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 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 $= 1 \,atm$).

Small drops of liquid of the same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is:

$A$ spherical soap bubble has internal pressure $P_0$ and radius $r_0$ and is in equilibrium in an enclosure with pressure $P_1 = \frac{8P_0}{9}$. The enclosure is gradually evacuated. Assuming temperature and surface tension of the soap bubble to be fixed, find the value of $\frac{\text{final radius}}{\text{initial radius}}$ of the soap bubble.

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

The excess pressure inside a spherical soap bubble of radius $1 \,cm$ is balanced by a column of oil (specific gravity $= 0.8$), $2 \,mm$ high. The surface tension of the bubble is: (in $\,N/m$)