Work done in splitting a drop of water of $1 \, mm$ radius into $10^6$ droplets is (Surface tension of water $= 72 \times 10^{-3} \, J/m^2$).

$A$ number of small water droplets of surface tension '$T$',each of radius '$r$',are combined to form a single drop of radius '$R$'. If the released energy is converted into kinetic energy,then the velocity acquired by the bigger drop is . . . . . . ($\rho$ - density of water)

The excess pressure inside the first soap bubble of radius $R_1$ is three times that inside the second soap bubble of radius $R_2$. The ratio of volumes of the first bubble to second 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:

$A$ soap bubble in vacuum has a radius $3 \, cm$ and another soap bubble in vacuum has radius $4 \, cm$. If two bubbles coalesce under isothermal conditions,then the radius of the new bubble will be .............. $cm$.

The excess pressure inside a soap bubble $A$ in air is half the excess pressure inside another soap bubble $B$ in air. If the volume of the bubble $A$ is $n$ times the volume of the bubble $B$,then the value of $n$ is . . . . . . .