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

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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

$a \cos 45=b \cos 45$

$a=b$

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.

Is a and b which you taken as tangential acceleration ?

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