$A$ 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 $\pi T r^2$)

The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is ($T=$ surface tension of solution).

The work done in splitting a water drop of radius $R$ into $64$ droplets is ($T$ is the surface tension of water) (in $\pi R^2 T$)

If $W_1$ is the work done in increasing the radius of a soap bubble from $r$ to $2r$ and $W_2$ is the work done in increasing the radius of the soap bubble from $2r$ to $3r$, then $W_1: W_2=$

The surface energy required to create a liquid surface of area $0.04 \text{ m}^2$ for a liquid of surface tension $75 \text{ N/m}$ will be ....... $\text{J}$

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: