Three small lead spheres of radius 3 cm, 4 cm and 5 cm respectively, are melted into a single sphere. The diameter of the new sphere is

Volume of new single sphere

$=\frac{4}{3} \pi\left(r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right)=\frac{4}{3} \pi\left(3^{3}+4^{3}+5^{3}\right)$ cu. $\mathrm{cm}$

$=\frac{4}{3} \pi(27+64+125)$ cu.cm.

$=\frac{4}{3} \pi \times 216$ cu.cm.

$\therefore \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi \times 216$

where R = radius of new sphere

$\Rightarrow \mathrm{R}^{3}=216$

$\Rightarrow \mathrm{R}=\sqrt[3]{216}=\sqrt[3]{6 \times 6 \times 6}=6 \mathrm{~cm}$

∴ Diameter of new sphere

= 2 × 6 = 12