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(B) Consider a fluid particle of mass $m$ on the free surface. In the non-inertial frame of the container,the effective forces acting on the particle

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$	an^{-1}\left[\frac{a+g\sin\alpha}{g\cos\alpha}ight]$

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(B) Consider a fluid particle of mass $m$ on the free surface. In the non-inertial frame of the container,the effective forces acting on the particle are gravity ($mg$ downwards),pseudo force ($ma$ directed down the incline),and the normal force from the fluid below.The free surface of the liquid must be perpendicular to the effective acceleration vector $\vec{g}_{eff}$.The effective acceleration $\vec{g}_{eff} = \vec{g} - \vec{a}$.Resolving components along and perpendicular to the incline:Component of $\vec{g}$ perpendicular to the incline = $g\cos\alpha$ (downwards).Component of $\vec{g}$ parallel to the incline = $g\sin\alpha$ (down the incline).Component of $\vec{a}$ parallel to the incline = $a$ (up the incline).Total effective acceleration parallel to the incline $a_{||} = g\sin\alpha + a$ (downwards along the incline).Total effective acceleration perpendicular to the incline $a_{\perp} = g\cos\alpha$ (downwards).The angle $	heta$ that the free surface makes with the horizontal (or the incline,depending on the diagram definition) is given by the ratio of the components of the effective acceleration.From the geometry,$	an	heta = \frac{a_{||}}{a_{\perp}} = \frac{a + g\sin\alpha}{g\cos\alpha}$.Thus,$	heta = 	an^{-1}\left[\frac{a+g\sin\alpha}{g\cos\alpha}ight]$.

Column-$I$	Column-$II$
$(A)$ Hydraulic lift	$(P)$ Archimedes' principle
$(B)$ Razor blade can be made to float on the surface of water	$(Q)$ Pascal's law
$(C)$ The dam of a reservoir is made thick at the bottom	$(R)$ Surface tension
$(D)$ Ship is floating on the ocean	$(S)$ Pressure

$A$ fluid container containing a liquid of density $\rho$ is accelerating upward with acceleration $a$ along an inclined plane of inclination $\alpha$ as shown. Then the angle of inclination $\theta$ of the free surface is:

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