Metallic spheres of radii 6 cm, 8 cm and 10 cm,

Solution:

Radii of the given spheres are :

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

$\Rightarrow$ Volume of the given spheres are :

$V_{1}=\frac{4}{3} \pi r_{1}{ }^{3}, V_{2}=\frac{4}{3} \pi r_{2}{ }^{3}$ and $V_{3}=\frac{4}{3} \pi r_{3}{ }^{3}$

$\therefore$ Total volume of the given spheres $=\mathrm{V}_{1}+\mathrm{V}_{2}+\mathrm{V}_{3}$

$=\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}+\frac{4}{3} \pi r_{3}^{3}=\frac{4}{3} \pi\left[r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right]$

$=\frac{4}{3} \times \frac{22}{7} \times\left[6^{3}+8^{3}+10^{3}\right] \mathrm{cm}^{3}$

$=\frac{\mathbf{4}}{\mathbf{3}} \times \frac{\mathbf{2 2}}{\mathbf{7}} \times[216+512+1000] \mathrm{cm}^{3}$

$=\frac{\mathbf{4}}{\mathbf{3}} \times \frac{\mathbf{2 2}}{\mathbf{7}} \times[1728] \mathrm{cm}^{3}$

Let the radius of the new big sphere be R.

$\therefore$ Volume of the new sphere $=\frac{\mathbf{4}}{\mathbf{3}} \times \pi \times \mathbf{R}^{\mathbf{3}}$

Since, the two volume must be equal.

$\therefore \frac{\mathbf{4}}{\mathbf{3}} \times \frac{\mathbf{2 2}}{\mathbf{7}} \times \mathrm{R}^{3}=\frac{\mathbf{4}}{\mathbf{3}} \times \frac{\mathbf{2 2}}{\mathbf{7}} \times 1728 \mathrm{~cm}^{3}$

$\Rightarrow \mathrm{R}^{3}=1728 \Rightarrow \mathrm{R}=12 \mathrm{~cm}$

Thus, the required radius of the resulting sphere $=12 \mathrm{~cm}$.