A bacterium can be approximated as a cylinder with a hemisphere capping each end, as shown in the figure. The cylinder has a height of 1 \(\mu m\) and a diameter of 1 \(\mu m\), so the two end hemispheres share the same radius as the cylinder.
Assuming that the density of the bacterium equals that of water, what is the approximate mass of this bacterium?
Given: density of water \(=10^3\ \text{kg/m}^3\); \(1\ \mu m = 10^{-6}\ m\).
Volume of a cylinder \(= \pi r^2 h\), where \(r\) is the radius and \(h\) is the height of the cylinder.
Volume of a sphere \(= \dfrac{4}{3}\pi r^3\), where \(r\) is the radius of the sphere.
Show Hint
The two end hemispheres together make exactly one full sphere of the same radius as the cylinder; add that sphere's volume to the cylinder's volume before converting to mass.
Step 1: Work in centimetres and grams instead of metres and kilograms.
Since $1\ \mu m = 10^{-4}\ cm$, the cylinder's radius is $r = 0.5\times10^{-4}\ cm = 5\times10^{-5}\ cm$, and its height is $h = 10^{-4}\ cm$. Water's density in these units is a clean number, $\rho = 1\ g/cm^3$.
Step 3: Add the two hemispherical caps as one sphere.
\[ V_{sph} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(5\times10^{-5})^3 = \frac{4}{3}\pi(1.25\times10^{-13}) \approx 5.24\times10^{-13}\ cm^3 \]
Step 4: Add the two volumes.
\[ V_{total} \approx 7.85\times10^{-13}+5.24\times10^{-13} = 1.31\times10^{-12}\ cm^3 \]
Step 5: Convert straight to mass, since $\rho=1\ g/cm^3$.
\[ m = \rho V \approx 1.31\times10^{-12}\ g \]
Step 6: Convert grams back to kilograms to compare with the options.
\[ m \approx 1.31\times10^{-12}\ g = 1.31\times10^{-15}\ kg \]
This confirms the same answer reached by working directly in SI units, the choice of unit system does not change the physical mass, only how the number looks partway through.
