The excess pressure inside a soap bubble of radius $2.5 \text{ cm}$ is $48 \text{ dyne/cm}^2$. The surface tension of the soap solution in $\text{dyne/cm}$ is:

The excess pressure inside a first spherical drop of water is three times that of a second spherical drop of water. Then the ratio of the mass of the first spherical drop to that of the second spherical drop is

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

$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

The excess pressure inside a soap bubble $A$ in air is half the excess pressure inside another soap bubble $B$ in air. If the volume of the bubble $A$ is $n$ times the volume of the bubble $B$,then the value of $n$ is . . . . . . .