Let $R_1, R_2$ and $R_3$ be the radii of three mercury drops. $A$ big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is

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

$A$ 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.

$Assertion :$ Smaller drops of liquid resist deforming forces better than the larger drops.
$Reason :$ Excess pressure inside a drop is directly proportional to its surface area.

The pressure of air in a soap bubble of $0.7 \ cm$ diameter is $8 \ mm$ of water above the pressure outside. The surface tension of the soap solution is ........ $dyne/cm$.

$8$ mercury drops coalesce to form one mercury drop. The energy changes by a factor of: