System shown in figure is released from rest. Pulley and spring is massless and friction is absent everywhere. The speed of $5\text{kg}$ block when $2\text{kg}$ block leaves contact with ground is (Take force constant of $5\text{kg}$ spring $k=40\frac{\text{N}}{\text{m}}$ and $g=10\frac{\text{m}}{{\text{s}}^2}$

The correct option is B $2\sqrt{2}\text{m/s}$
Let $x$ be the extension in the spring when $2\text{k}g$ block leaves the contact with ground.
FBD of $2\text{kg}$ block : (just leaving the ground )

Then $T=mg=kx$
$kx=2g\text{or}x=\frac{2g}{k}=\frac{2\times 10}{40}=\frac{1}{2}\text{m}$
Now from conservation of mechanical energy loss of gravitational potential energy of mass $5\text{kg}$ causes gain in elastic potential energy of spring and Kinetic energy of mass $5\text{kg}$
$mgx=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2(m=5\text{kg})$
$v=\sqrt{2gx-\frac{k{x}^2}{m}}.$
Substituting the values
$v=\sqrt{2\times 10\times \frac{1}{2}-\frac{\left(40\right)}{4\times 5}}v=2\sqrt{2}\text{m/s}$