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Question -

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

(i) radius r’ of the new sphere,

(ii) ratio of Sand S’.



Answer -

Volume of the solidsphere = (4/3)πr3

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

(i) New solid ironsphere radius = r’

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

(4/3)π(r’)=36 π r3

(r’)=27r3

r’= 3r

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

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

Surface area of ironsphere 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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