First slide

Fluid statics

Question

A rectangular box containing water is accelerated upwards at 3 m / s² on a inclined plane making 30" to the horizontal as shown in fig. The side of the free liquid surface is

Moderate

A

0.23
B

$1 / \sqrt{3}$
C

$\sqrt{3}$
D

0.32

Solution

see fig
The slope is given by
$\tan θ = \frac{a_{x}}{a_{y} + g} = \frac{acos 30^{\circ}}{asin 30^{\circ} + g} = \frac{3 \times (\sqrt{3} / 2)}{(3 / 2) + 9 \cdot 8} = \frac{3 \sqrt{3}}{22 \cdot 6} = 0 \cdot 23$

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