A rectangular tank of length 2.0 m contains water and is accelerated horizontally toward the right with constant acceleration 7.5 m/s². No liquid spills out. Taking g = 10 m/s², what is the difference between the free-surface heights at the two ends of the tank?
Options

Question

A rectangular tank of length 2.0 m contains water and is accelerated horizontally toward the right with constant acceleration 7.5 m/s². No liquid spills out. Taking g = 10 m/s², what is the difference between the free-surface heights at the two ends of the tank?

Options

G1 Physics · Accepted Answer

In the frame of the accelerating tank, the liquid is in hydrostatic equilibrium under effective gravity. For horizontal acceleration a, tanθ = a/g, where θ is the angle made by the free surface with the horizontal. For a tank of horizontal length L, Δh/L = tanθ. Therefore, Δh = La/g = 2.0×7.5/10 = 1.5 m. The free surface is higher at the left end and lower at the right end. Correct option: C.