Correct Answer - Option 2 : 0.125
Given that :
Radial clearance, λ = 50 μm
Load, F = 2 kN, N = 20 rps
diameter, d = 50 mm
viscosity, μ = 20 mPa-s
length, L = 50 mm
\(P = \frac{F}{A} = \frac{F}{{L \times d}} \\= \frac{{2000}}{{50 \times 50}} = 0.8\;\frac{N}{{m{m^2}}}\)
sommerfield number
\(Z = \frac{{\mu N}}{P}{\left( {\frac{r}{\lambda}} \right)^2}\)
\(= \frac{{\left( {20 \times {{10}^{ - 3}}} \right) \times 20}}{{\left( {0.8 \times {{10}^6}} \right)}} \times {\left( {\frac{{25\times10^{-3}}}{{50\times10^{-6}}}} \right)^2}\)
= 0.125