The velocity field in a fluid is given to be $\overrightarrow{V}=(4xy)\hat{i}+2(x^2-y^2)\hat{j}$ Which of the following statement(s) is/are correct?

The correct statement is: The acceleration experienced by a fluid particle is zero at (𝑥 = 0, 𝑦 = 0).

Let us analyze the velocity field and calculate the acceleration of a fluid particle step-by-step to understand why this statement is correct.

The given velocity field is:
$$\overrightarrow{V}=u\hat{i}+v\hat{j}=(4xy)\hat{i}+2(x^2-y^2)\hat{j}$$
From this expression, the horizontal velocity component ($u$) and the vertical velocity component ($v$) are:
$$u=4xy$$
$$v=2(x^2-y^2)$$

For a steady two-dimensional flow, the acceleration components in the $x$ and $y$ directions ($a_x$ and $a_y$) are given by the convective acceleration formulas:
$$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$$
$$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$$

First, let us compute the required partial derivatives:
$$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(4xy)=4y$$
$$\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}(4xy)=4x$$
$$\frac{\partial v}{\partial x}=\frac{\partial }{\partial x}[2(x^2-y^2)]=4x$$
$$\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}[2(x^2-y^2)]=-4y$$

Now, substitute these derivatives into the acceleration equations:
$$a_x=(4xy)(4y)+2(x^2-y^2)(4x)=16x{y}^2+8x^3-8x{y}^2=8x^3+8x{y}^2$$
$$a_y=(4xy)(4x)+2(x^2-y^2)(-4y)=16x^{2}y-8x^{2}y+8y^3=8x^{2}y+8y^3$$

Evaluating the acceleration components at the origin $(x=0,y=0)$:
$$a_x{|}_{(0,0)}=8{(0)}^3+8(0){(0)}^2=0$$
$$a_y{|}_{(0,0)}=8{(0)}^2(0)+8{(0)}^3=0$$
Since both the $x$ and $y$ components of the acceleration vector are zero at the origin, the total acceleration experienced by a fluid particle at $(x=0,y=0)$ is indeed zero.