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A rectangular container carrying liquid moves down an inclined plane, of inclination \mathrm{30^{\circ}} with the horizontal, with a constant acceleration \mathrm{3 \mathrm{~m} / \mathrm{s}^{\wedge} 2} down the incline. The angle made by the free surface of the liquid with the incline is

Option: 1

30^{\circ}


Option: 2

\mathrm{\tan ^{-1} \frac{2}{\sqrt{3}}}


Option: 3

\mathrm{\tan ^{-1} \frac{2}{5 \sqrt{3}}}


Option: 4

\mathrm{\tan ^{-1} \frac{10}{\sqrt{3}}}


Answers (1)

best_answer

\mathrm{\perp} inclined \mathrm{ \mathrm{g} \cos 30^{\circ}=5 \sqrt{3}}

Along inclined \mathrm{g \sin \theta \sqcup 3=2}

For surface \mathrm{\tan \theta=\frac{2}{5 \sqrt{3}}}, from inclined.

Posted by

Ritika Jonwal

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