A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s}$ while falling under gravity through an oil of density $925 \mathrm{~kg} \mathrm{~m}^{-3}$. Take density of steel as $7825 \mathrm{~kg} \mathrm{~m}^{-3}$ and g as $9.8 \mathrm{~m} / \mathrm{s}^2$. The viscosity of the oil in SI unit is

<p>To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid:</p> <p>$ v_T = \frac{2}{9} \frac{(\rho_s - \rho_f) r^2 g}{\eta} $</p> <p>Where:</p> <p><p>$ v_T $ is the terminal velocity,</p></p> <p><p>$ \rho_s $ is the density of the steel ball,</p></p> <p><p>$ \rho_f $ is the density of the fluid,</p></p> <p><p>$ r $ is the radius of the ball,</p></p> <p><p>$ g $ is the acceleration due to gravity,</p></p> <p><p>$ \eta $ is the viscosity of the fluid.</p></p> <p>Given:</p> <p><p>Diameter of the steel ball = 3.6 mm, therefore radius $ r = 1.8 $ mm = $ 1.8 \times 10^{-3} $ m,</p></p> <p><p>Terminal velocity $ v_T = 2.45 \times 10^{-2} $ m/s,</p></p> <p><p>Density of oil $ \rho_f = 925 $ kg/m$^3$,</p></p> <p><p>Density of steel $ \rho_s = 7825 $ kg/m$^3$,</p></p> <p><p>Acceleration due to gravity $ g = 9.8 $ m/s$^2$.</p></p> <p>Rearranging the formula to solve for $\eta$:</p> <p>$ \eta = \frac{2}{9} \cdot \frac{(\rho_s - \rho_f) \cdot r^2 \cdot g}{v_T} $</p> <p>Substituting the known values into the equation:</p> <p>$ \eta = \frac{2}{9} \cdot \frac{(7825 - 925) \cdot (1.8 \times 10^{-3})^2 \cdot 9.8}{2.45 \times 10^{-2}} $</p> <p>Calculating this, we approximate the viscosity $\eta$ to be:</p> <p>$ \eta \approx 1.99 \, \text{Pa}\cdot\text{s} $</p> <p>Thus, the viscosity of the oil is approximately 1.99 Pa·s.</p>