Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively are melted to form a single solid sphere.

Solution:

We have,

the radii $r_{1}=6 \mathrm{~cm}, r_{2}=8 \mathrm{~cm}$ and $r_{3}=10 \mathrm{~cm}$

Let the radius of the resulting sphere be $R$.

As,

Volume of resulting sphere $=$ Volume of three metallic spheres

$\Rightarrow \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}+\frac{4}{3} \pi r_{3}{ }^{3}$

$\Rightarrow \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi\left(r_{1}{ }^{3}+r_{2}{ }^{3}+r_{3}{ }^{3}\right)$

$\Rightarrow R^{3}=r_{1}{ }^{3}+r_{2}{ }^{3}+r_{3}{ }^{3}$

$\Rightarrow R^{3}=6^{3}+8^{3}+10^{3}$

$\Rightarrow R^{3}=216+512+1000$

$\Rightarrow R^{3}=1728$

$\Rightarrow R=\sqrt[3]{1728}$

$\Rightarrow R=12 \mathrm{~cm}$

So, the radius of the resulting sphere is 12 cm.