The radius of one soap bubble is four times that of another. What is the ratio of the excess pressure inside the two bubbles?

What is the pressure inside the drop of mercury of radius $3.00 \; mm$ at room temperature? Surface tension of mercury at that temperature $(20 \; ^{\circ}C)$ is $4.65 \times 10^{-1} \; N m^{-1}$. The atmospheric pressure is $1.01 \times 10^{5} \; Pa$. Also,calculate the excess pressure inside the drop.

The excess pressure in a soap bubble is double that in another one. The ratio of their volumes is .............

$A$ cylinder with a movable piston contains air under a pressure $p_1$ and a soap bubble of radius $r$. The pressure $p_2$ to which the air should be compressed by slowly pushing the piston into the cylinder for the soap bubble to reduce its size by half will be: (The surface tension is $\sigma$, and the temperature $T$ is maintained constant)

$A$ spherical liquid drop splits into $729$ identical spherical drops. If $E$ is the surface energy of the original drop and $U$ is the total surface energy of the resulting drops,then $\frac{E}{U} = \frac{1}{x}$. The value of $x$ is

The force required to separate two glass plates of area $10^{-2} \, m^2$ with a film of water $0.05 \, mm$ thick between them is ...... $N$. (Surface tension of water is $70 \times 10^{-3} \, N/m$)