Fluid MechanicsHardBloom L3

Question

A beaker filled with water is accelerated at a m/s² in the +x direction. The surface of the water shall make an angle

Options

A.tan⁻¹(a/g) backwards
B.tan⁻¹(a/g) forwards
C.cot⁻¹(g/a) backwards
D.cot⁻¹(g/a) forwards

Solution

For a liquid in a horizontally accelerated container tan θ = a/g, so θ = tan⁻¹(a/g) measured backwards, which is the same angle as cot⁻¹(g/a) backwards.

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