Question

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.

• A. \(10^{-12}\ \text{kg}\)
• B. \(10^{-18}\ \text{g}\)
• C. \(10^{-15}\ \text{kg}\)
• D. \(10^{-15}\ \text{g}\)

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.

Answer 1

The Correct Option is C

Step 1: Set up the dimensions.
The cylinder has diameter \(1\ \mu m\), so its radius is
\[ r = 0.5\ \mu m = 0.5\times10^{-6}\ m = 5\times10^{-7}\ m \]
Its height is
\[ h = 1\ \mu m = 1\times10^{-6}\ m \]
The two hemispherical caps have the same radius \(r\) as the cylinder, and together they make up exactly one full sphere of radius \(r\).

Step 2: Find the volume of the cylindrical part.
\[ V_{cyl} = \pi r^2 h = \pi (5\times10^{-7})^2 (1\times10^{-6}) \]
\[ = \pi (2.5\times10^{-13})(1\times10^{-6}) = \pi \times 2.5\times10^{-19}\ m^3 \]
\[ \approx 7.85\times10^{-19}\ m^3 \]

Step 3: Find the volume of the two hemispherical caps together, one full sphere.
\[ V_{sph} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (5\times10^{-7})^3 \]
\[ = \frac{4}{3}\pi (1.25\times10^{-19}) \approx 5.24\times10^{-19}\ m^3 \]

Step 4: Add the two volumes.
\[ V_{total} = V_{cyl}+V_{sph} \approx 7.85\times10^{-19}+5.24\times10^{-19} = 1.31\times10^{-18}\ m^3 \]

Step 5: Convert volume to mass using the density of water.
\[ m = \rho \times V = (10^3\ kg/m^3)(1.31\times10^{-18}\ m^3) \]
\[ m \approx 1.31\times10^{-15}\ kg \]

Step 6: Match to the closest option.
The calculated mass, about \(1.3\times10^{-15}\ kg\), is closest in order of magnitude to \(10^{-15}\ kg\). The other options are off, either by many orders of magnitude, or use the wrong unit, grams instead of kilograms, for that same order of magnitude.

Step 7: Final answer.
\[ \boxed{10^{-15}\ kg} \]