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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 a horizontal plane in circular path with O as centre. If the velocity of the outermost particle v0, then the ratio of tensions in the three sections of the string is

A
3:5:7
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B
3:4:5
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C
7:11:6
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D
3:5:6
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Solution

The correct option is D 3:5:6
Let ω is the angular speed of revolution

From FBD of C
T3=mω23l
From FBD of B
T2T3=mω22lT2=mω25l
From FBD of A
T1T2=mω2lT1=mω26l

Ratio of tension in the strings is
T3:T2:T1=mω23l:mω25l:mω26l=3:5:6

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