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A beaker containing a liquid of density p moveswith an acceleration 'a'. The pressure due toliquid at a depth h below free surface of the liquids(1) hpgthe(2) hp (g -a)(4) 2hpg1 tag +(3) hp (g + a) |
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Answer» As the beaker is moving up with acceleration, so we have to consider pseudo force on the pressure element. Or simply in P= hdg , g will change to its apparent value g'=(g+a) So pressure at depth h will be, P= h d (g+a) |
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