Gujarati
Hindi
9-1.Fluid Mechanics
normal

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

A

$g/3$

B

$2g/3$

C

$3g/2$

D

None

Solution

$A=$ 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 $L^{3}$

$\Rightarrow \frac{\tan \theta}{2}=\frac{1}{3} \Rightarrow \tan \theta=2 / 3=\mathrm{a} / \mathrm{g}$

$a=2 g / 3$

Standard 11
Physics

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