Question

Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the 
(i) radius r' of the new sphere, 
(ii) ratio of S and S'

Answer 1

(i) Radius of 1 solid iron sphere = r 
Volume of 1 solid iron sphere = \(\frac{4}{3}\pi\)r³
Volume of 27 solid iron spheres =27 × [\(\frac{4}{3}\pi\)r³]
27 solid iron spheres are melted to form 1 iron sphere. 
Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres.
Let the radius of this new sphere be r'.
Volume of new solid iron sphere = \(\frac{4}{3}\pi\)r³
(\(\frac{4}{3}\)) \(\pi\)r'³ = 36\(\pi\)r³
\(⇒\) r'³ = 36\(\pi\)r³ ×\( \frac{3}{4}\pi\)
\(⇒\) r'³ = 27r³
\(⇒\) r' = \(^3\sqrt{27}\) r³
r' = 3r

Radius of the new sphere, r' = 3r

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(ii) Surface area of 1 solid iron sphere of radius r =\( 4\pi r^2 \)
Surface area of iron sphere of radius r' = 4\(\pi\) (r')² 
= 4 \(\pi\) (3r)²
= 36 \(\pi r^2\)
\(\frac{S}{S’}\) = \(\frac{4\pi r^2}{36\pi r^2} = \frac{1}{9}=1:9.\)