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Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’.

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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 Sand S’.

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Volume of the solid sphere = \(\frac{4}{3}\)πr3

Volume of twenty seven solid sphere = 27 × \(\frac{4}{3}\)πr3 

= 36 π r3

(i) New solid iron sphere radius = r’

Volume of this new sphere = \(\frac{4}{3}\)π(r’)3

\(\frac{4}{3}\)π(r’)= 36 π r3

(r’)= 27r3

r’= 3r

Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S =4πr2

Surface area of iron sphere of radius r’= 4π (r’)2

Now

S/S’ = (4πr2)/( 4π (r’)2)

S/S’ = r2/(3r’)2 = \(\frac{1}{9}\)

The ratio of S and S’ is 1:9.

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