A container of liquid release from the rest, on a smooth inclined plane as shown in the figure. Length of at the inclined plane is sufficient, and assume liquid finally equilibrium. Finally liquid surface makes an angle with horizontal ...... $^o$
Medium
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Acceleration of the container on smooth inclined plane is a $=g \sin \theta=5 \mathrm{ms}^{-2}$
Consider a particle of liquid on the liquid surface
$\tan \alpha=\frac{\operatorname{acos} \theta}{g-a \sin \theta}=\frac{5 \cos 30}{10-5 \sin 30}=\frac{5(\sqrt{3} / 2)}{10-5(1 / 2)}$
$=\frac{5 \sqrt{3} / 2}{15 / 2}$
$\tan \alpha=\frac{1}{\sqrt{3}}=\tan 30, \alpha=30$
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