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Question

Three identical particles are joined together by a thread as shown in figure.
All the three particles are moving in circular path in a horizontal plane. If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is
281292_8f4aa031abae4daeba23e70a45088a66.png
  1. 3:5:7
  2. 3:4:5
  3. 7:11:6
  4. 3:5:6

A
3:5:7
B
3:4:5
C
7:11:6
D
3:5:6
Solution
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Step 1: Drawing free body diagrams [Ref. Fig 1 & 2]
Angular velocity (ω) will be same for all

Step 2: Writing force equation for A, B and C
Solving in ground frame:
Applying Newton's Second Law on A, B, and C along the centripetal direction
(Considering direction towards center as positive)
F=mac

Where, ac= Centripetal acceleration =ω2r, r= radius
For C–––––
T1=maC=mω23l ....(1)

For B–––––
T2T1=maB

T2=mω23l+mω22l=mω25l ....(2)

For A–––––
T3T2=maA

T3=mω25l+mω2l=mω26l ....(3)

Step 3: Ratio calculation
From eqn. (1), (2) and (3)

T1:T2:T3=3:5:6

Hence option D is correct

2109446_281292_ans_68be45833bc44a779d3b57cd1f358980.png

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