(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
- An organization allows its employees to work independently on consultancy projects but charges an overhead on the consulting fee. The overhead is 20% of the consulting fee, if the fee is up to . 5,00,000. For higher fees, the overhead is . 1,00,000 plus 10% of the amount by which the fee exceeds . 5,00,000. The government charges a Goods and Services Tax of 18% on the total amount (the consulting fee plus the overhead). An employee of the organization charges this entire amount, i.e., the consulting fee, overhead, and tax, to the client. If the client cannot pay more than . 10,00,000, what is the maximum consulting fee that the employee can charge?
Three frictionless pulleys with rope attachment are in a static equilibrium as shown in the figure. The mass $m_1$ and $m_2$, in kg, respectively are
If the in-situ density of coal is 1320 kg/m$^3$ and the density of blasted coal is 952 kg/m$^3$, the swell factor is _____________ (rounded off to 3 decimal places)
A five-member truss system is shown in the figure. The maximum vertical force P in kN that can be applied so that loads on the member CD and BC do NOT exceed 50 kN and 30 kN, respectively is _____________(rounded off to 2 decimal places)
The Fourier transform and its inverse transform are respectively defined as $\tilde{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(x)e^{i\omega x}dx$ and $f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \tilde{f}(\omega)e^{-i\omega x}d\omega$. Consider two functions $f$ and $g$. Another function $f * g$ is defined as
$(f * g)(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(y)g(x - y)dy$
Which of the following relation is/are true?
Note: Tilde ($\sim$) denotes the Fourier transform.
