(Round off to one decimal place)
To determine the number of polymeric particles in 1 mL of the solution, we need to calculate the volume and mass of a single particle, then use the concentration to find the total mass of particles in 1 mL, and finally divide by the mass of a single particle.
Calculate Particle Volume
The diameter of the particles is given as $20 \, \mu\text{m}$. The radius ($r$) is half the diameter:
$r = \frac{20 \, \mu\text{m}}{2} = 10 \, \mu\text{m}$
Convert the radius to centimeters (cm), knowing that $1 \, \mu\text{m} = 10^{-4} \, \text{cm}$:
$r = 10 \times 10^{-4} \, \text{cm} = 10^{-3} \, \text{cm}$
The volume of a single spherical particle ($V_p$) is calculated using the formula $V = \frac{4}{3} \pi r^3$:
$V_p = \frac{4}{3} \pi (10^{-3} \, \text{cm})^3 = \frac{4}{3} \pi \times 10^{-9} \, \text{cm}^3$
Calculate Particle Mass
The mass density of the polymeric particle ($\rho_p$) is given as $1 \, \text{g/cm}^3$. The mass of a single particle ($m_p$) is:
$m_p = \rho_p \times V_p = (1 \, \text{g/cm}^3) \times (\frac{4}{3} \pi \times 10^{-9} \, \text{cm}^3) = \frac{4}{3} \pi \times 10^{-9} \, \text{g}$
Determine Total Particle Mass in Solution
The concentration is 1% weight/volume (w/v). This means there is 1 gram of solute (polymeric particles) for every 100 mL of solution.
For 1 mL of solution, the total mass of polymeric particles ($m_{total}$) is:
$m_{total} = \frac{1 \, \text{g}}{100 \, \text{mL}} \times 1 \, \text{mL} = 0.01 \, \text{g} = 10^{-2} \, \text{g}
Calculate Number of Particles
The number of particles ($N$) is the total mass of particles divided by the mass of a single particle:
$N = \frac{m_{total}}{m_p} = \frac{10^{-2} \, \text{g}}{\frac{4}{3} \pi \times 10^{-9} \, \text{g}}
Simplify the expression:
$N = \frac{3 \times 10^{-2}}{4 \pi \times 10^{-9}} = \frac{3}{4 \pi} \times 10^7
Now, calculate the numerical value:
$N \approx \frac{3}{4 \times 3.14159} \times 10^7 \approx \frac{3}{12.56636} \times 10^7 \approx 0.2387 \times 10^7
Express the result in the required format ($ \text{______} \times 10^6$):
$N \approx 2.387 \times 10^6
Rounding to one decimal place, the number of particles is $2.4 \times 10^6$.
Important Questions from Numerical Computation
In an engineering college of 10,000 students, 1,500 like neither their core branches nor other branches. The number of students who like their core branches is 1/4th of the number of students who like other branches. The number of students who like both their core and other branches is 500.
The number of students who like their core branches is$A$ 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 $t$?
