A spherical ball of radius 3 cm is melted and recast into three spherical balls.
Question:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.

Solution:

The radius of the big spherical ball is 3cm. Therefore, the volume of the big spherical ball is

$V=\frac{4}{3} \pi \times(3)^{3}$ cubic $\mathrm{cm}$

The radii of the 1st and 2nd small spherical balls are 1.5 cm and 2 cm respectively. Therefore, the volumes of the 1st and 2nd spherical balls are respectively

$V_{1}=\frac{4}{3} \pi \times(1.5)^{3}$ cubic $\mathrm{cm}$,

$V_{2}=\frac{4}{3} \pi \times(2)^{3}$ cubic cm

$V_{2}=\frac{4}{3} \pi \times(2)^{3}$ cubic $\mathrm{cm}$

Let, the radius of the 3rd small spherical ball is cm. Then, its volume is

$V_{3}=\frac{4}{3} \pi \times(r)^{3}$ Cubic $\mathrm{cm}$

Since, the big spherical ball is melted to produce the three small spherical balls; the volume of the big spherical ball is same as the sum of the volumes of the three small spherical balls. Therefore, we have

$V=V_{1}+V_{2}+V_{3}$

$\Rightarrow \frac{4}{3} \pi \times(3)^{3}=\frac{4}{3} \pi \times(1.5)^{3}+\frac{4}{3} \pi \times(2)^{3}+\frac{4}{3} \pi \times(r)^{3}$

$\Rightarrow \quad(3)^{3}=(1.5)^{3}+(2)^{3}+(r)^{3}$

$\Rightarrow \quad(r)^{3}=(3)^{3}-(1.5)^{3}-(2)^{3}$

$\Rightarrow \quad r^{3}=27-3.375-8$

$\Rightarrow \quad=15.625$

$\Rightarrow \quad=2.5$

Therefore, the diameter of the $3^{\text {rd }}$ ball is $2 r=5 \mathrm{~cm}$

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