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)
$45$
$100$
$90$
None of these
What is the excess pressure inside a bubble of soap solution of radius $5.00 \;mm$, given that the surface tension of soap solution at the temperature ($20\,^{\circ} C$) is $2.50 \times 10^{-2}\; N m ^{-1}$ ? If an air bubble of the same dimension were formed at depth of $40.0 \;cm$ inside a container containing the soap solution (of relative density $1.20$), what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \;Pa$ ).
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 the another single soap bubble of radius $r\,cm$. The value of $r$ is.......
A spherical drop of water has radius $1\, mm$ If surface tension of water is $70 \times {10^{ - 3}}\,N/m$ difference of pressures between inside and out side of the spherical drop is ........ $N/{m^{ - 2}}$
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.
Two soap bubbles of radii $3r$ and $4r$ in contact with each other. The radius of curvature of the interface between bubbles is