Air is pushed into a soap bubble to increase its radius from $R$ to $2R$. In this case, the pressure inside the bubble

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

One end of a uniform glass capillary tube of radius $r = 0.025 \ cm$ is immersed vertically in water to a depth $h = 1 \ cm$. The excess pressure in $N/m^2$ required to blow an air bubble out of the tube is: (Surface tension of water $T = 7 \times 10^{-2} \ N/m$,Density of water $\rho = 10^3 \ kg/m^3$,Acceleration due to gravity $g = 10 \ m/s^2$)

$A$ liquid column of height $0.04 \,cm$ balances excess pressure of a soap bubble of a certain radius. If the density of the liquid is $8 \times 10^3 \,kg \,m^{-3}$ and the surface tension of the soap solution is $0.28 \,N \,m^{-1}$, then the diameter of the soap bubble is . . . . . . $cm$.
$(g = 10 \,m \,s^{-2})$

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is . . . . . . . ($S$ is the surface tension of mercury). (in $\pi r^2 S$)

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is