Question

Water Is filled In a container. The container Is moving horizontally with acceleration of 5 m/s². The density of water Is 1000 kg/m³. Find the pressure difference between point A and C.

• A. 1500 Pa
• B. 1960 Pa
• C. 3460 Pa
• D. 4960 Pa

Answer 1

Answer: (C)

3460 Pa

Answer 2

The pressure difference in an accelerating fluid is given by $P_A - P_C = \rho a_x \Delta x + \rho g \Delta y$. Given: Density of water, $\rho = 1000 \, kg/m^3$ Horizontal acceleration, $a_x = 5 \, m/s^2$ Vertical acceleration due to gravity, $g \approx 9.8 \, m/s^2$ Horizontal distance, $\Delta x = 30 \, cm = 0.3 \, m$ (from C to A, so $\Delta x = x_A - x_C = -0.3$ but the formula uses the magnitude of the distance and the sign is accounted for in the direction of acceleration. Here, A is to the left of C, so the pressure increases towards the left in the direction of acceleration.) Vertical distance, $\Delta y = 20 \, cm = 0.2 \, m$ (from C to A, so $\Delta y = y_A - y_C = -0.2$. However, pressure increases downwards. Point A is below C, so this contributes positively to $P_A > P_C$).

The pressure difference calculation considers both the horizontal acceleration and gravity. Pressure difference due to horizontal acceleration: $\Delta P_x = \rho \cdot a_x \cdot \Delta x = 1000 \, kg/m^3 \times 5 \, m/s^2 \times 0.3 \, m = 1500 \, Pa$. Pressure difference due to gravity: $\Delta P_y = \rho \cdot g \cdot \Delta y = 1000 \, kg/m^3 \times 9.8 \, m/s^2 \times 0.2 \, m = 1960 \, Pa$. Total pressure difference $P_A - P_C = \Delta P_x + \Delta P_y = 1500 \, Pa + 1960 \, Pa = 3460 \, Pa$.