Two persons of masses $m_1$ and $m_2$ are standing on a smooth surface facing each other holding a massless rope in their hands. If they pull each other with a force $F$, they meet each other

$(d$ is the initial distance between them$)$

The correct option is D after $\sqrt{\frac{2m_1{m}_{2}d}{F(m_1+m_2)}}$ seconds at the centre of mass
Let $a_1$ and $a_2$ are the accelerations of $m_1$ and $m_2$ repectively.


When force is applied on both persons, we can write
$F=m_1{a}_1=m_2{a}_2...\left(1\right)$

Let their centre of mass is at distance, $x$ from $m_1$.

So, moment of mass about centre of mass,
$m_{1}x=m_2(d-x)$

$\Rightarrow (m_1+m_2)x=m_{2}d$

$\Rightarrow x=\left(\frac{m_2}{m_1+m_2}\right)d...\left(2\right)$

Let the time taken by them to meet at centre of mass is $t$.
For first person,
Applying $s=ut+\frac{1}{2}{a}_1{t}^2$ we get,

$x=0+\frac{1}{2}\left(\frac{F}{m_1}\right)t^2$

$\Rightarrow \left(\frac{m_2}{m_1+m_2}\right)d=\frac{1}{2}\left(\frac{F}{m_1}\right)t^2$

$\therefore t=\sqrt{\frac{2m_1{m}_{2}d}{F(m_1+m_2)}}$ second at the centre of mass.