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))`

Options

A. 60
B. 70
C. 80
D. 90

Answer

C

Explanation

Let thickness of layer be x.
So, volume V = Area `xx` thickness
`V = A xx x rArr x = V//A (because x = 2r)`
`:. 2r = (V)/(A) rArr r = (V)/(2A)` …(i)
and `Delta p = (S)/(r)`
We know that, `F = Delta p xx A = (S)/(r) xx A`
`F = (S)/((V)/(2A)) xx A` [from Eq. (i)]
`S = (F xx V)/(2A^(2))`
Given, `F = 16 xx 10^(5)` dyne, `V = 0.04 cm^(3), A = 20 cm^(2)`
`S = (16 xx 10^(5) xx 0.04)/(2 xx 20^(2)) = (8 xx 10^(5) xx 4)/(20^(2) xx 100)`
`= 8 xx 10^(5) xx 10^(-4) = 80 "dyne " cm^(-1)`