An open cubical tank was initially fully filled with water. When the tank was accelerated on a horizontal plane along one of its side it was found that one third of volume of water spilled out. The acceleration was
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$A=$ area of the base $\tan \theta=a / g$
Finally $1 / 3$ rd of the water spilled out So volume of water spilled out finally
$=V f=$$\frac{2 \tan \theta \times A}{2}=\frac{L^{3} \tan \theta}{2}$
this is $1 / 3$ volume of $L^{3}$
$\Rightarrow \frac{\tan \theta}{2}=\frac{1}{3} \Rightarrow \tan \theta=2 / 3=\mathrm{a} / \mathrm{g}$
$a=2 g / 3$
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