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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.

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Solution

(i)

Radius of 1 solid iron sphere =r

Volume of 1 solid iron sphere =43πr3

Volume of 27 solid iron spheres =27×43πr3

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 =43π(r)3

43π(r)3=27×43πr3

(r)3=27r3

(r)3=(3r)3

r=3r

(ii)

Surface area of 1 solid iron sphere of radius (S)=4πr2

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

S=4π(3r)2

S=36πr2

Now,

SS=4πr236πr2

SS=19

S:S=1:9


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