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

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 $= 1 \,atm$).

The excess pressure inside a soap bubble $A$ in air is half the excess pressure inside another soap bubble $B$ in air. If the volume of the bubble $A$ is $n$ times the volume of the bubble $B$,then the value of $n$ is . . . . . . .

An air bubble rises from the bottom to the top of a water tank in which the temperature of the water is uniform. The surface area of the bubble at the top of the tank is $125 \%$ more than its surface area at the bottom of the tank. If the atmospheric pressure is equal to the pressure of $10 \ m$ water column, then the depth of water in the tank is (in $m$)

$A$ spherical drop of radius $r$ is divided into $8$ equal droplets. If the surface tension is $S$, then the work done in the process will be

$A$ 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