Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).

The radius of one soap bubble is four times that of another. What is the ratio of the excess pressure inside the two bubbles?

There is an air bubble of radius $1.0\,mm$ in a liquid of surface tension $0.075\,N\,m^{-1}$ and density $1000\,kg\,m^{-3}$ at a depth of $10\,cm$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is $....Pa \left( g = 10\,m\,s^{-2} \right)$

The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal conditions,they meet together in a vacuum. Then the radius of the resultant bubble is given by

An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true?

Two soap bubbles of radius $2 \ cm$ and $4 \ cm$,respectively,are in contact with each other. The radius of curvature of the common surface,in $cm$,is . . . . . . .