Question

A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameter of two balls be $\frac{3}{2}$ cm and 2 cm, find the diameter of the third ball.

Answer 1

Let the diameter of the third ball be $d\text{cm}$

Then,

$\frac{4}{3}\pi \times {\left(\frac{3}{2}\right)}^3=\frac{4}{3}\pi \times {\left(\frac{3}{4}\right)}^3+\frac{4}{3}\pi \times {\left(\frac{2}{2}\right)}^3+\frac{4}{3}\pi \times {\left(\frac{d}{2}\right)}^3$

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

$\frac{d^3}{8}=\frac{125}{64}$

$d^3=\frac{125}{8}$

$d=\frac{5}{2}$ $\text{cm}$

$d=2.5$ $\text{cm}$