A drop of some liquid of volume 0.04 cm^3 is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area 20cm^2 between the surface of the two slides. To separate the slides , force of 16xx10^5 dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyn. cm^-1)

A drop of some liquid of volume 0.04 cm^3 is placed on the surface of

1.	A drop of some liquid of volume 0.04 cm^3 is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area 20cm^2 between the surface of the two slides. To separate the slides , force of 16xx10^5 dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyn. cm^-1)
Answer» 60 70 80 90 Solution :Thickness of layer, `h=("VOLUME")/("area")=(0.04)/(20)=0.002 cm` RADIUS of curvature of concave SURFACE, `r=(h)/(2)=0.001 cm.` Force, `F=T/rxx`area `THEREFORE` Surface tension. `T=(Fxxr)/("area")=(16xx10^5xx0.001)/(20) DYN. cm^-1`