Question

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?

Short answer

(a)$\frac{3}{r_0}$

(b)$\frac{6}{l}$

(c)$\frac{2.38}{r_0}$

Step-by-step solution

Find the surface-to-volume ratio of the sphere

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$\frac{3}{r_0}$ .

Find the surface-to-volume ratio of the cube

(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$\frac{6}{l}$

Finding the surface-to-volume ratio of a sphere of twice the volume

(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 $\frac{2.38}{r_0}$.