Right option is (b) 1.29 × 10^-6 m^2s^-1
Easiest explanation: Damage due to eddies is avoided if the \(K \ddot{o}\) 1mogorov scale remains greater than 2/3-1/2 the diameter of the beads. Let us determine the stirrer speed required to create eddies with size, λ = 2/3 (120 μm) = 80 μm = 8 x 10^-5 rn. The stirrer power producing eddies of this dimension can be estimated:
Kinematic viscosity, v = \(\frac{\mu}{\rho} = \frac{1.3×10^{-3} kg \,m^{-1} s^{-1}}{1010 \,kgm^{-3}}\) = 1.29 × 10^-6 m^2s^-1.