Practice question
Question
Water (ρ\=1000kg/m3) flows horizontally at 4.2m/s with pressure 1.95×105Pa. If the speed increases to 6.5m/s, what is the new pressure?
Explanation
Bernoulli’s equation: P1+12ρv12 = P2+12ρv22. P1 = 1.95×105Pa, v1 = 4.2m/s, v2 = 6.5m/s, ρ = 1000kg/m3. P2 = 1.95×105+12×1000(17.64−42.25) = 1.95×105−12305 = 1.82695×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.83 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
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