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

What is the excess pressure inside a bubble of soap solution of radius $5.00 \;mm$, given that the surface tension of soap solution at the temperature ($20\,^{\circ} C$) is $2.50 \times 10^{-2}\; N m ^{-1}$ ? If an air bubble of the same dimension were formed at depth of $40.0 \;cm$ inside a container containing the soap solution (of relative density $1.20$), what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \;Pa$ ).

A soap bubble in vacuum has a radius of $3 \,cm$ and another soap bubble in vacuum has a radius of $4 \,cm$. If the two bubbles coalesce under isothermal condition, then the radius of the new bubble is ....... $cm$

A bubble of $8$ mm diameter is formed in the air. The surface tension of soap solution is $30$ dynes/cm. The excess pressure inside the bubble is ........ $dynes/cm^2$

A spherical drop of water has radius $1\, mm$ If surface tension of water is $70 \times {10^{ - 3}}\,N/m$ difference of pressures between inside and out side of the spherical drop is ........ $N/{m^{ - 2}}$

If a section of soap bubble (of radius $R$) through its center is considered, then force on one half due to surface tension is