Practice question
Question
Water (ρ\=1000kg/m3) flows horizontally at 4.5m/s with pressure 2.0×105Pa. If the speed increases to 7.0m/s, what is the new pressure?
Explanation
Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 2.0×105Pa, v1 = 4.5m/s, v2 = 7.0m/s, ρ = 1000kg/m3. P2 = 2.0×105+12×1000(20.25−49) = 2.0×105−14375 = 1.85625×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.86 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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