Question

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

Answer 1

Diameter of spherical ball of lead = 3 cm
$\therefore Radius\left(r\right)=\frac{3}{2}cm\therefore Volume=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times \frac{3}{2}\times \frac{3}{2}\times \frac{3}{2}c{m}^3=\frac{9\pi }{2}c{m}^3$
Diameter of smaller ball $=\frac{3}{2}cm$
$\therefore$ Radius $\left(r_1\right)=\frac{3}{4}$
and volume $=\frac{4}{3}\pi \times \frac{3}{4}\times \frac{3}{4}\times \frac{3}{4}c{m}^3=\frac{9}{16}\pi c{m}^3$
Diameter of second smaller ball = 2 cm
and radius $\left(r_2\right)=\frac{2}{2}=1cm$
$\therefore Volume=\frac{4}{3}\pi \times 1\times 1\times 1=\frac{4}{3}\pi c{m}^3$
$\therefore$ Volume of third smaller ball
$\frac{9}{2}\pi -(\frac{9}{16}\pi +\frac{4}{3}\pi )=(\frac{9}{2}-\frac{9}{16}-\frac{4}{3})\pi c{m}^3=\frac{216-27-64}{48}=\frac{125}{48}\pi c{m}^3$
$\therefore$ Radius of the third ball $={\left(\frac{Volume\times 3}{4\pi }\right)}^{\frac{1}{3}}={\left(\frac{125\pi \times 3}{48\times 4\pi }\right)}^{\frac{1}{3}}={\left(\frac{125}{64}\right)}^{\frac{1}{3}}={\left[{\left(\frac{5}{4}\right)}^3\right]}^{\frac{1}{3}}=\frac{5}{4}=1.25cm$
$\therefore$ Diameter of third ball $=2\times radius$
$=2\times 1.25=2.5cm$