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 should be the radius of a water drop so that the excess pressure inside it is $72 \ Nm^{-2}$ (in $mm$)? (The surface tension of water is $7.2 \times 10^{-2} \ Nm^{-1}$)

$A$ 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)

$A$ big liquid drop splits into $n$ similar small drops under isothermal conditions,then in this process

The excess pressure inside the first soap bubble of radius $R_{1}$ is two times that inside the second soap bubble of radius $R_{2}$. The ratio of the volumes of the first bubble to that of the second bubble is:

$A$ soap bubble is blown with the help of a mechanical pump at the mouth of a tube. The pump produces a constant increase per minute in the volume of the bubble,irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time $t$ will be-