Question

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

• A. ${\tan }^{-1}\left[\frac{a}{g\cos \alpha }\right]$
• B. ${\tan }^{-1}\left[\frac{a+g\sin }{g\cos \alpha }\right]$
• C. ${\tan }^{-1}\left[\frac{a-g\sin \alpha }{g(1+\cos \alpha )}\right]$
• D. ${\tan }^{-1}\left[\frac{a-g\sin \alpha }{g(1-\cos \alpha }\right]$

Answer 1

The correct option is B ${\tan }^{-1}\left[\frac{a+g\sin }{g\cos \alpha }\right]$

Angle of inclination of water surface for such case is given by

$\tan \theta =\frac{a_h}{g_{eff}}$

where $a_h$ is net acceleration in direction of motion and $g_{eff}$ is

effective gravity .

so from diagram $a_h=a+g\sin \alpha$

and $g_{eff}=g\cos \alpha$

putting values $\tan \theta =\frac{a+g\sin \alpha }{g\cos \alpha }$

so $\theta ={\tan }^{-1}\left[\frac{a+g\sin \alpha }{g\cos \alpha }\right]$