# Twenty seven solid iron spheres, each of radius $r$ and surface area $\mathrm{S}$ are melted to form a sphere with surface area $S^{\prime}$. Find the(i) radius $r^{\prime}$ of the new sphere,(ii) ratio of $\mathrm{S}$ and $\mathrm{S}^{\prime}$.

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Given:

Twenty seven solid iron spheres, each of radius $r$ and surface area $\mathrm{S}$ are melted to form a sphere with surface area $S^{\prime}$.

To do:

We have to find the

(i) radius $r^{\prime}$ of the new sphere,
(ii) ratio of $\mathrm{S}$ and $\mathrm{S}^{\prime}$.

Solution:

(i) Radius of each solid iron sphere $=r$

This implies,

Volume of each solid iron sphere $=\frac{4}{3} \pi r^{3}$

Volume of each 27 solid iron sphere $=27\times\frac{4}{3} \pi r^{3}$

$=36 \pi r^{3}$

Radius of the new sphere $=r^{\prime}$

Therefore,

Volume of the new sphere $=\frac{4}{3} \pi r^{\prime 3}$

$\frac{4}{3} \pi r^{\prime 3}=36 \pi r^{3}$

$r^{\prime 3}=\frac{36 \times 3 \times r^{3}}{4}$

$=27 r^{3}$

$=(3 r)^{3}$

$r^{\prime}=3 r$

The radius $r^{\prime}$ of the new sphere is $3r$.

(ii) Surface area of each iron sphere of radius $r$ is $S =4\pi r^2$

Surface area of the iron sphere of radius $r^{\prime}= 4 \pi r^{\prime 2}$

Ratio of $S$ and $S^{\prime}=4\pi r^2:4 \pi r^{\prime 2}$

$=r^2:(3r)^2$

$=r^2:9r^2$

$=1:9$

The ratio of $S$ and $S^{\prime}$ is $1: 9$.

Updated on 10-Oct-2022 13:46:39