$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

$A$ big drop is formed by coalescing $1000$ small droplets of water. The surface energy will become :

An air bubble rises from the bottom to the top of a water tank in which the temperature of the water is uniform. The surface area of the bubble at the top of the tank is $125 \%$ more than its surface area at the bottom of the tank. If the atmospheric pressure is equal to the pressure of $10 \ m$ water column, then the depth of water in the tank is (in $m$)

The excess pressure inside the first soap bubble is three times that inside the second soap bubble. The ratio of the volume of the first soap bubble to the volume of the second soap bubble is

The surface tension of a soap bubble is $0.03 \text{ N/m}$. The work done in increasing the diameter of the bubble from $2 \text{ cm}$ to $6 \text{ cm}$ is $\alpha \times 10^{-4} \text{ J}$. The value of $\alpha$ is . . . . . . (Take $\pi = 3.14$).

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).