Two soap bubbles coalesce to form a single bubble. If $V$ is the subsequent change in volume of contained air and $S$ is the change in total surface area, $T$ is the surface tension and $P$ is atmospheric pressure, then which of the following relations is correct?

When a large bubble rises from the bottom of a water lake to its surface,its radius doubles. If the atmospheric pressure is equal to the pressure of a water column of height $H$,then the depth of the lake will be

What should be the diameter of a soap bubble,in order that the excess pressure inside it is $25.6 \ Nm^{-2}$ (in $cm$)? [surface tension of soap solution $= 3.2 \times 10^{-2} \ Nm^{-1}$]

Pressure inside two soap bubbles are $1.02 \ atm$ and $1.05 \ atm$ respectively. The ratio of their surface area is

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 . . . . . . .

$A$ small soap bubble of radius $4 \ cm$ is trapped inside another bubble of radius $6 \ cm$ without any contact. Let $P_2$ be the pressure inside the inner bubble and $P_0$ be the pressure outside the outer bubble. The radius of another bubble with a pressure difference $P_2 - P_0$ between its inside and outside would be....... $cm$.