Question

When a loop of wire is dipped in a wetting liquid and is taken out. A liquid film is formed and a loop of light inextensible thread is gently put on the liquid film. A hole is pricked inside the loop of thread. Due to property of surface tension, free surface of the liquid tries to minimize its surface area and hence area of the hole will be maximized. In the above description, loop wire is square of side 4 units. If length of thread is 15 units, The final surface area for soap film on one side of wire frame is $\frac{1}{K}(\frac{1}{4-\frac{\pi}{4}})$, Find K. (Neglect the friction everywhere)

Answer 1

K = 16(4-π)^2 / ((16-π)(124-33π))

Answer 2

The wire frame (area 16) loses its central film leaving a “hole” whose boundary is a thread of fixed length 15. By symmetry the thread detaches from the square along four equal segments and in the corners leaves four quarter–circles (of radius $x$). One finds that

$x=\frac{1}{8-2\pi}$

by writing the total length as

$16-8x+2\pi x=15$.

Then the hole’s area is

$(4-2x)^2+\pi x^2$

so that the remaining film area is $16$ minus that. Writing that answer in the form

$\dfrac{1}{K}\cdot\dfrac{1}{4-\pi/4}$

leads after algebra to

$K=\dfrac{16(4-\pi)^2}{(16-\pi)(124-33\pi)}$.