Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to -

$A$ spherical drop of water has a radius of $1\, mm$. If the surface tension of water is $70 \times 10^{-3}\, N/m$,the difference of pressure between the inside and outside of the spherical drop is ........ $N/m^2$.

$A$ spherical soap bubble of radius $3\,cm$ is formed inside another spherical soap bubble of radius $6\,cm$. If the internal pressure of the smaller bubble of radius $3\,cm$ in the above system is equal to the internal pressure of another single soap bubble of radius $r\,cm$,then the value of $r$ is:

$A$ big drop of radius $R$ is formed by $729$ small drops of water of radius $r$. Then the radius of each small drop will be:

$Assertion :$ 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 excess pressure inside the first soap bubble is three times that inside the second soap bubble. The ratio of the volume of the first soap bubble to the volume of the second soap bubble is