Pressure inside two soap bubbles are $1.02 \ atm$ and $1.05 \ atm$ respectively. The ratio of their surface area is

The excess pressure inside an air bubble of radius $r$ just below the surface of water is $p_1$. The excess pressure inside a drop of the same radius just outside the surface is $p_2$. If $T$ is surface tension,then

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 $g = 10 \ m/s^2$)

Fill in the blanks:
$(i)$ $A$ bubble in water has .......... free surface.
$(ii)$ $A$ bubble in air has .......... free surface.
$(iii)$ $A$ rain drop has .......... free surface.

$A$ 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 . . . . . . $cm$.
$(g = 10 \,m \,s^{-2})$

$A$ drop of water with a volume of $0.05 \ cm^3$ is pressed between two glass plates,causing it to spread between the plates. The area of contact with each plate is $40 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,the minimum normal force required to separate the two glass plates in newtons is approximately: (assuming the angle of contact is zero)