$A$ big liquid drop splits into $n$ similar small drops under isothermal conditions, then in this process

$A$ soap bubble of radius $1.0 \text{ cm}$ is formed inside another soap bubble of radius $2.0 \text{ cm}$. The radius of another soap bubble which has the same pressure difference as that between the inside of the smaller and outside of the large soap bubble, in metres is:

An air bubble doubles its radius on rising from the bottom of a water reservoir to the surface. If the atmospheric pressure is equal to $10 \, m$ of water,the height of the water in the reservoir is ..... $m$.

The pressure inside a soap bubble $A$ is $1.01 \text{ atm}$ and that in a soap bubble $B$ is $1.02 \text{ atm}$. The ratio of the volume of $A$ to that of $B$ is

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

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 $r$.