picorants03p

2023-03-06

If the ratio of volumes of two spheres is 1 : 8 then the ratio of their surface areas is?

popescy91

Beginner2023-03-07Added 5 answers

Violume of first sphere $=\frac{4}{3}\pi {r}_{1}^{2}$

and volume of second sphere $=\frac{4}{3}\pi {r}_{2}^{3}$

But ratio in their volumes = 1 : 8

$\therefore \frac{\frac{4}{3}\pi {r}_{1}^{3}}{\frac{4}{3}\pi {r}_{2}^{3}}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}^{3}}{{r}_{2}^{3}}=\frac{1}{8}=(\frac{1}{2}{)}^{3}\phantom{\rule{0ex}{0ex}}=(\frac{1}{2}{)}^{3}\Rightarrow \frac{{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Now ratio in their surfaces areas $=\frac{4\pi {r}_{1}^{2}}{4\pi {r}_{2}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{r}_{1}^{2}}{{r}_{2}^{2}}\phantom{\rule{0ex}{0ex}}=(\frac{{r}_{1}}{{r}_{2}}{)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}=\frac{1}{4}$

and volume of second sphere $=\frac{4}{3}\pi {r}_{2}^{3}$

But ratio in their volumes = 1 : 8

$\therefore \frac{\frac{4}{3}\pi {r}_{1}^{3}}{\frac{4}{3}\pi {r}_{2}^{3}}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}^{3}}{{r}_{2}^{3}}=\frac{1}{8}=(\frac{1}{2}{)}^{3}\phantom{\rule{0ex}{0ex}}=(\frac{1}{2}{)}^{3}\Rightarrow \frac{{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Now ratio in their surfaces areas $=\frac{4\pi {r}_{1}^{2}}{4\pi {r}_{2}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{r}_{1}^{2}}{{r}_{2}^{2}}\phantom{\rule{0ex}{0ex}}=(\frac{{r}_{1}}{{r}_{2}}{)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}=\frac{1}{4}$