(f) (i) What is meant by Stokes law? (1 mark) (ii) An iron sphere of radius 0.3 cm drops in oil at 0.5cm/s . If the densities of oil and iron are 0.94g/cm^3 and 7.8g/cm^3 respectively , find the viscosity of the oil. (3 marks)

(i) Stokes' law, also known as Stokes' equation, describes the drag force experienced by a small spherical object moving through a viscous fluid. It states that the drag force is directly proportional to the radius of the sphere, the velocity of the sphere, and the viscosity of the fluid, and inversely proportional to the radius of the sphere.<br /><br />(ii) To find the viscosity of the oil, we can use Stokes' law:<br /><br />\[ F = 6 \pi \eta r v \]<br /><br />Where:<br />- \( F \) is the drag force<br />- \( \eta \) is the viscosity of the oil<br />- \( r \) is the radius of the iron sphere<br />- \( v \) is the velocity of the sphere<br /><br />Given:<br />- Radius of the iron sphere, \( r = 0.3 \, \text{cm} = 0.003 \, \text{m} \)<br />- Velocity of the sphere, \( v = 0.5 \, \text{cm/s} = 0.005 \, \text{m/s} \)<br />- Density of iron, \( \rho_{\text{iron}} = 7.8 \, \text{g/cm}^3 = 7800 \, \text{kg/m}^3 \)<br />- Density of oil, \( \rho_{\text{oil}} = 0.94 \, \text{g/cm}^3 = 940 \, \text{kg/m}^3 \)<br /><br />First, we need to calculate the buoyant force acting on the sphere:<br /><br />\[ F_{\text{buoyant}} = \rho_{\text{oil}} \cdot g \cdot V_{\text{sphere}} \]<br /><br />Where:<br />- \( V_{\text{sphere}} = \frac{4}{3} \pi r^3 \)<br /><br />\[ V_{\text{sphere}} = \frac{4}{3} \pi (0.003 \, \text{m})^3 = 3.14 \times 10^{-11} \, \text{m}^3 \]<br /><br />\[ F_{\text{buoyant}} = 940 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \cdot 3.14 \times 10^{-11} \, \text{m}^3 = 2.92 \times 10^{-3} \, \text{N} \]<br /><br />Next, we calculate the drag force using Stokes' law:<br /><br />\[ F = 6 \pi \eta r v \]<br /><br />Since the sphere is in equilibrium, the drag force equals the buoyant force:<br /><br />\[ 2.92 \times 10^{-3} \, \text{N} = 6 \pi \eta \cdot 0.003 \, \text{m} \cdot 0.005 \, \text{m/s} \]<br /><br />Solving for \( \eta \):<br /><br />\[ \eta = \frac{2.92 \times 10^{-3} \, \text{N}}{6 \pi \cdot 0.003 \, \text{m} \cdot 0.005 \, \text{m/s}} = 0.098 \, \text{Pa s} \]<br /><br />Therefore, the viscosity of the oil is \( 0.098 \, \text{Pa s} \).