Question

A spherical ball of lead $5cm$ in diameter is melted and recast into three spherical balls. The diameters of two of these balls are $2cm$ and $2(14.5{)}^{1/3}cm$. Find the diameter of the third ball.

• A. $8$ cm
• B. $5$ cm
• C. $4$ cm
• D. $1$ cm

Answer 1

The correct option is D $1$ cm

Radius of first spherical ball $=\frac{2}{2}cm=1cm$

Radius of second spherical ball $=\frac{2\sqrt[3]{314.5}}{2}cm=\sqrt[3]{314.5}cm$

Let the radius of the third ball be $r$ cm.

Radius of the main spherical ball $=\frac{5}{2}cm=2.5cm$

Volume of main spherical ball $=$ Volume of spherical ball $1+$Volume of spherical ball $2+$ Volume of spherical ball $3$

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

So, $\frac{4}{3}\pi \times {2.5}^3=\frac{4}{3}\pi 1^3+\frac{4}{3}\pi {\left(\sqrt[3]{314.5}\right)}^3+\frac{4}{3}\pi r^3$

$\Rightarrow {2.5}^3=1^3+{\left(\sqrt[3]{314.5}\right)}^3+r^3$

$\Rightarrow r^3=15.625-1-14.5$

$\Rightarrow r^3=0.125=0.5\times 0.5\times 0.5$

$\Rightarrow r=0.5$

The radius of the third spherical ball is $0.5cm$
Hence, the diameter of the third spherical ball is $2\times 0.5cm=1cm$.