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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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I don’t know how to solve this problem of Surface Areas and Volumes of exercise 13.8 of class 9th 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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1. Volume of the solid sphere = (4/3)πr3

Volume of twenty seven solid sphere = 27×(4/3)πr3 = 36 π r3

(i) New solid iron sphere radius = r’

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

(4/3)π(r’)3 = 36 π r3

(r’)3 = 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 = 1/9

The ratio of S and S’ is 1: 9.

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