When a mercury drop of radius $R$ breaks into $n$ droplets of equal size,the radius $r$ of each droplet is

The excess pressure inside a spherical soap bubble of radius $1 \,cm$ is balanced by a column of oil (specific gravity $= 0.8$), $2 \,mm$ high. The surface tension of the bubble is: (in $\,N/m$)

$A$ spherical soap bubble inside an air chamber at pressure $P_0 = 10^5 \ Pa$ has a certain radius $R$ such that the excess pressure inside the bubble is $\Delta P = 144 \ Pa$. Now,the chamber pressure is reduced to $8P_0 / 27$ so that the bubble radius and its excess pressure change. In this process,all temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P'$ in $Pa$ is:

Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \ N/m^2$. The radii of bubbles $A$ and $B$ are $2 \ cm$ and $4 \ cm$,respectively. The surface tension of the soap solution is $0.04 \ N/m$. Find the ratio $n_B / n_A$,where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$,respectively. [Neglect the effect of gravity.]

The excess pressure in a soap bubble is thrice that in another one. Then the ratio of their volume is

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