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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 mg2
Just 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 T2 = ma

Similarly for bead B:


mgTcos45=ma(2)

Substituting from equation (1) in equation (2),
mgma=ma
or a=g2(3)
Substituting Eq.(3) in Eq. (1), we get:

T=ma2=mg2

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