A beaker filled with water is given a horizontal acceleration a (in the +x direction). The free surface of the water tilts and makes an angle with the horizontal equal to:

tan⁻¹(a/g) backwards
tan⁻¹(a/g) forwards
cot⁻¹(g/a) backwards
cot⁻¹(g/a) forwards

Correct answer: tan⁻¹(a/g) backwards

Solution

In the frame of the beaker the effective gravity is the vector sum of g (down) and the pseudo-acceleration a (in -x, opposite to the real acceleration). The free surface is perpendicular to g_eff, so it makes angle theta with horizontal where tan(theta) = a/g. The surface is higher at the rear (backwards). Note cot⁻¹(g/a) = tan⁻¹(a/g) numerically, but the standard stated answer is tan⁻¹(a/g) backwards.

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