3 \pi r_2^2$
$\Rightarrow r_2^2=\frac{4}{3} r_1^2$
$\Rightarrow r_2=\frac{2}{\sqrt{3}} r_1$
Volume of sphere, $v_1=\frac{4}{3} \pi r_1^3$
Volume of hemisphere, $v_2=\frac{2}{3} \pi r_2^3$
$v_2=\frac{2}{3} \pi r_2^3$
$\Rightarrow v_2=\frac{2}{3} \pi\left(\frac{r_1 2}{3 \sqrt{3}}\right)^3$
$\Rightarrow v_2=\frac{2}{3} \pi \frac{r_2^3 8}{3 \sqrt{3}}$
Dividing $v_1$ by $ v_2$
$\frac{v_1}{v_2}=\frac{\frac{4}{3} \pi r_1^3}{\frac{2}{3} \pi \frac{8}{3 \sqrt{3}} r_1^3}$
$\Rightarrow \frac{v_1}{v_2}=\frac{\frac{4}{3}}{\frac{2}{3} \frac{8}{3 \sqrt{3}}}$
$\Rightarrow \frac{v_1}{v_2}=\frac{4}{3} \times \frac{9 \sqrt{3}}{16}$
$\Rightarrow \frac{v_1}{v_2}=\frac{3 \sqrt{3}}{4}$
Let the radius of the hemisphere be $I r_2 I$
$\text{TSA }$ of sphere $=4 \pi r_1^2$
$\text{TSA }$ of hemisphere $=3 \pi r_2^2 e$
$\text{TSA }$ of sphere $=\text{TSA }$ of hemi $-$ sphre
$4 \pi r_1^2=3 \pi r_2^2$
$\Rightarrow r_2^2=\frac{4}{3} r_1^2$
$\Rightarrow r_2=\frac{2}{\sqrt{3}} r_1$
Volume of sphere, $v_1=\frac{4}{3} \pi r_1^3$
Volume of hemisphere, $v_2=\frac{2}{3} \pi r_2^3$
$v_2=\frac{2}{3} \pi r_2^3$
$\Rightarrow v_2=\frac{2}{3} \pi\left(\frac{r_1 2}{3 \sqrt{3}}\right)^3$
$\Rightarrow v_2=\frac{2}{3} \pi \frac{r_2^3 8}{3 \sqrt{3}}$
Dividing $v_1$ by $ v_2$
$\frac{v_1}{v_2}=\frac{\frac{4}{3} \pi r_1^3}{\frac{2}{3} \pi \frac{8}{3 \sqrt{3}} r_1^3}$
$\Rightarrow \frac{v_1}{v_2}=\frac{\frac{4}{3}}{\frac{2}{3} \frac{8}{3 \sqrt{3}}}$
$\Rightarrow \frac{v_1}{v_2}=\frac{4}{3} \times \frac{9 \sqrt{3}}{16}$
$\Rightarrow \frac{v_1}{v_2}=\frac{3 \sqrt{3}}{4}$
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