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

Solution(By Examveda Team)

Let the radius of the third ball be R cm
Then,
$$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$   $$ + \frac{4}{3}\pi \times {R^3}$$   $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$
$$\eqalign{ & \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr & \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow R = \frac{5}{4} \cr} $$
∴ Diameter of the third ball :
$$ = 2R = \frac{5}{2}cm = 2.5\,cm$$