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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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Objects A and B each of mass m are connected by light inextensible cord. They are constrained to move on a frictionless ring in a vertical plane. The object is released from rest at the positions as shown. The tension T in the cord just after release will be

\[(a)\;mg \sqrt 2\quad (b)\;\frac{mg}{\sqrt 2} \quad (c)\;\frac{mg}{2} \quad (d)\;\frac{mg}{4} \]

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Let mass A move with acceleration 'b' and mass B move with acceleration 'a'.
directions of 'a' and 'b' are different but magnitude of 'a' and 'b' are same.
Since the length of the rope AB is same
$a \cos 45=b \cos 45$
Also $T \sin 45=mb$------(1) for A
$mg-T\sin 45=ma$-----(2) for B
$2 ma =mg=>a=g/2$
Substituting in (1)
$\large\frac{T}{\sqrt 2}=\large\frac{mg}{2}$
$T=\large\frac{mg}{\sqrt 2}$
Hence b is the correct answer.
answered Jul 16, 2013 by meena.p
edited May 27, 2014 by lmohan717
Is a and b which you taken as tangential acceleration ?

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