Show on a diagram the forces acting on a sphere moving in a viscous liquid, and calculate the volume of an oil drop given its diameter.

Diameter of drop, $D_{drop} = 0.05 \text{ cm} = 0.05 \times 10^{-2} \text{ m} = 5 \times 10^{-4} \text{ m}$  
Radius of drop, $r_{drop} = \frac{D_{drop}}{2} = \frac{5 \times 10^{-4} \text{ m}}{2} = 2.5 \times 10^{-4} \text{ m}$

$$
V_{drop} = \frac{4}{3}\pi (2.5 \times 10^{-4} \text{ m})^3
$$
$$
V_{drop} = \frac{4}{3}\pi (15.625 \times 10^{-12} \text{ m}^3)
$$
$$
V_{drop} \approx 6.54 \times 10^{-11} \text{ m}^3
$$
The volume of the drop is $\boxed{6.54 \times 10^{-11} \text{ m}^3}$.

Diameter of patch, $D_{patch} = 15 \text{ cm} = 15 \times 10^{-2} \text{ m} = 0.15 \text{ m}$  
Radius of patch, $R_{patch} = \frac{D_{patch}}{2} = \frac{0.15 \text{ m}}{2} = 0.075 \text{ m}$

$$
A_{patch} = \pi (0.075 \text{ m})^2
$$
$$
A_{patch} = \pi (0.005625 \text{ m}^2)
$$
$$
A_{patch} \approx 1.77 \times 10^{-2} \text{ m}^2
$$
The area of the patch is $\boxed{1.77 \times 10^{-2} \text{ m}^2}$.

$$
h = \frac{V_{drop}}{A_{patch}}
$$
$$
h = \frac{6.54498 \times 10^{-11} \text{ m}^3}{1.76714 \times 10^{-2} \text{ m}^2}
$$
$$
h \approx 3.70 \times 10^{-9} \text{ m}
$$
The size of the oil molecule is $\boxed{3.70 \times 10^{-9} \text{ m}}$.