Twenty-seven solid iron spheres,each of radiu | vedclass

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(C) Let the radius of each small sphere be $r$ and the radius of the new large sphere be $r^{\prime}$.The volume of one small sphere is given by $V =

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$3$

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(C) Let the radius of each small sphere be $r$ and the radius of the new large sphere be $r^{\prime}$.The volume of one small sphere is given by $V = \frac{4}{3} \pi r^{3}$.The total volume of $27$ small spheres is $27 	imes \frac{4}{3} \pi r^{3}$.When these are melted to form a new sphere of radius $r^{\prime}$,the volume of the new sphere is $\frac{4}{3} \pi (r^{\prime})^{3}$.Since the volume remains constant during melting and recasting,we have:$\frac{4}{3} \pi (r^{\prime})^{3} = 27 	imes \frac{4}{3} \pi r^{3}$Dividing both sides by $\frac{4}{3} \pi$,we get:$(r^{\prime})^{3} = 27 r^{3}$Taking the cube root on both sides:$r^{\prime} = \sqrt[3]{27 r^{3}} = 3r$.Thus,the radius of the new sphere is $3r$.

Twenty-seven solid iron spheres,each of radius $r$ and surface area $S$,are melted to form a single sphere with surface area $S^{\prime}$. Find the radius $r^{\prime}$ of the new sphere. (in $r$)

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