$A$ soap bubble has radius $R$ and thickness $d$ $(d \ll R)$ as shown. It collapses into a spherical drop. The ratio of excess pressure in the drop to the excess pressure inside the bubble is

Two soap bubbles having radii $r_1$ and $r_2$ have inside pressures $P_1$ and $P_2$ respectively. If $P_0$ is the external pressure,then the ratio of their volumes is:

Two soap bubbles of radii $x$ and $y$ coalesce to form a single bubble of radius $z$. Then $z$ is equal to

The pressures inside two soap bubbles $A$ and $B$ are $1.02 \text{ atm}$ and $1.04 \text{ atm}$ respectively. The ratio of the volume of bubble $A$ to that of bubble $B$ is (outside pressure = $1 \text{ atm}$)

An ice cube of edge $1 \ cm$ melts in a gravity-free container. The approximate surface area of the water formed is (water is in the form of a spherical drop)

There is an air bubble of radius $1.0\,mm$ in a liquid of surface tension $0.075\,N\,m^{-1}$ and density $1000\,kg\,m^{-3}$ at a depth of $10\,cm$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is $....Pa \left( g = 10\,m\,s^{-2} \right)$