A particle of mass 1 kg is subjected to a force which depends on the position as $\overrightarrow{F}=-k\left(x\hat{i}+y\hat{j}\right)$ ${\text{kg ms}}^{-2}$ with $k=1{\text{kg s}}^{-2}$. At time $t=0$, the particle's position is $\overrightarrow{r}=\frac{1}{\sqrt{2}}\hat{i}+\sqrt{2}\hat{j}$ m and its velocity is $\overrightarrow{v}=-\sqrt{2}\hat{i}+\sqrt{2}\hat{j}+\sqrt{2}\hat{k}$ ${\text{ms}}^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5\text{m}$, the value of $(x{v}_y-y{v}_x)$ is _____ ${\text{m}}^2{\text{s}}^{-1}$.

The correct answer is 3.

Step 1: Understand the given physical parameters
We are given:
- Mass of the particle, $m=1\text{kg}$
- Force vector, $\overrightarrow{F}=-k\left(x\hat{i}+y\hat{j}\right)$ with $k=1{\text{kg s}}^{-2}$
- At $t=0$, position components: $x_0=\frac{1}{\sqrt{2}}\text{m}$, $y_0=\sqrt{2}\text{m}$
- At $t=0$, velocity components: $v_{x0}=-\sqrt{2}{\text{ms}}^{-1}$, $v_{y0}=\sqrt{2}{\text{ms}}^{-1}$

Step 2: Relate the expression to angular momentum conservation
The expression $(x{v}_y-y{v}_x)$ represents the $z$-component of angular momentum per unit mass, as:

$L_z={\left(\overrightarrow{r}\times m\overrightarrow{v}\right)}_z=m\left(x{v}_y-y{v}_x\right)$

Step 3: Check torque about the z-axis
The torque $\overrightarrow{\tau }$ acting on the particle is given by:

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\left(x\hat{i}+y\hat{j}+z\hat{k}\right)\times \left(-kx\hat{i}-ky\hat{j}\right)$

Calculating the $z$-component of torque (${\tau }_z$):

${\tau }_z=x{F}_y-y{F}_x=x\left(-ky\right)-y\left(-kx\right)=-kxy+kxy=0$

Since the torque along the $z$-axis is zero (${\tau }_z=0$), the $z$-component of angular momentum is conserved and remains constant throughout the motion.

Step 4: Compute the conserved value
Since $(x{v}_y-y{v}_x)$ is constant at all times, we can evaluate it using initial conditions at $t=0$:

$x{v}_y-y{v}_x=x_0{v}_{y0}-y_0{v}_{x0}$

Substitute the initial values into the expression:

$x{v}_y-y{v}_x=\left(\frac{1}{\sqrt{2}}\right)\left(\sqrt{2}\right)-\left(\sqrt{2}\right)\left(-\sqrt{2}\right)$

$x{v}_y-y{v}_x=1-\left(-2\right)=1+2=3$

Thus, when $z=0.5\text{m}$, the value of $(x{v}_y-y{v}_x)$ is 3 ${\text{m}}^2{\text{s}}^{-1}$.