In an assembly of fissionable material. The larger the surface area per fissioning nucleus (i.e. per unit volume), the more likely is the escape of valuable neutrons.

(a) What is the surface-to-volume ratio of a sphere of radius r?

(b) What is the surface-to-volume ratio of a cube of the same volume?

(c) What is the surface-to-volume ratio of a sphere of twice the volume?

We need to find the surface-to-volume ratio of a sphere of radius after that, the surface-to-volume ratio of a cube of the same volume, and finally, the surface-to-volume ratio of a sphere of twice the volume. Surface area to volume ratio can be found easily for several simple shapes, like, for example, a cube or a sphere:

(a)

For a sphere, the equation for the surface area is $S=4\pi r_0^2$where ${\text{r}}_0$is the radius of the sphere. The volume of a sphere is $V=\frac{4}{3}\pi r_0^3$.

So for a cube, the ratio of surface area to volume is $\frac{S}{V}$.

Therefore,

.$\begin{array}{rcl}\frac{S}{V}& =& \frac{4\pi r_0^{}}{\frac{4}{3}\pi r_0^3}\\ & =& \frac{3}{r_0}\\ & & \end{array}$

Thus, the surface-to-volume ratio of the sphere is$\frac{3}{r_0}$ .

(b)

For a cube, the equation for the surface area is$s=6l^2$ where l is the length of a side. Similarly, the volume of a cube is$V=l^3$ .

So for a cube, the ratio of the surface area to volume is$\frac{S}{V}$ .

Therefore,

$\begin{array}{rcl}\frac{S}{V}& =& \frac{6l^2}{l^3}\\ & =& \frac{6}{l}\\ & & \end{array}$

Thus, the surface-to-volume ratio of the cube is$\frac{6}{l}$

(c)

For a sphere, the surface area is$S=4\pi r_0^2$where is the radius of the sphere and pi is 3.14.

The volume of the sphere is $V\text{'}=2\frac{4}{3}\pi r_0^3$. Since the sphere is twice the volume, we have 2V. the radius$r\text{'}={\left(2\right)}^{1/3}{r}_0$

So, for a sphere, the ratio of surface area to volume is $\frac{S}{V\text{'}}$.

Therefore,

$\begin{array}{rcl}\frac{S}{V\text{'}}& =& \frac{4\pi r_0^2}{\frac{4}{3}\pi {r\text{'}}^3}\\ & =& \frac{3}{{\left(2\right)}^{1/3}{r}_0}\\ & =& \frac{2.38}{r_0}\\ & & \end{array}$

Thus, the surface-to-volume ratio of a sphere of twice the volume is $\frac{2.38}{r_0}$.