Two bubbles $A$ and $B$ $(r_A > r_B)$ are joined through a narrow tube. Then

If the radius of a soap bubble is four times that of another,then the ratio of their excess pressures will be

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

The force required to separate two glass plates of area $10^{-2} \, m^2$ with a film of water $0.05 \, mm$ thick between them is ...... $N$. (Surface tension of water is $70 \times 10^{-3} \, N/m$)

The excess pressure inside the first soap bubble of radius $R_1$ is three times that inside the second soap bubble of radius $R_2$. The ratio of volumes of the first to second bubble is

$A$ spherical drop of oil of radius $1\, cm$ is broken into $1000$ droplets of equal radii. If the surface tension of oil is $50\, dynes/cm$,the work done is