Two soap bubbles of radii $x$ and $y$ coalesce to form a single bubble of radius $z$. Then $z$ is equal to

What is the excess pressure inside a bubble of soap solution of radius $5.00 \; mm$,given that the surface tension of soap solution at the temperature $(20 \; ^{\circ}C)$ is $2.50 \times 10^{-2} \; N m^{-1}$? If an air bubble of the same dimension were formed at a depth of $40.0 \; cm$ inside a container containing the soap solution (of relative density $1.20$),what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \; Pa$).

$A$ 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 $N$)

An air bubble doubles in radius when it rises from the bottom of the sea to the surface. If the atmospheric pressure is equal to the pressure exerted by a $10 \, m$ column of water,then the depth of the sea is $... \, m$. (Assume surface tension is negligible.)

Let $n$ be the number of liquid drops,each with surface energy $E$. These drops join to form a single drop. In this process:

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is