91Ó°ÊÓ

The slope \(\theta\) can be calculated using the formula for tan, which is opposite/hypotenuse. Here, the opposite is the acceleration due to gravity \(0.25g\), and the hypotenuse is actual acceleration due to gravity \(g\), hence \(tan(\theta) = 0.25g / g\). So, \(\theta = atan(0.25) = 14.0^{\circ}\). So the slope of the free surface is \(14.0^{\circ}\).

The pressure at a depth \(h\) in a fluid is given by \( P = \rho gh \) where \(\rho\) is the density, \(g\) is acceleration due to gravity and \(h\) is the depth. Here, we need to calculate the additional pressure due to added acceleration using the same formula but with different parameters due to different orientation of the forces. So, \( P_{accel} = \rho a h = \rho (0.25g) h \) where \(a\) is the added acceleration. From the two points furthest apart in the free surface, i.e., depth difference is \(2h = 80 cm\), or \(h = 40 cm\). Now, pressure at the deepest point from the added acceleration \( P_{accel} = \rho (0.25g) h = 0.8 * 1000 kg/m^3 * 9.81 m/s^2 * 0.25 * 0.4m = 784.8 Pa\). The pressure from the fluid column is due to the height of the column and gravity, \( P_{gravity} = \rho gh = 0.8 * 1000kg/m^3 * 9.81m/s^2 * 0.8m = 6272 Pa\). Then, the total pressure at the deepest point is the sum of these 2 pressure i.e. \( P_{total} = P_{accel} + P_{gravity} = 784.8 Pa + 6272 Pa = 7056.8 Pa \).

On the other side of the box, where the fluid depth is lower, the pressure from the underneath fluid column is lower and so is the pressure from additional acceleration. So, Pressure \( P_{accel} = \rho a h = 0.8 * 1000kg/m^3 * 9.81m/s^2 * 0.25 * 0.4m = 784.8 Pa\). Pressure due to gravity \( P_{gravity} = \rho gh = 0.8 * 1000kg/m^3 * 9.81 * 0.4m = 3136 Pa \). So, the total pressure at the shallowest point is \( P_{accel} + P_{gravity} = 784.8 Pa + 3136 Pa = 3920.8 Pa\).