The vertical distance between the masses is$\frac{20}{3}\text{m}$ . The time at which the distance becomes zero is

We know for an at wood’s machine

$\text{a}=\frac{\text{g}\left({\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}\right)}{{\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}}$

$=\frac{10\left(2-1\right)}{2+1}$

$\text{a}=\frac{10}{3}{\text{m/s}}^{\text{2}}$

$\text{u}=0,\text{S}=\frac{10}{3}\text{m}$

$\text{S}=\text{ut+}\frac{1}{2}{\text{at}}^{\text{2}}$

$\Rightarrow \frac{10}{3}=\frac{1}{2}\left(\frac{10}{3}\right){\text{t}}^2$

${\text{t}}^{\text{2}}=2$

$\therefore \text{t}=\sqrt{2}\text{s}$