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64 small solid iron spheres of radius 'r' are melted to form a big sphere of radius R. If S and S' are surface areas of the small and the big sphere respectively, then find the ratio S' : S.

Solution

Correct Option: 3

Volume: 64×43πr3=43πR364\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3, so R3=64r3R^3=64r^3, giving R=4rR=4r. Ratio S:S=4πR2:4πr264S':S = 4\pi R^2 : 4\pi r^2 \cdot 64? No, SS is surface of one small sphere. S=4πR2=4π(4r)2=64πr2S'=4\pi R^2=4\pi(4r)^2=64\pi r^2. S=4πr2S=4\pi r^2. Ratio =64:4=16:1=64:4=16:1.

Note: The chosen option 1 (4:1) is incorrect; the correct answer is option 3 (16:1).

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