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 $\pi = 3.14$).

Under isothermal conditions,two soap bubbles of radii $r_{1}$ and $r_{2}$ coalesce to form a single larger bubble. The radius of the larger bubble is

$A$ large number of water droplets each of radius '$r$' combine to form a large drop of radius '$R$'. If the surface tension of water is '$T$' and the mechanical equivalent of heat is '$J$',then the rise in temperature due to this process is:

An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)

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 $T = 70 \times 10^{-3} \, N/m$ and atmospheric pressure $P_0 = 1.013 \times 10^5 \, N/m^2$]

If $W$ amount of work is required to form a bubble of volume $V$,then how much work is required to form a bubble of volume $2V$?