The velocity field of a certain two dimensional flow is given by V (x, y )= k $$(x_i− y_j)$$​ where k = 2 / s and x and y are in meters. If the fluid density is 100 kg/$$m^3$$​ and pressure at origin is 10 kPa, what is the pressure at point (3,3)?

Correct option is B

​$$\bullet \ \text{Velocity field: } \vec{V}(x, y) = k(x\hat{i} - y\hat{j}) \\\bullet \ k = 2 \ \text{s}^{-1} \\\bullet \ \text{Fluid density: } \rho = 100 \ \text{kg/m}^3 \\\bullet \ \text{Pressure at origin } (0, 0): \ P_0 = 10 \ \text{kPa} = 10000 \ \text{Pa} \\\bullet \ \text{Find pressure at point } (3, 3)$$

$$\text{Assuming \textbf{steady, incompressible, inviscid, irrotational} flow, we use Bernoulli's equation:} \\P + \frac{1}{2} \rho V^2 = \text{constant} \\\text{Apply this between the origin and point } (3, 3): \\P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \\\\\textbf{Compute Velocities} \\\text{At origin } (0, 0): \\\vec{V}_1 = k(0\hat{i} - 0\hat{j}) = 0 \Rightarrow V_1 = 0 \\\text{At point } (3, 3): \\\vec{V}_2 = k(3\hat{i} - 3\hat{j}) = 2(3\hat{i} - 3\hat{j}) = 6\hat{i} - 6\hat{j} \\V_2 = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \ \text{m/s} \\\\\textbf{Apply Bernoulli Equation} \\P_1 = 10000 \ \text{Pa}, \quad V_1 = 0, \quad V_2 = 6\sqrt{2} \\10000 + 0 = P_2 + \frac{1}{2} \cdot 100 \cdot (6\sqrt{2})^2 \\10000 = P_2 + 50 \cdot 72 = P_2 + 3600 \\P_2 = 10000 - 3600 = 6400 \ \text{Pa} = \boxed{6.4 \ \text{kPa}}$$​​​