The surface tension of a soap bubble is $0.03 \text{ N/m}$. The work done in increasing the diameter of the bubble from $2 \text{ cm}$ to $6 \text{ cm}$ is $\alpha \times 10^{-4} \text{ J}$. The value of $\alpha$ is . . . . . . (Take $\pi = 3.14$).

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

$A$ certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid,which one of the following statements is true for the energy $(E)$ in the process?

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

Two soap bubbles $A$ and $B$ are kept in a closed chamber where the air is maintained at pressure $8 \ N/m^2$. The radii of bubbles $A$ and $B$ are $2 \ cm$ and $4 \ cm$,respectively. The surface tension of the soap solution is $0.04 \ N/m$. Find the ratio $n_B / n_A$,where $n_A$ and $n_B$ are the number of moles of air in bubbles $A$ and $B$,respectively. [Neglect the effect of gravity.]

The radius of a soap bubble is increased from $R$ to $2R$. The work done in this process in terms of surface tension $S$ is: (in $\pi R^2 S$)