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(A) When a ring is pulled from the surface of a liquid,the surface tension acts on both the inner and outer circumferences of the ring.Total length of

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$72$

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(A) When a ring is pulled from the surface of a liquid,the surface tension acts on both the inner and outer circumferences of the ring.Total length of contact $L = 2\pi r_1 + 2\pi r_2 = \pi(D_1 + D_2)$,where $D_1$ and $D_2$ are the internal and external diameters.Given: $D_1 = 8.5\, cm$,$D_2 = 8.7\, cm$,and mass $m = 3.97\, g$.The force required to pull the ring is $F = mg = L 	imes \sigma$,where $\sigma$ is the surface tension.$F = \pi(D_1 + D_2) 	imes \sigma = mg$.Substituting the values: $\pi(8.5 + 8.7) 	imes \sigma = 3.97 	imes 980$.$\pi(17.2) 	imes \sigma = 3890.6$.$\sigma = \frac{3890.6}{3.14159 	imes 17.2} \approx 72\, dyne\, cm^{-1}$.

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