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Worked Examples · Example 2.11

Q.The given figure shows a streamline flow of a non-viscous liquid having density 1000 kg/m³. The cross sectional area at point A is 2 cm² and at point B is 1 cm². The speed of liquid at the point A is 5 cm/s. Both points A and B are at the same horizontal level. Calculate the difference in pressure at A and B.

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Figure — Horizontal streamline channel narrowing from point A to point B
FigureHorizontal streamline channel narrowing from point A to point B

A₁v₁=A₂v₂ then (P1−P2)=12ρ(v22−v12)(P_1-P_2) = \frac{1}{2}\rho(v_2^2-v_1^2).

From continuity v2=A1v1/A2=2×5/1=10v_2 = A_1v_1/A_2 = 2\times5/1 = 10 cm/s. With h2=h1h_2 = h_1, Bernoulli gives (P1−P2)=12ρ(v22−v12)=12×1000×[(0.10)2−(0.05)2]=12×1000×(100−25)×10−4=3.75 Pa(P_1-P_2) = \frac{1}{2}\rho(v_2^2-v_1^2) = \frac{1}{2}\times1000\times\left[(0.10)^2-(0.05)^2\right] = \frac{1}{2}\times1000\times(100-25)\times10^{-4} = 3.75\ \text{Pa}

Note

The printed Solution carries the squared speeds as plain (100−25) without the 10−410^{-4} from (cm/s)² → (m/s)² and so prints P1−P2=37500P_1-P_2 = 37500 Pa =3.75×104= 3.75\times10^4 Pa. With the speeds correctly converted (5 cm/s = 0.05 m/s, 10 cm/s = 0.10 m/s), the difference is 3.753.75 Pa. The printed value corresponds to speeds of 5 m/s and 10 m/s.

✓Final answer

P1−P2=12ρ(v22−v12)=12×1000×[(0.10)2−(0.05)2]=3.75P_1 - P_2 = \frac{1}{2}\rho(v_2^2-v_1^2) = \frac{1}{2}\times1000\times\left[(0.10)^2-(0.05)^2\right] = 3.75 Pa (the book's printed 3.75×1043.75\times10^4 Pa comes from a cm→m slip — see the note in the full solution).

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