A two-dimensional incompressible frictionless flow field is given by   If ρ is the density of the fluid, the expression for pressure gradient vector at any point in the flow field is given as 

The correct answer is:
$$-\rho (x\hat{i}+y\hat{j})$$

Step-by-step Explanation:

1. Identify the Velocity Field:
Based on the problem description and visual details in the provided images (including the derivation in the fifth image), the two-dimensional velocity field is given by:
$$\overrightarrow{V}=u\hat{i}+v\hat{j}=x\hat{i}-y\hat{j}$$
Thus, the velocity components in the $x$ and $y$ directions are:
$$u=x$$
$$v=-y$$

2. Euler's Equation of Motion:
For a steady, incompressible, and frictionless (inviscid) flow with no external body forces (meaning body forces $f_x=0$ and $f_y=0$), the momentum equations in the $x$ and $y$ directions simplify to:
$$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho }\frac{\partial p}{\partial x}$$
$$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=-\frac{1}{\rho }\frac{\partial p}{\partial y}$$

3. Compute the Acceleration Components:
Evaluate the partial derivatives of the velocity components:
$$\frac{\partial u}{\partial x}=1,\frac{\partial u}{\partial y}=0$$
$$\frac{\partial v}{\partial x}=0,\frac{\partial v}{\partial y}=-1$$
Now, substitute these derivatives into the acceleration equations:
$$a_x=(x)(1)+(-y)(0)=x$$
$$a_y=(x)(0)+(-y)(-1)=y$$

4. Determine the Pressure Gradient Components:
Rearranging the Euler equations to isolate the partial derivatives of pressure:
$$$$ ∂p∂x = − ρ ax = − ρ x
$$\frac{\partial p}{\partial y}=-\rho a_y=-\rho y$$

5. Write the Pressure Gradient Vector:
The gradient of pressure $\nabla p$ is expressed as:
$$\nabla p=\frac{\partial p}{\partial x}\hat{i}+\frac{\partial p}{\partial y}\hat{j}$$
Substituting our computed partial derivatives:
$$\nabla p=-\rho x\hat{i}-\rho y\hat{j}=-\rho (x\hat{i}+y\hat{j})$$