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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

A particle of mass m 1 is fastened to one end of a string and another particle of mass m 2 is attached to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. If the particles are made to revolve in a horizontal circular path as shown in the figure, then the ratio of tension in the part of the string between the centre and m 2 to the tension in the part of the st

Options

  1. Am 1 m 1 + m 2
  2. Bm 1 + m 2 m 1
  3. C2 m 1 + m 2 2 m 1
  4. D2 m 1 m 1 + m 2

Correct answer

C. 2 m 1 + m 2 2 m 1

Step-by-step solution

for m 1 , T 1 = m 1 ω 2 2 r ......(i) for m 2 , T 2 - T 1 = m 2 ω 2 r ....(ii) Dividing both the equations: T 1 T 2 - T 1 = 2 m 1 m 2 ⇒ T 2 T 1 = 2 m 1 + m 2 2 m 1

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