A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls.

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 3/2cm and 2 cm, find the diameter of the third ball.

Solution:

Volume of lead ball $=4 / 3 \pi r^{3}$

$=4 / 3 \times 22 / 7 \times(3 / 2)^{3}$

Diameter of first ball d1 = 3/2 cm

Radius of first ball $r_{1}=\frac{\frac{3}{2}}{2}=\frac{3}{4} \mathrm{~cm}$

Diameter of second ball d2 = 2 cm

Radius of second ball r2 = 2/2cm = 1 cm

Diameter of third ball d3 = d

Radius of third ball r3 = d/2 cm

Volume of lead ball $=\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}+\frac{4}{3} \pi r_{3}^{3}$

Volume of lead ball $=\frac{4}{3} \times \pi \times\left(\frac{3}{4}\right)^{3}+\frac{4}{3} \times \pi \times\left(\frac{2}{2}\right)^{3}+\frac{4}{3} \times \pi \times\left(\frac{\mathrm{d}}{2}\right)^{3}$

$\frac{4}{3} \times \frac{22}{7} \times\left(\frac{3}{2}\right)^{3}=\frac{4}{3} \times \pi \times\left(\frac{3}{4}\right)^{3}+\frac{4}{3} \times \pi \times\left(\frac{2}{2}\right)^{3}+\frac{4}{3} \times \pi \times\left(\frac{d}{2}\right)^{3}$

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

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

$\mathrm{d}^{3}=8\left[\frac{27}{8}-\frac{27}{64}-1\right]$

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

$\frac{d}{2}=\frac{5}{4}$

$\mathrm{d}=\frac{10}{4}$

d = 2.5 cm

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