The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

$A$ water drop of radius '$r$' and volume '$V$' is kept between two identical glass plates such that it forms a thin layer of area '$A$' between the plates. $A$ force '$F$' is applied such that the two plates separate from each other. The surface tension '$T$' of the liquid is:

The excess pressure inside a soap bubble of radius $2 \ cm$ is $50 \ dyne/cm^2$. The surface tension is

The radius of a soap bubble is increased from $R$ to $2R$. The work done in this process in terms of surface tension $S$ is: (in $\pi R^2 S$)

The radius of a soap bubble is $r$ and the surface tension of the soap solution is $S$. The electric potential to which the soap bubble must be raised by charging it so that the pressure inside the bubble becomes equal to the pressure outside the bubble is $(\varepsilon_0 = \text{permittivity of the free space})$

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3} \, N/m$,then the difference of pressure between the inside and outside of the spherical drop is: