Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops of the same liquid combine, then the excess pressure inside the bigger drop is (in $units$)

$8$ mercury drops coalesce to form one mercury drop. The energy changes by a factor of:

The pressure inside a small air bubble of radius $0.1 \, mm$ situated just below the surface of water will be equal to. [Take surface tension of water $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

Two spherical soap bubbles formed in vacuum have diameters $3.0 \, mm$ and $4.0 \, mm$. They coalesce to form a single spherical bubble. If the temperature remains unchanged,find the diameter of the bubble so formed in $mm$.

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 -

An air bubble doubles in radius when it rises from the bottom of the sea to the surface. If the atmospheric pressure is equal to the pressure exerted by a $10 \, m$ column of water,then the depth of the sea is $... \, m$. (Assume surface tension is negligible.)