A thin circular ring of radius 5 cm is slowly lifted from a perfectly wetting liquid having surface tension 0.072 N/m. Neglect the weight of the ring. If the required surface-tension force is approximately N × 10⁻³ N, find N. Take π = 3.14.
Physics · Properties of solid and fluid · properties of fluid and surface tension
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Question

A thin circular ring of radius 5 cm is slowly lifted from a perfectly wetting liquid having surface tension 0.072 N/m. Neglect the weight of the ring. If the required surface-tension force is approximately N × 10⁻³ N, find N. Take π = 3.14.

Physics · Properties of solid and fluid · properties of fluid and surface tension

G1 Physics · Accepted Answer

For a thin ring, surface tension acts along both inner and outer circumferences. Therefore, F = 4πRS. Substitute R = 0.05 m, S = 0.072 N/m, π = 3.14. Thus, F = 4×3.14×0.05×0.072 ≈ 0.0452 N = 45.2×10⁻³ N. Hence, to the nearest integer, N = 45.