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 $\,N/m$)

Pressure inside two soap bubbles are $1.01 \, atm$ and $1.02 \, atm$. The ratio between their volumes is:

The excess pressure due to surface tension in a spherical liquid drop of radius $r$ is directly proportional to

Two bubbles $A$ and $B$ $(r_A > r_B)$ are joined through a narrow tube. Then

What is the pressure inside the drop of mercury of radius $3.00 \; mm$ at room temperature? Surface tension of mercury at that temperature $(20 \; ^{\circ}C)$ is $4.65 \times 10^{-1} \; N m^{-1}$. The atmospheric pressure is $1.01 \times 10^{5} \; Pa$. Also,calculate the excess pressure inside the drop.

Two soap bubbles with radii $r_1$ and $r_2$ $(r_1 > r_2)$ come in contact. Their common surface has a radius of curvature $r$. Find $r$.