$A$ container,whose bottom has round holes with diameter $0.1 \ mm$ is filled with water. The maximum height in $cm$ up to which water can be filled without leakage will be ........ $cm$. (Surface tension $= 75 \times 10^{-3} \ N/m$ and $g = 10 \ m/s^2$)

Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops of the same liquid combine, then the excess pressure inside the bigger drop is (in $units$)

Two soap bubbles of radii $3r$ and $4r$ are in contact with each other. The radius of curvature of the interface between the bubbles is:

When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

The excess pressure inside a spherical drop of water $A$ is four times that of another drop $B$. Then the ratio of mass of drop $A$ to that of drop $B$ is

$A$ spherical soap bubble inside an air chamber at pressure $P_0 = 10^5 \ Pa$ has a certain radius $R$ such that the excess pressure inside the bubble is $\Delta P = 144 \ Pa$. Now,the chamber pressure is reduced to $8P_0 / 27$ so that the bubble radius and its excess pressure change. In this process,all temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure $\Delta P$ in both cases to be much smaller than the chamber pressure. The new excess pressure $\Delta P'$ in $Pa$ is: