A liquid drop has principal radii of curvature R1 = 1 m and R2 = 2 m in two mutually perpendicular directions. Taking the surface tension S = 0.07 N/m, the excess pressure inside the drop (in N/m²) is:

0.035
0.0525
0.105
0.07

Correct answer: 0.105

Solution

For a liquid surface with two principal radii, the Young-Laplace excess pressure is S*(1/R1 + 1/R2) = 0.07*(1/1 + 1/2) = 0.07*1.5 = 0.105 N/m².

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