A vessel containing water has a constant acceleration of 19.6m/s2 along horizontal direction. The free surface of water will be inclined with the horizontal at an angle of : [Take g=9.8m/s2]
A
tan−1[12]
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B
sin−1[1√3]
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C
tan−1[√2]
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D
sin−1[2√5]
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Solution
The correct option is Dsin−1[2√5] The free surface of the liquid will get tilted and become perpendicular to the net force (of pseudo force & weight) on the fluid mass.
For horizontally accelerated fluid, tan θ=mamg=ag...(i) ⇒tanθ=19.69.8=2 ∴tanθ=2