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 the first soap bubble of radius $R_1$ is three times that inside the second soap bubble of radius $R_2$. The ratio of volumes of the first bubble to second bubble is

An air bubble of radius $0.1 \ cm$ lies at a depth of $20 \ cm$ below the free surface of a liquid of density $1000 \ kg/m^3$. If the pressure inside the bubble is $2100 \ N/m^2$ greater than the atmospheric pressure,then the surface tension of the liquid in $SI$ unit is (use $g=10 \ m/s^2$)

An air bubble of radius $1 \ mm$ is at a depth of $8 \ cm$ below the free surface of a liquid column. If the surface tension and density of the liquid are $0.1 \ N \ m^{-1}$ and $2000 \ kg \ m^{-3}$ respectively, by what amount is the pressure inside the bubble greater than the atmospheric pressure (in $N \ m^{-2}$)? (Take $g = 10 \ m \ s^{-2}$)

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

If work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing another bubble of volume $2V$ will be