A Problem from my book -4
Two matchsticks of mass 10 g each and length = 3 cm each are floating on the surface of a certain liquid of surface tension = 25/6 N/m. A drop of a certain chemical is dropped in the gap between sticks and the vessel which destroys the surface tension of that part of liquid completely and don't effect the remaining part of liquid.The sticks start coming closer to each other.The time taken by them to collide with each other is 1/(k√b) s. Find .
Details and Assumptions :
- The volume of matchsticks are to be negligible.
- neglect any viscous forces.
- The matchstics keep on floating all the time.
The answer is 8.
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1 solution
Vertical force=0.1N and net force=0.125N. So horizontal force on 1 matchstick=0.075. Distance 1 matchstick has to travel=0.05cm. Use s=ut+0.5at^2 where u=0, s=0.05 and a=7.5 and get the answer=1/5√3.