Question Detail

When the ring is on the verge of being pulled out, the spring force equals the weight of the ring plus the force due to surface tension. The surface tension force acts along the circumference of both the inner and outer circles. The total length of the liquid film in contact with the ring is 2π(r_outer + r_inner) because the film has two sides (inside and outside). So F_surface = γ × 2 × 2π(r_outer + r_inner) = 4πγ(r_outer + r_inner). The spring force F_spring = kx = 0.7 × 0.034 = 0.0238 N. Weight = mg = 0.0007 kg × 9.8 = 0.00686 N. Then F_spring = mg + F_surface => 0.0238 = 0.00686 + 4πγ(0.03 + 0.01) = 0.00686 + 4πγ(0.04) = 0.00686 + 0.50265γ. So 0.50265γ = 0.01694 => γ = 0.01694/0.50265 = 0.0337 N/m. That is not among the options. If we consider only one side of the film (sometimes for a ring pulled out, the film has two surfaces, so factor 2 is correct). Using g=10: mg=0.007 N, then 0.0238-0.007=0.0168, γ=0.0168/(0.50265)=0.0334. None match. If we use only outer circumference: 2πr_outer, and one surface: then F_surface = γ × 2πr_outer, then 0.0238 = 0.00686 + γ × 2π×0.03 = 0.00686 + 0.1885γ => γ = (0.01694)/0.1885 = 0.0899 N/m, still not an option. If we use both circumferences but one surface: γ × 2π(r_outer+r_inner) = 0.0238-0.00686=0.01694 => γ = 0.01694/(2π×0.04)=0.01694/0.2513=0.0674 N/m. Option B is 0.076, close to 0.067. Possibly with g=9.8 and exact values. Let's use exact: kx=0.7×0.034=0.0238, mg=0.0007×9.8=0.00686, difference=0.01694. If we use both circumferences and two surfaces (as is correct for a ring pulled out of a liquid, the film has two sides, so the total length is 4π(r_outer+r_inner) = 4π×0.04 = 0.50265, γ=0.01694/0.50265=0.0337. That's 0.0337, not 0.076. If we use both circumferences and one surface (maybe they consider the ring as having only one film), then length = 2π(r_outer+r_inner)=0.2513, γ=0.01694/0.2513=0.0674, which rounds to 0.067. Option B is 0.076, which is a bit higher. Possibly they used r in cm? If they used r in cm, then r_outer=3 cm, r_inner=1 cm, but then kx is in N, so units would be inconsistent. Maybe they forgot to convert mm to m? If they used r=30 mm = 0.03 m, r_inner=0.01 m, then 4π(0.04)=0.50265 m, that's correct. Their answer 0.076 might come from using diameter instead of radius? Let's check: if they used diameters: outer diameter=60 mm=0.06 m, inner=0.02 m, sum=0.08 m, times 2π = 0.50265, same. So not that. Perhaps they used the formula with only one surface and used the sum of radii but forgot the 2? I'll go with B as it is the closest.