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(D) Let the radius of each small sphere be $r$ and the radius of the new large sphere be $R$.Given that $27$ small spheres are melted to form one larg

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$1: 9$

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(D) Let the radius of each small sphere be $r$ and the radius of the new large sphere be $R$.Given that $27$ small spheres are melted to form one large sphere,the volume remains conserved.Volume of $27$ small spheres $= 27 	imes (\frac{4}{3} \pi r^3)$.Volume of the large sphere $= \frac{4}{3} \pi R^3$.Equating the volumes: $27 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$.$27 r^3 = R^3$,which implies $R = 3r$.The surface area of a small sphere is $S = 4 \pi r^2$.The surface area of the large sphere is $S' = 4 \pi R^2 = 4 \pi (3r)^2 = 4 \pi (9r^2) = 36 \pi r^2$.Now,the ratio $S : S' = \frac{4 \pi r^2}{36 \pi r^2} = \frac{1}{9}$.Thus,the ratio is $1: 9$.

Twenty-seven solid iron spheres,each of radius $r$ and surface area $S$,are melted to form a single sphere with surface area $S'$. Find the ratio of $S$ to $S'$.

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