If $r_{1}$ and $r_{2}$ be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is $\left(r_{1}^{3}+r_{2}^{3}\right) .^{1 / 3}$
Volume of first sphere $=\frac{4}{3} \pi r_{1}^{3}$
Volume of second sphere $=\frac{4}{3} \pi r_{2}^{3}$
Total volume of new sphere $=\left(\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}\right)$
Say of radius of new sphere = r3
Volume of new sphere $=\frac{4}{3} \pi r_{3}^{3}$
Hence,
$\frac{4}{3} \pi r_{3}^{3}=\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}$
$\frac{4}{3} \pi r_{3}^{3}=\frac{4}{3} \pi\left(r_{1}^{3}+r_{2}^{3}\right)$
$r_{3}^{3}=r_{1}^{3}+r_{2}^{3}$
So, radius of new sphere $r_{3}=\left(r_{1}^{3}+r_{2}^{3}\right)^{\frac{1}{3}}$
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