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Question

Angle of inclination of the water surface from horizontal for the given accelerated container shown in the figure is [Assume g=10 m/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 A 45
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=tan11=45
Angle of inclination of the free surface of water with the horizontal is 45.

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