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
$\frac{T}{\sqrt{\rho R}}$
$\sqrt{\frac{2 T}{\rho R}}$
$\sqrt{\frac{4 T}{\rho R}}$
$\sqrt{\frac{8 T}{\rho R}}$
A drop of water volume $0.05\ cm^3$ is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is $40\ cm^2$ . If the surface tension of water is $70 \ dyne/cm$ , the minimum normal force required to seperate out the two glass plate in newton is approximately...... $N$ (assuming angle of contact is zero)
A small soap bubble of radius $4\,cm$ is trapped inside another bubble of radius $6\,cm$ without any contact. Let $P_2$ be the pressure inside the inner bubble and $P_0$ the pressure outside the outer bubble. Radius of another bubble with pressure difference $P_2 - P_0$ between its inside and outside would be....... $cm$
What is the pressure inside the drop of mercury of radius $3.00 \;mm$ at room temperature? Surface tension of mercury at that temperature $\left(20\,^{\circ} C \right)$ is $4.65 \times 10^{-1}\; N m ^{-1} .$ The atmospheric pressure is $1.01 \times 10^{5}\; Pa$. Also give the excess pressure inside the drop.
If the excess pressure inside a soap bubble is balanced by oil column of height $2\; mm$, then the surface tension of soap solution will be ($r = 1 \,cm$ and density $d = 0.8\, gm/cc$)
Two bubbles $A$ and $B$ $(r_A > r_B)$ are joined through a narrow tube. Then