Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

$8000$ identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

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 $T = 0.465\,J/m^2$)

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

Let $W_1$ be the work done in blowing a soap bubble of radius $r$ from a soap solution at room temperature. The soap solution is now heated and a second soap bubble of radius $2r$ is blown from the heated soap solution. If $W_2$ is the work done in forming this bubble,then: