Chapter 18, Problem 50A

Consider eight little spheres of mercury, each with a diameter of 1 millimeter. When they coalesce to form a single sphere, how big will it be? How does its surface area compare to the total surface area of the previous eight little spheres?

To determine

Size of the combined singlesphere and total surface area in compare to all littlesphere.

Answer to Problem 50A

The combined single sphere is twice the diameter of the little sphere and the total surface area is half of the surface area of eight little spheres.

Explanation of Solution

Given:

Number of littlespheres, $n=8$

Diameter of a littlesphere, $d_1=1\text{mm}$

Formula Used:

The volume of sphere is, $V=\frac{4}{3}\pi r^3$

Where, $r$ is radius of sphere.

The total surface area of sphere is, $S=4\pi r^2$

Where, $r$ is radius of sphere.

Calculation:

Volume of a littlesphere, $V_1=\frac{4}{3}\pi r_1{}^3$

And volume of a combined single sphere, $V_2=\frac{4}{3}\pi r_2{}^3$

The volume of the combined singlesphere will be equal to the volume of the all littlespheres.

So,

  $\begin{array}{l}{V}_2=n{V}_1\\ \Rightarrow \frac{4}{3}\pi r_2{}^3=8\times \frac{4}{3}\pi r_1{}^3\\ \Rightarrow r_2=2r_1\\ \text{or }{d}_2=2d_1=2\times 1\text{mm}=2\text{mm}\end{array}$

It means the single sphere will be twice of the diameter of the little sphere.

Now,

The total surface area of littlesphere is,

  $S_1=nS=8\times 4\pi r_1{}^2$

While the total surface area of the combined singlesphere is,

  $S_2=4\pi r_2{}^2$

So, $\begin{array}{l}\frac{S_2}{S_1}=\frac{4\pi r_2{}^2}{8\times 4\pi r_1{}^2}=\frac{1}{8}{\left(2\right)}^2\\ \Rightarrow S_2=\frac{1}{2}{S}_1\end{array}$

Conclusion:

Therefore, the combined single sphere is twice the diameter of the little sphere and the total surface area is half of the eight littlespheres.