91Ó°ÊÓ

The problem involves rod-like particles of length \( L = 0.5 \, \mu \mathrm{m} \) and diameter \( d = 1 \, \mathrm{nm} \). The viscosity of water is given as \( 1 \, \mathrm{mPa} \cdot \mathrm{s} \). We need to estimate the viscosity of this solution for volume fractions \( \phi = 0.01\%, 0.1\%, 1\% \), and \( 10\% \).

The viscosity of a suspension \( \eta \) can be approximately calculated using the Einstein equation for dilute suspensions: \[ \eta = \eta_0 (1 + 2.5 \phi) \] where \( \eta_0 \) is the viscosity of the pure fluid (water in this case), and \( \phi \) is the volume fraction of the particles.

Convert the volume fractions from percentages to decimal form: \( \phi = 0.01\% = 0.0001, \phi = 0.1\% = 0.001, \phi = 1\% = 0.01 \), \( \phi = 10\% = 0.1 \).

Using the formula from Step 2, calculate the solution viscosity for each volume fraction.For \( \phi = 0.0001 \): \[ \eta_{0.0001} = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times (1 + 2.5 \times 0.0001) = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times 1.00025 \approx 1.00025 \, \mathrm{mPa} \cdot \mathrm{s} \]For \( \phi = 0.001 \): \[ \eta_{0.001} = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times (1 + 2.5 \times 0.001) = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times 1.0025 \approx 1.0025 \, \mathrm{mPa} \cdot \mathrm{s} \]For \( \phi = 0.01 \): \[ \eta_{0.01} = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times (1 + 2.5 \times 0.01) = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times 1.025 \approx 1.025 \, \mathrm{mPa} \cdot \mathrm{s} \]For \( \phi = 0.1 \): \[ \eta_{0.1} = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times (1 + 2.5 \times 0.1) = 1 \, \mathrm{mPa} \cdot \mathrm{s} \times 1.25 \approx 1.25 \, \mathrm{mPa} \cdot \mathrm{s} \]