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A fluid container is containing a liquid of density $\rho $ is accelerating upward with acceleration a along the inclined place of inclination $\alpha$ as shown. Then the angle of inclination $ \theta $ of free surface is :
${\tan ^{ - 1}}\left[ {\frac{a}{{g\cos \alpha }}} \right]$
${\tan ^{ - 1}}\left[ {\frac{{a + g\sin \alpha }}{{g\cos \alpha }}} \right]$
${\tan ^{ - 1}}\left[ {\frac{{a - g\sin \alpha }}{{g(1 + \cos \alpha )}}} \right]$
${\tan ^{ - 1}}\left[ {\frac{{a - g\sin \alpha }}{{g(1 - \cos \alpha )}}} \right]$
Solution
Apply pseudo force on a particle of mass $m .$ Net force along the surface is zero.
$m a \cos \theta=m g \cos [90-(\theta-\alpha)]$
$\frac{a}{g} \cos \theta=g \sin (\theta-\alpha)$
$\frac{a}{g}=\frac{\sin \theta \cos \alpha}{\cos \theta}-\frac{\cos \theta \sin \alpha}{\cos \theta}$
$\tan \theta=\frac{\frac{a}{g}+\sin \alpha}{\cos \alpha}=\frac{a+g \sin \alpha}{g \cos \alpha}$
