Angle of inclination of the water surface from horizontal for the given accelerated container shown in the figure is [Assume g=10m/s2]
A
45∘
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B
30∘
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C
60∘
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D
70∘
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Solution
The correct option is A45∘ For a horizontally accelerated fluid, the inclination of the free surface of fluid is given by, tan θ=ag...(i)
The above result comes after applying equilbrium condition on the liquid mass with respect to container, due to the reason that free surface of a liquid is always oriented perpendicular to the net force on fluid mass.
⇒tan θ=1010 ⇒tan θ=1 ∴θ=tan−11=45∘
Angle of inclination of the free surface of water with the horizontal is 45∘.