Question

2000 identical drops of density 2.8 have a mass of 4.2*10^-8 kg. If 80 of the drops form a thin film where thickness is 9.4Pm on water, calculate the area of the film over water

Answer 1

$$V=N\frac{m}{\rho}=Ah\\2000\frac{(4.2\cdot10^{-8})}{2800}=(9.4\cdot10^{-12})A\\A=3200\ m^2$$