The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T=$ surface tension of water). (in $\pi TR^2$)

In air,a charged soap bubble of radius $R$ breaks into $27$ small soap bubbles of equal radius $r$. Then,the ratio of mechanical force acting per unit area of the big soap bubble to that of a small soap bubble is

Due to surface tension, the excess pressure inside a smaller drop is $9 \text{ units}$. If $27$ smaller drops combine, then the excess pressure inside the bigger drop is: (in $\text{ units}$)

$Assertion :$ 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.

$A$ large number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

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 $= 1 \,atm$).