Question

Sol. Key concept: Free surface of a liquid in a horizontally accelerating container.

Use L = 4 m for the cube side and a = 1.96 m s⁻² for its acceleration. In the accelerating frame, the free surface is perpendicular to the resultant of gravity and the horizontal pseudo-acceleration. Thus its slope has magnitude a/g.

$$\Delta\ h = L\frac{a}{g} = 4 \times \frac{1.96}{9.8} = 0.8\ m$$

After spilling, the higher end of the surface touches the rim. The empty space is a triangular prism of width L, base L and height Δh.

$$V_{spill} = \frac{1}{2}L^{2}\Delta\ h = \frac{1}{2}(4)^{2}(0.8) = 6.4\ m^{3}$$

Correction / clarification: Using different symbols for side length and acceleration removes the conflicting use of a in the previous working. The answer remains (c).