A drop of water of volume 0.05cm³ is pressed between two glass plates, as a consequence of which it spreads and occupies an area of 40cm². If the surface tension of water is 70 dyne/cm, then the normal force required to separate out the two glass plates will be in Newton

Given,

${\mathrm{r}}_1=\frac{\mathrm{t}}{2}, {\mathrm{r}}_2=\infty$

There is a difference in pressure between within and outside the film.

$\mathrm{P}=\mathrm{T}\left[\frac{1}{{\mathrm{r}}_1}+\frac{1}{{\mathrm{r}}_2}\right]$

therefore the amount of force needed to separate the two glass plates that the enclosed liquid film is sandwiched between is

$\mathrm{F}=\mathrm{P}\times \mathrm{A}=\frac{2\mathrm{AT}}{\mathrm{t}}$

$\mathrm{r}=\frac{\mathrm{t}}{2}$

$\mathrm{F}=\frac{2{\mathrm{A}}^2\mathrm{T}}{\mathrm{At}}=\frac{2{\mathrm{A}}^2\mathrm{T}}{\mathrm{V}}$

$\mathrm{F}=\frac{2\times {(40\times {10}^{-4})}^2\times (70\times {10}^{-3})}{0.05\times {10}^{-6}}=45\mathrm{N}$

Hence the correct answer is 45.