The excess pressure in a soap bubble is thrice that in other one. Then the ratio of their volume is
$1:3$
$1:9$
$27:1$
$1:27$
Air (density $\rho$ ) is being blown on a soap film (surface tension $T$ ) by a pipe of radius $R$ with its opening right next to the film. The film is deformed and a bubble detaches from the film when the shape of the deformed surface is a hemisphere. Given that the dynamic pressure on the film due to the air blown at speed $v$ is $\frac{1}{2} \rho v^{2}$, the speed at which the bubble formed is
An air bubble of radius $r$ in water is at depth $h$ below the water surface at same instant. If $P$ is atmospheric pressure and $d$ and $T$ are the density and surface tension of water respectively. The pressure inside the bubble will be
Two soap bubbles coalesce to form a single bubble. If $V$ is the subsequent change in volume of contained air and $S$ change in total surface area, $T$ is the surface tension and $P$ atmospheric pressure, then which of the following relation is correct?
Fill in the Blank :
$(i)$ Bubble in water have .......... free surface.
$(ii)$ Bubble in air have .......... free surface.
$(iii)$ Rain drop have .......... free surface.
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 $70 \times {10^{ - 3}}N{m^{ - 1}}$ and atmospheric pressure = $1.013 \times {10^5}N{m^{ - 2}}$]