Question

Volume of one drop = 3.0*10-5mm3
Diameter of oil patch=0.2m.
Determine the size of the molecule
Describe how the same information can be used to determine the extent of spillage of the know volume of the same oil at the sea?

Answer 1

Volume of drop is given by $\dfrac{4}{3}\pi r^3$ where r is the radius of the drop

$3*10^{-5}=\dfrac{4}{3}\pi r^3 \newline \dfrac{90}{4\pi}*10^{-6}=r^3$

$r=1.92mm$

$diameter = 2*1.92=3.84 mm$ =$3.84*10^{-3}m$

area covered by this oil patch = $(\dfrac{diameter \ of \ oil\ patch}{diameter\ of \ oil \ drop })^2$ = $({\dfrac{0.2}{3.84*10^{-3}}})^2$ = $(\dfrac{200}{3.84})^2$ =$(78.125m)^2=6104m^2$

the minimum volume of the oil covering this surface can be calculated in meters:

$6104*10^{-6}m^3=0.61*10^{-2} m^3$