Pressure inside two soap bubbles is $1.01 \, atm$ and $1.03 \, atm$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \, atm$). (in $ : 1$)

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 $70 \times 10^{-3} \, N/m$)

When two soap bubbles of radii $4 \, cm$ and $5 \, cm$ coalesce,what is the radius of the common interface surface in $cm$?

$A$ capillary type tube $AB$ is connected to a manometer $M$ as shown in the figure. Stopper $S$ controls the flow of air. $AB$ is dipped into a soap solution where surface tension is $T$. On opening the stopper for a while,a bubble is formed at $B$ end of the manometer and the level difference in manometer limbs is $h$. If $\rho$ is the density of the soap solution and $r$ is the radius of curvature of the bubble,then the surface tension $T$ of the liquid is given by:

Two soap bubbles of radii $2 \, cm$ and $4 \, cm$ join to form a double bubble in air. The radius of curvature of the interface is .......... $cm$.

$A$ water drop of $0.01 \ cm^3$ is squeezed between two glass plates and spreads into an area of $10 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,then the normal force required to separate the glass plates from each other will be: (in $N$)