Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
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 $=12 \mathrm{~cm}$.
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