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Two identical balls are interconnected with a massless and inextensible thread. The system is in gravity free space with the thread just taut. Each ball is imparted a velocity $v$, one towards the other ball and the other perpendicular to the first, at $t = 0$. Then,
the thread will become taut at $t = (L/v)$
the thread will become taut at some time $t < (L/v).$
the thread will always remain taut for $t > (L/v).$
Both $(A)$ and $(C)$
Solution
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 $y>L$ $L=\sqrt{x^{2}+(L-x)^{2}}$
$L^{2}=x^{2}+L^{2}+x^{2}-2 l x$
$2 x^{2}-2 L x=0$
$2 x(x-L)=0$
$x=L$
time taken to travel $x=t=\frac{x}{V}=\left(\frac{L}{V}\right)$ thread will taught at $x=L$ and time $t=\frac{1}{2}$ and after that thread always taught because $y=\sqrt{x^{2}-\left(L-x^{2}\right)}$
$y>L d \quad f x>L$
