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 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: \(10^{-15} \text{ kg}\).
Important Questions from Numerical Computation
$A$ is an ($n \times n$) matrix. Consider the following two statements
Statement 1: Columns of matrix $A$ are linearly independent
Statement 2: Inverse of matrix $A$ exists
Which one of the following statements is TRUE?
- Consider the function $G(x, y, z) = 0$. This function allows us to implicitly define each of three variables as a function of the other two variables. Assume that all partial derivatives of the function $G(x, y, z)$ exist everywhere. Then, the value of $\left(\frac{\partial z}{\partial x} \times \frac{\partial x}{\partial y} \times \frac{\partial y}{\partial z}\right)$ is _________ (in integer)
- Levenshtein distance is used to measure the minimum edit distance between two strings by counting the minimum number of editing operations (such as, insertions, deletions, substitutions) required to transform one string to another. Consider an alternative version of the Levenshtein distance in which each insertion and each deletion has a cost of 1 and substitutions are not allowed.
Based on this alternative version, the Levenshtein distance between the strings, word and work is ______ (Answer in integer). - In a given collection of documents, let $N$ be the total number of documents and let $d$ be the number of documents in which the term $t$ occurs.
Which ONE of the following fractions is used to define the inverse document frequency (idf) of the term $t$? - The total number of unique character bigrams in the string thoughtsoftheghosts (which contains 19 characters) is ______.
