Question

If $r_1$ and $r_2$ be the radii of two metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is ${(r_1^3+r_2^3)}^{1/3}$.

Answer 1

Given,

If $r_1$ and $r_2$ be the radii of the two metallic spheres.

Volume of sphere having radius $r_1$ is $V_1=\frac{4}{3}\pi r_1^3$

Volume of sphere having radius $r_2$ is $V_2=\frac{4}{3}\pi r_2^3$

If the two spheres are melted to form one solid sphere of radius $R$ then,

Volume of sphere of radius $R=$ Volume of sphere of radius $r_1+$ Volume of sphere of radius $r_2$

$\to \frac{4}{3}\pi R^3=\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^3$

$\to \frac{4}{3}\pi R^3=\frac{4}{3}\pi (r_1^3+r_2^3)$

$\to R^3=(r_1^3+r_2^3)$

$\to R=(r_1^3+r_2^3{)}^{1/3}$

Hence, radius of new sphere is $(r_1^3+r_2^3{)}^{1/3}$