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 $= 0.03\, N/m$)

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

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is:

Surface tension of a soap bubble is $2.0 \times 10^{-2} \; N m^{-1}$. Work done to increase the radius of the soap bubble from $3.5 \; cm$ to $7 \; cm$ will be $......... \times 10^{-4} \; J$ [Take $\pi = \frac{22}{7}$]

Consider a soap film on a rectangular frame of wire of area $4 \times 4 \text{ cm}^2$. If the area of the soap film is increased to $4 \times 5 \text{ cm}^2$,the work done in the process will be (The surface tension of the soap film is $3 \times 10^{-2} \text{ N/m}$).

$A$ big drop is formed by coalescing $1000$ small droplets of water. The ratio of surface energy of $1000$ droplets to that of the energy of the big drop is $\frac{10}{x}$. The value of $x$ is . . . . . . .