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Question

A bacterium can be approximated as a cylinder with a hemisphere at each end, as shown in the figure. The cylinder has a height of $1 \mu\text{m}$ and diameter of $1 \mu\text{m}$. Assuming that the density of the bacterium is equal to that of water, what is the approximate mass of this bacterium? 

Given: density of water = $10^3 \text{ kg/m}^3$; $1 \mu\text{m} = 10^{-6}\text{ m}$ 

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 = $\frac{4}{3} \pi r^3$, where $r$ is the radius of the sphere

The correct answer is
$10^{-15}\text{ kg}$

The bacterium can be approximated as a cylinder with a hemisphere at each end. We need to calculate the volume of this shape and subsequently find its mass, assuming the density is equal to that of water.

Step 1: Calculate the volume of the cylindrical part.

  • The height of the cylinder, \(h = 1 \, \mu\text{m}\, = 10^{-6} \text{ m}\).
  • The diameter = \(1 \, \mu\text{m}\, = 10^{-6} \text{ m}\), so the radius is \(r = 0.5 \times 10^{-6} \text{ m}\).
  • Volume of the cylinder, \(V_{\text{cylinder}} = \pi r^2 h = \pi (0.5 \times 10^{-6})^2 \times 10^{-6} = 0.25 \pi \times 10^{-18} \text{ m}^3\).

Step 2: Calculate the volume of the hemispherical parts.

  • Volume of a full sphere is \(V_{\text{sphere}} = \frac{4}{3} \pi r^3\).
  • Volume of one hemisphere is half of this: \(V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3\).
  • Total volume of both hemispheres: \(V_{\text{hemispheres}} = 2 \times \frac{2}{3} \pi (0.5 \times 10^{-6})^3 = \frac{4}{3} \pi \times 0.125 \times 10^{-18} \text{ m}^3 = \frac{\pi}{6} \times 10^{-18} \text{ m}^3\).

Step 3: Calculate the total volume of the bacterium.

  • Total volume, \(V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemispheres}} = 0.25 \pi \times 10^{-18} + \frac{\pi}{6} \times 10^{-18} = \frac{\pi}{3} \times 10^{-18} \text{ m}^3\).

Step 4: Calculate the mass of the bacterium.

  • Use the formula \(\text{mass} = \text{density} \times \text{volume}\).
  • \(\text{density} = 10^3 \text{ kg/m}^3\).
  • Mass = \(10^3 \times \frac{\pi}{3} \times 10^{-18} \text{ kg} = \frac{\pi}{3} \times 10^{-15} \text{ kg}\).
  • Approximating \(\pi \approx 3.14\), the mass is around \(10^{-15} \text{ kg}\).

Conclusion: The approximate mass of the bacterium is \(10^{-15} \text{ kg}\). Thus, the correct answer is: \(10^{-15} \text{ kg}\).

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