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As shown in figure $A$ and $B$ are two identical mass. Suppose $A$ move $x$ and $B$ move $x$ because velocity is same. Length taught when $y>L$

$L=x_{2}+(L−x)_{2} L_{2}=x_{2}+L_{2}+x_{2}−2lx2x_{2}−2Lx=02x(x−L)=0x=L$

time taken to travel $x=t=Vx =(VL )$

thread will taught at $x=L$ and time $t=21 $

and after that thread always taught because

$y=x_{2}−(L−x_{2}) $

$y>Ldfx>L$

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