A drop of some liquid of volume `0.04 cm^(3)` is placed on the surface of a glass slide. Then, another glass forms a thin layer of area `20 cm^(2)` between the surfaces of the two slides. To separate the slides a force of `16 xx 10^(5)` dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne `cm^(-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 20 cm^2 between the surfaces of the two slides. To separate the slides a force of 16 xx 10^5 dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne- cm^(-1) )

The surface tension of a liquid is 10^(8) dyne cm^(-1) . It is equivalent to

The surface tension of a liquid is 10^(8) dyne cm^(-1) . It is equivalent to

The surface tension of a liquid is 10^(8) dyne/cm .It is equivalent to

. The surface tension of liquid is 70 dynes/cm. It may be expressed in MKS as

IF the surface tension of liquid is 70 dyn/cm, what is the total energy of the free surface of the liquid drop of radius 0.1 cm ?