The work done in blowing a soap bubble of radius $0.2\, m$ is (the surface tension of soap solution being $0.06\, N/m$).

What will be the work done in increasing the radius of a soap bubble from $\frac{r}{2}$ to $2r$, if the surface tension of the soap solution is $T$ (in $\pi r^2 T$)?

The surface energy of a liquid drop is $V$. It is sprayed into $1000$ equal droplets. The surface energy of all the droplets is

$A$ soap bubble of surface tension $0.04 \ N/m$ is blown to a diameter of $7 \ cm$. If $(15000 - x) \ \mu J$ of work is done in blowing it further to make its diameter $14 \ cm$,then the value of $x$ is . . . . . . . (Take $\pi = 22/7$)

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

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 $\pi R^2 T$)