The surface tension of soap solution is $25 \times {10^{ - 3}}\,N{m^{ - 1}}$. The excess pressure inside a soap bubble of diameter $1 \,cm$ is ....... $Pa$

The pressure inside a small air bubble of radius $0.1\, mm$ situated just below the surface of water will be equal to   [Take surface tension of water $70 \times {10^{ - 3}}N{m^{ - 1}}$ and atmospheric pressure = $1.013 \times {10^5}N{m^{ - 2}}$]

A drop of water volume $0.05\ cm^3$ is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is $40\ cm^2$ . If the surface tension of water is $70 \ dyne/cm$ , the minimum normal force required to seperate out the two glass plate in  newton is approximately...... $N$ (assuming angle of contact is zero) 

When two soap bubbles of radii $a$ and $b ( b > a )$ coalesce, the radius of curvature of common surface is

A drop of water of volume $0.05\, cm^3$ is pressed between two glass plates, as a consequence of which it spreads and occupies an area of $40\, cm^2$. 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

Write the equation of excess pressure for liquid drop.