The radius of a soap bubble is $r$. The surface tension of the soap solution is $T$. Keeping the temperature constant,the radius of the soap bubble is doubled. The energy necessary for this process will be: (in $\pi r^2 T$)

Work done in increasing the size of a soap bubble from a radius of $3\, cm$ to $5\, cm$ is nearly (Surface tension of soap solution $= 0.03\, N/m$)

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 $\pi R^2 T$)

The surface tension of a soap solution is $3.5 \times 10^{-2} \text{ N/m}$. The work required to increase the radius of a soap bubble from $1 \text{ cm}$ to $2 \text{ cm}$ is $\alpha \times 10^{-6} \text{ J}$. The value of $\alpha$ is . . . . . . . $(\pi = 22/7)$

If $1000$ droplets of water of surface tension $0.07\,N/m$,each having the same radius $1\,mm$,combine to form a single drop,the released surface energy in the process is: (Take $\pi = \frac{22}{7}$)

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