Question

twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with radius r' and surface area S'.find the radius r' of new sphere and ratio of S and S'.​

Answer 1

Answer:

(i) Radius of 1 solid iron sphere = r

Volume of 1 solid iron sphere = $\sf\dfrac{4}{3}\:\pi\:r^3$

Volume of 27 solid iron spheres = $\sf{27\:\times}$$\sf\dfrac{4}{3}\:\pi\:r^3$

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Let the radius of the new sphere be x

Volume of new solid iron sphere = $\sf\dfrac{4}{3}\:\pi\:r^3$

$\\ \sf \: \frac{4}{3} \: \pi \: {r}^{3} = 27 \times \frac{4}{3} \: \pi \: {r}^{3} \\ \\ \\ \implies \sf \: {r}^{3} = 27 \: \: {r}^{3} \\ \\ \\ \implies \sf \blue{r = 3r}$

• Surface area of 1 solid iron sphere R = 4πr³

Surface area of iron sphere of Radius r' = 4π (r)²

$\\ \sf \: - 4\pi \: (3r) {}^{2} - 36\pi \: {r}^{2} \\ \\ \\ \sf \: \frac{S}{S'} = \frac{4 \: \pi \: {r}^{3} }{36 \: \pi \: {r}^{3} } = \frac{1}{9} \\ \\ \\ \implies \sf \blue{1 \: \colon \: 9}$