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Question

# Two beads A and B of equal mass m are connected by a light rod, which acts as a chord for the ring. They are constrained to move on the frictionless ring in a vertical plane. The beads are released from rest as shown in figure. The tension in the chord just after release will be:

A
mg4
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B
2 mg
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C
mg2
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D
mg2
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Solution

## The correct option is D mg√2Just after the release, bead B moves vertically downwards and bead A moves leftwards horizontally with the same acceleration i.e a From FBD of A, applying equation of circular motion along tangential direction (i.e direction of acceleration) Tcos45∘=ma−(1) i.e T√2 = ma Similarly for bead B: mg−Tcos45∘=ma−(2) Substituting from equation (1) in equation (2), mg−ma=ma or a=g2−(3) Substituting Eq.(3) in Eq. (1), we get: T=ma√2=mg√2

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