There is a small hole in hollow sphere. Water enters in sphere when it is taken at depth of $40\,cm$ in water. Diameter of hole is ....... $mm$ (Surface tension of water $= 0.07\, N/m$):

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) 

Write the equation of excess pressure (pressure difference) for the bubble in air and bubble in water.

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 = $1.013 \times {10^5}N{m^{ - 2}}$]

Pressure inside a soap bubble is greater than the pressure outside by an amount :

(given : $\mathrm{R}=$ Radius of bubble, $\mathrm{S}=$ Surface tension of bubble)

A liquid column of height $0.04 \mathrm{~cm}$ balances excess pressure of soap bubble of certain radius. If density of liquid is $8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and surface tension of soap solution is $0.28 \mathrm{Nm}^{-1}$, then diameter of the soap bubble is . . . . . . .. . $\mathrm{cm}$.

$\text { (if } g=10 \mathrm{~ms}^{-2} \text { ) }$