The steady velocity field in an inviscid fluid of density 1.5 is given to be $\overrightarrow{V}=(y^2-x^2)\hat{i}+(2xy)\hat{j}$ . Neglecting body forces, the pressure gradient at (x =1, y = 1) is ______.

The correct option is -6î - 6ĵ.

Here is the detailed, step-by-step derivation of the pressure gradient:

Step 1: Identify the given values and components of velocity
The density of the inviscid fluid is:
$\rho =1.5$
The steady velocity field is given as:
$\overrightarrow{V}=(y^2-x^2)\hat{i}+\left(2xy\right)\hat{j}$
From this velocity field, the horizontal velocity component ($u$) and the vertical velocity component ($v$) are:
$u=y^2-x^2$
$v=2xy$

Step 2: State Euler's equation of motion
For a steady flow of an inviscid fluid, neglecting body forces, Euler's equation relates the acceleration to the pressure gradient as follows:
$\nabla p=-\rho \overrightarrow{a}$
where $\overrightarrow{a}$ is the acceleration vector, and its components in the x and y directions are $a_x$ and $a_y$ respectively.

Step 3: Determine the acceleration components
For a steady 2D flow, the acceleration components are given by:
$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$
$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$

Now, let's calculate the required partial derivatives:
$\frac{\partial u}{\partial x}=-2x$
$\frac{\partial u}{\partial y}=2y$
$\frac{\partial v}{\partial x}=2y$
$\frac{\partial v}{\partial y}=2x$

Step 4: Evaluate the velocity and derivative values at the point (x = 1, y = 1)
Substituting $x=1$ and $y=1$:
$u=1^2-1^2=0$
$v=2\left(1\right)\left(1\right)=2$
$\frac{\partial u}{\partial x}=-2$
$\frac{\partial u}{\partial y}=2$
$\frac{\partial v}{\partial x}=2$
$\frac{\partial v}{\partial y}=2$

Now, calculate the acceleration components at this point:
$a_x=\left(0\right)(-2)+\left(2\right)\left(2\right)=4$
$a_y=\left(0\right)\left(2\right)+\left(2\right)\left(2\right)=4$
This gives the acceleration vector:
$\overrightarrow{a}=4\hat{i}+4\hat{j}$

Step 5: Compute the pressure gradient
Substituting the acceleration vector and the density into Euler's equation:
$\nabla p=-1.5(4\hat{i}+4\hat{j})$
$\nabla p=-6\hat{i}-6\hat{j}$
Thus, the pressure gradient at (1, 1) is $-6\hat{i}-6\hat{j}$.