Correct Answer: 2.5 cm

Solution : Given:
A spherical ball of lead, 3 cm in diameter, is melted and recast into three spherical balls.
Radius = $\frac{3}{2}$ cm
So, the volume of the sphere = $\frac{4}{3}\pi r^3=\frac{4}{3}\pi (\frac{3}{2})^3$ cm ³
The diameters of two of the new balls are $\frac{3}{2}$ cm and 2 cm, respectively.
Radii of these two balls are $\frac{\frac{3}{2}}{2},\frac{2}{2}$ respectively.
Let the radius of the third ball be $a$.
So, $\frac{4}{3}\pi (\frac{3}{2})^3=\frac{4}{3}\pi (\frac{\frac{3}{2}}{2})^3+\frac{4}{3}\pi(\frac{2}{2}^3)+\frac{4}{3}\pi a^3$
⇒ $\frac{4}{3}\pi (\frac{3}{2})^3=\frac{4}{3}\pi(\frac{27}{64}+1+a^3)$
⇒ $\frac{27}{8}=\frac{91}{64}+a^3$
⇒ $a^3=\frac{125}{64}$
⇒ $a=\frac{5}{4}$
So, the diameter of the third ball $= 2a=2×\frac{5}{4}=2.5$ cm
Hence, the correct answer is 2.5 cm.