A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical balls.
The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third
ball is

Let radius of third ball be $$r$$ cm

Volume of a sphere = $$\frac{4}{3}\pi R^3$$

=> $$[\frac{4}{3}\times\pi\times(\frac{3}{4})^3]+[\frac{4}{3}\times\pi\times(1)^3]+[\frac{4}{3}\times\pi\times r^3]=\frac{4}{3}\times\pi\times(\frac{3}{2})^3$$

=> $$\frac{27}{64}+1+r^3=\frac{27}{8}$$

=> $$r^3=\frac{27}{8}-\frac{27}{64}-1$$

=> $$r^3=\frac{216-27-64}{64}$$

=> $$r=\sqrt[3]{\frac{125}{64}}=\frac{5}{4}$$

$$\therefore$$ Diameter of third ball = $$2\times\frac{5}{4}=2.5$$ cm

=> Ans - (A)