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Consider for simplicity a wire frame suspending a liquid film. If one bar of the frame is movable it is found that a force 5N per unit length must be applied to maintain the bar in position. If this force moves a small distance so that the total area of the film is increased by 0.5m2. Interfacial free energy per unit area is given as 4 J/m2. Calculate the change in interfacial energy? (Take the area of interface as 2m2)

Consider for simplicity a wire frame suspending a liquid film. If one bar of the frame is movable it is found that a force 5N per unit length must be applied to maintain the bar in position. If this force moves a small distance so that the total area of the film is increased by 0.5m2. Interfacial free energy per unit area is given as 4 J/m2. Calculate the change in interfacial energy? (Take the area of interface as 2m2) Correct Answer 8J

If one bar of the frame is movable it is found that a force F per unit length must be applied to maintain the bar in position. If this force moves a small distance so that the total area of the film is increased by dA the work done by the force is FdA. Then the force is given as F = ᵞ + dᵞ/dA (where ᵞ is the interfacial free energy). That is 4*2=8J.

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(i) Increased if drawbar HP of the tractor is increased for a given hauling distance
(ii) Decreased if drawbar HP of the tractor is increased for a given hauling distance
(iii) Increased if the hauling distance is increased for a given drawbar HP of the tractor
(iv) Decreased if the hauling distance is increased for a given drawbar HP of the tractor
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