A solid steel ball of diameter $3.6\ \text{mm}$ acquired a terminal velocity of $2.45\times10^{-2}\ \text{m/s}$ while falling under gravity through an oil of density $925\ \text{kg m}^{-3}$. Take the density of steel as $7825\ \text{kg m}^{-3}$ and $g$ as $9.8\ \text{m/s}^2$. The viscosity of the oil in SI units is:
Answer: (D) $1.99$
At terminal velocity, weight = buoyancy + Stokes drag, which gives
$$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta} \Rightarrow \eta = \frac{2r^2(\rho - \sigma)g}{9v_t}$$
With $r = 1.8\times10^{-3}\ \text{m}$:
$$\eta = \frac{2\times(1.8\times10^{-3})^2\times(7825 - 925)\times9.8}{9\times2.45\times10^{-2}} = \frac{2\times3.24\times10^{-6}\times6900\times9.8}{0.2205} \approx 1.99\ \text{Pa s}$$
Solution by Sreeraj P, M.Sc Physics