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Smooth Gaurav semi Vertical angle Alpha is kept on a horizontal surface e flexible ring of radius R and mass M is slipped on to the cone as shown find tension in the ring take equals to 9.8 M per second square alpha equals to tan inverse 1 by pi and mass is 2 kg?
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Smooth Gaurav semi Vertical angle Alpha is kept on a horizontal surfac...



Given Data:
  • Vertical angle Alpha = tan inverse(1/pi)
  • Acceleration due to gravity g = 9.8 m/s^2
  • Mass of the ring M = 2 kg



Analysis:
  • The ring is placed on a smooth cone with a semi-vertical angle Alpha.
  • The ring will experience a centripetal force due to its circular motion on the cone.
  • The tension in the ring can be calculated using the equation: Tension = Mass * Acceleration



Solution:
  • Since the angle Alpha is given as tan inverse(1/pi), we can calculate its numerical value.
  • Substitute the values of Alpha, Mass, and gravity into the equation to find the tension in the ring.
  • Tension = 2 kg * 9.8 m/s^2 = 19.6 N



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Smooth Gaurav semi Vertical angle Alpha is kept on a horizontal surface e flexible ring of radius R and mass M is slipped on to the cone as shown find tension in the ring take equals to 9.8 M per second square alpha equals to tan inverse 1 by pi and mass is 2 kg?
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