The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at A is 1.5 cm², and B is 25 mm², if the speed of liquid at B is 60 cm/s then (PA - PB) is: (Given PA and PB are liquid pressures at A and B points. density ρ = 1000 kg m⁻³ A and B are on the axis of tube (a) 27 Pa (b) 175 Pa (c) 135 Pa (d) 36 Pa

Answer: (b). AA = 1.5 cm² = 150 mm², AB = 25 mm². Continuity: vA = 25/150× 0.6 = 0.1 m/s Bernoulli: PA - PB = 1/2×1000×(0.36 - 0.01) = 500×0.35 PA - PB = 175 Pa So the answer is (b).

The figure shows a liquid of given density flowing steadily in… | JEE Main 2023 (13 Apr, Shift 1)

Q17JEE Main 2023 · 13 APR · SHIFT 1Medium

Bernoulli's Equation

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at $A$ is $1.5\text{ cm}^2$ , and $B$ is $25\text{ mm}^2$ , if the speed of liquid at $B$ is $60\text{ cm/s}$ then $(P_A - P_B)$ is:

(Given $P_A$ and $P_B$ are liquid pressures at $A$ and $B$ points.

density $\rho = 1000\text{ kg m}^{-3}$

$A$ and $B$ are on the axis of tube

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Answer: (b)

$A_A = 1.5\text{ cm}^2 = 150\text{ mm}^2$ , $A_B = 25\text{ mm}^2$ .
Continuity: $v_A = \dfrac{25}{150}\times 0.6 = 0.1\text{ m/s}$
Bernoulli: $P_A - P_B = \frac{1}{2}\times1000\times(0.36 - 0.01) = 500\times0.35$
$P_A - P_B = 175\text{ Pa}$
So the answer is (b).

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Bernoulli's Equation question from JEE Main 2023 (13 Apr, Shift 1)

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