$A$ soap bubble (surface tension $= T$) is charged to a maximum surface density of charge $= \sigma$. When it is just about to burst,its radius $R$ is given by:

$A$ soap bubble in vacuum has a radius of $3 \, cm$ and another soap bubble in vacuum has a radius of $4 \, cm$. If the two bubbles coalesce under isothermal condition,then the radius of the new bubble is ....... $cm$.

The excess pressure inside a soap bubble is $1.5$ times the excess pressure inside a second soap bubble. The volume of the second bubble is '$x$' times the volume of the first bubble. The value of '$x$' is

The pressure inside a small air bubble of radius $0.1 \, mm$ situated just below the surface of water will be equal to. [Take surface tension of water $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

The excess pressure inside a bubble in water is $P_1$. The excess pressure inside a drop of the same radius is $P_2$. Then:

An air bubble rises from the bottom to the top of a water tank in which the temperature of the water is uniform. The surface area of the bubble at the top of the tank is $125 \%$ more than its surface area at the bottom of the tank. If the atmospheric pressure is equal to the pressure of $10 \ m$ water column, then the depth of water in the tank is (in $m$)