Practice question
Question
Water (ρ\=1000kg/m3) flows horizontally at 3.2m/s with pressure 1.85×105Pa. If the speed increases to 5.8m/s, what is the new pressure?
Explanation
Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 1.85×105Pa, v1 = 3.2m/s, v2 = 5.8m/s, ρ = 1000kg/m3. P2 = 1.85×105+12×1000(10.24−33.64) = 1.85×105−11700 = 1.733×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.73 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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