I don’t know how to solve this problem of Surface Areas and Volumes of exercise 13.8 of class 9^{th} math give me the easiest and simple way for solving this tough question of question no.9 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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# 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’. Q.9

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Volume of the solid sphere = (4/3)πr

^{3}Volume of twenty seven solid sphere = 27×(4/3)πr

^{3}= 36 π r^{3}(i) New solid iron sphere radius = r’

Volume of this new sphere = (4/3)π(r’)

^{3}(4/3)π(r’)

^{3 }= 36 π r^{3}(r’)

^{3 }= 27r^{3}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 =4πr

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

^{2}Now

S/S’ = (4πr

^{2})/( 4π (r’)^{2})S/S’ = r

^{2}/(3r’)^{2}= 1/9The ratio of S and S’ is 1: 9.