Diameter of oil drop, d_drop = 0.05 cm

Here are the calculations for the oil drop experiment: Given: Diameter of oil drop, $d_{drop} = 0.05 \text{ cm}$ Diameter of circular patch, $d_{patch} = 20 \text{ cm}$ b) (i) The volume of the oil drop Step 1: Calculate the radius of the oil drop. $$r_{drop} = \frac{d_{drop}}{2}$$ $$r_{drop} = \frac{0.05 \text{ cm}}{2}$$ $$r_{drop} = 0.025 \text{ cm}$$ Step 2: Calculate the volume of the oil drop using the formula for the volume of a sphere. $$V_{drop} = \frac{4}{3} \pi r_{drop}^3$$ $$V_{drop} = \frac{4}{3} \pi (0.025 \text{ cm})^3$$ $$V_{drop} = \frac{4}{3} \pi (0.000015625 \text{ cm}^3)$$ $$V_{drop} \approx 6.54 \times 10^{-5} \text{ cm}^3$$ The volume of the oil drop is $\boxed{\text{6.54 \times 10^{-5} cm}^3}$. b) (ii) The area of the patch covered by the oil Step 1: Calculate the radius of the circular patch. $$r_{patch} = \frac{d_{patch}}{2}$$ $$r_{patch} = \frac{20 \text{ cm}}{2}$$ $$r_{patch} = 10 \text{ cm}$$ Step 2: Calculate the area of the patch using the formula for the area of a circle. $$A_{patch} = \pi r_{patch}^2$$ $$A_{patch} = \pi (10 \text{ cm})^2$$ $$A_{patch} = 100 \pi \text{ cm}^2$$ $$A_{patch} \approx 314.16 \text{ cm}^2$$ The area of the patch covered by the oil is $\boxed{\text{314.16 cm}^2}$. That's 2 down. 3 left today — send the next one.