$A$ mercury drop of $1\, cm$ radius is broken into ${10^6}$ small drops. The energy used will be (surface tension of mercury is $35 \times {10^{ - 3}} \, N/cm$).

On the surface of the liquid in equilibrium,molecules of the liquid possess

The surface tension of a soap solution is $2 \times 10^{-2} \ N/m$. The work done in producing a soap bubble of radius $2 \ cm$ is:

If $8$ identical droplets are formed from a single drop of radius $2\,mm$,what is the change in energy in $\mu J$? (Surface tension $T = 0.465\,J/m^2$)

The radius $R$ of the soap bubble is doubled under isothermal conditions. If $T$ is the surface tension of the soap bubble, the work done in doing so is given by (in $\pi R^2 T$)

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $u_i$ is the surface energy of the original drop and $u_f$ is the total surface energy of the resulting drops (ignoring evaporation),and $\frac{u_f}{u_i} = \left(\frac{10}{x}\right)$,then the value of $x$ is $......$