A liquid of density 600 kg/m³ flowing steadily in a tube of varying cross-section. The cross-section at a point A is 1.0 cm² and that at B is 20 mm². Both the points A and B are in same horizontal plane, the speed of the liquid at A is 10 cm/s. The difference in pressures at A and B points is ______ Pa. (a) 18 (b) 144 (c) 36 (d) 72

Answer: (d). AA = 1.0 cm² = 100 mm², AB = 20 mm², so vB = 5vA = 50 cm/s = 0.5 m/s Bernoulli: PA - PB = 1/2ρ(vB² - vA²) = 1/2×600×(0.25 - 0.01) PA - PB = 300 × 0.24 = 72 Pa So the answer is (d) 72.

A liquid of density 600 kg/m³ flowing steadily in a tube of varying… | JEE Main 2026 (8 Apr, Shift 2)

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Bernoulli's Equation

A liquid of density $600\text{ kg/m}^3$ flowing steadily in a tube of varying cross-section. The cross-section at a point $A$ is $1.0\text{ cm}^2$ and that at $B$ is $20\text{ mm}^2$ . Both the points $A$ and $B$ are in same horizontal plane, the speed of the liquid at $A$ is $10\text{ cm/s}$ . The difference in pressures at $A$ and $B$ points is ______ Pa.

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Answer: (d)

$A_A = 1.0\text{ cm}^2 = 100\text{ mm}^2$ , $A_B = 20\text{ mm}^2$ , so $v_B = 5v_A = 50\text{ cm/s} = 0.5\text{ m/s}$
Bernoulli: $P_A - P_B = \frac{1}{2}\rho(v_B^2 - v_A^2) = \frac{1}{2}\times600\times(0.25 - 0.01)$
$P_A - P_B = 300 \times 0.24 = 72\text{ Pa}$
So the answer is (d) 72.

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