The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

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

Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to

When a large bubble rises from the bottom of a lake to the surface,its radius doubles. If atmospheric pressure is equal to that of a column of water of height $H$,then the depth of the lake is:

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

Under isothermal conditions,two soap bubbles of radii $r_1$ and $r_2$ coalesce to form a single soap bubble of radius $R$. The radius of the new bubble is