The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

A drop of water of volume $0.05\, cm^3$ is pressed between two glass plates, as a consequence of which it spreads and occupies an area of $40\, cm^2$. If the surface tension of water is $70\, dyne/cm$, then the normal force required to separate out the two glass plates will be in Newton

Fill in the Blank :

$(i)$ Bubble in water have .......... free surface.

$(ii)$ Bubble in air have .......... free surface.

$(iii)$ Rain drop have .......... free surface.

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

A soap bubble in a form of circular tube having radius of curvature $R$ and radius of curvature perpendicular to it is $5R$ . Find the excess pressure in the bubble :

When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\frac{{5r}}{4}$.Taking the atmospheric pressure to be equal to $10\,m$ height of water column, the depth of the lake would approximately be ....... $m$ (ignore the surface tension and the effect of temperature)