The volumes of two spheres are in the ratio 6 | vedclass

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Question

(A) Let $V_{1}$ and $V_{2}$ be the volumes of two spheres.$	herefore \frac{V_{1}}{V_{2}} = \frac{64}{27}$$\Rightarrow \frac{\frac{4}{3} \pi r_{1}^{3}

Vedclass · Accepted Answer

$16: 9$

Vedclass · Answer

(A) Let $V_{1}$ and $V_{2}$ be the volumes of two spheres.$	herefore \frac{V_{1}}{V_{2}} = \frac{64}{27}$$\Rightarrow \frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}} = \frac{64}{27}$$\Rightarrow \frac{r_{1}^{3}}{r_{2}^{3}} = \frac{64}{27}$$\Rightarrow \left(\frac{r_{1}}{r_{2}}ight)^{3} = \left(\frac{4}{3}ight)^{3}$$\Rightarrow \frac{r_{1}}{r_{2}} = \frac{4}{3} \quad ...(1)$Now,let $SA_{1}$ and $SA_{2}$ be the surface areas of the two spheres.$	herefore \frac{SA_{1}}{SA_{2}} = \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}$$\Rightarrow \frac{SA_{1}}{SA_{2}} = \left(\frac{r_{1}}{r_{2}}ight)^{2}$$\Rightarrow \frac{SA_{1}}{SA_{2}} = \left(\frac{4}{3}ight)^{2} \quad [	ext{Using } (1)]$$\Rightarrow \frac{SA_{1}}{SA_{2}} = \frac{16}{9}$$	herefore SA_{1} : SA_{2} = 16 : 9$Thus,the ratio of the surface areas of the two spheres is $16 : 9$.

The volumes of two spheres are in the ratio $64: 27$. Find the ratio of their surface areas.

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