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CBSE Class 9 Mathematics Question 43 of 44

Surface Areas and Volumes — Question 11

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Question

Question 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 S and S′.

Answer

By formula,

The volume of a sphere = 43πr3\dfrac{4}{3}πr^3

The volume of 27 solid spheres with radius r = 27×43πr3=36πr327 \times \dfrac{4}{3}πr^3 = 36πr^3 ......(1)

The volume of the new sphere with radius r' = 43π(r′)3\dfrac{4}{3}π(r')^3 .......(2)

(i) Volume of the new sphere = Volume of 27 solid spheres

⇒43π(r′)3=36πr3(From equation (1) and (2))⇒(r′)3=36r3×34⇒(r′)3=9r3×3⇒(r′)3=27r3⇒r′=27r33⇒r′=3r\Rightarrow \dfrac{4}{3}π(r')^3 = 36πr^3 (\text{From equation (1) and (2)}) \\[1em] \Rightarrow (r')^3 = 36r^3 \times \dfrac{3}{4} \\[1em] \Rightarrow (r')^3 = 9r^3 \times 3 \\[1em] \Rightarrow (r')^3 = 27r^3 \\[1em] \Rightarrow r' = \sqrt[3]{27r^3} \\[1em] \Rightarrow r' = 3r

Hence, radius of the new sphere r' = 3r.

(ii) By formula,

Surface area of each iron sphere (S) = 4πr2

Surface area of the new sphere (S') = 4π(r′)2=4π(3r)2=36πr24π(r')^2 = 4π(3r)^2 = 36πr^2

Required ratio = 4πr236πr2\dfrac{4πr^2}{36πr^2}

The ratio of the S and S’ = 19\dfrac{1}{9}

Hence, the ratio of S and S’ is 1 : 9.