A film of soap solution is formed on a loop frame loop of `6.28 cm` long thread is gently put on the film and the film is broken with a needle inside the loop. The tread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution `= 0.030 n//m`.

Answer

`3 xx 10^(-4) N`

Explanation

`r_(1) = 1.44 xx 10^(-3)m`., `r_(2) = 0.72 xx 10^(-3)m`.
Equating pressures at point `(B)` & `(C)`
`P_(A) - (2sigma)/(r_(2)) + (0.2) rhog = P_(C)`. and `P_(B) - (2sigma)/(r_(1)) = P_(C)`.
so `P_(B) - P_(A) = 2sigma((1)/(r_(1)) - (1)/(r_(2))) + 0.2 rhog`
`= 2 xx 72 xx 10^(-3)(N)/(m) [(10^(3))/(1.44) - (10^(3))/(0.72)] + (0.2) xx 10^(3) xx 9.8`
`= (144 xx (-0.72))/(1.44 xx 0.72) + 1960`
`= -100 + 1960 = 1860 N//m^(2)`.