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(D) Consider a fluid element of mass $m$ at the surface. The forces acting on it are gravity ($mg$ downwards),pseudo force ($ma$ backwards),and the no

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$A$ and $C$ both

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(D) Consider a fluid element of mass $m$ at the surface. The forces acting on it are gravity ($mg$ downwards),pseudo force ($ma$ backwards),and the normal force from the fluid below (perpendicular to the surface).For the surface to be in equilibrium in the non-inertial frame of the beaker,the net force must be perpendicular to the surface.The slope of the water surface is given by $	an 	heta = \frac{	ext{vertical force}}{	ext{horizontal force}} = \frac{ma}{mg} = \frac{a}{g}$.Thus,$	heta = 	an^{-1}(a/g)$.Since the beaker accelerates in the $+x$ direction,the water level rises at the back and falls at the front,making the angle $	heta$ with the horizontal towards the back of the beaker.Note that $	an^{-1}(a/g) = \cot^{-1}(g/a)$,so both options $A$ and $C$ represent the same physical angle.

$A$ beaker filled with water is accelerated at $a \ m/s^2$ in the $+x$ direction. The surface of the water shall make an angle:

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