Volume of the drop =0.24mm³
Area of the patch =4.2 x 10^4mm^2
v= Ah
h = V/A =0.24/(4.2×?10?^4 )
h=5.714×?10?^(-6)mm
Thickness= 5.714 x 10^-9 m.
Githiari answered the question on September 16, 2017 at 07:27
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