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When at rest a liquid stands at same level in the tubes shown . But as indicated a height difference of 'h' occurs when the system is given an acceleration towards right. then 'h' equals

$ a)\;\frac{aL}{2a} \\ b)\;\frac{gL}{2a} \\ c)\;\frac{gL}{a} \\d)\;\frac{aL}{g}$

Can you answer this question?
 
 

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The water surface while accelerated will lie as shown. with lines of constant pressure parallel to the surface.
 
Take an element of water between lines of constant pressure.
 
$A \Delta P \sin \theta =ma$-------(1)
$A \Delta P \cos \theta =mg$---------(2)
dividing (1) by (2)
$\tan \theta =\large\frac{a}{g}$
$\qquad=\large\frac{h}{L}$
$\therefore h=\large\frac{aL}{g}$
Hence d is the correct answer.

 

answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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