The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at A is $1.5$ cm$^2$, and B is $25$ mm$^2$, if the speed of liquid at B is $60$ cm/s then $(P_A - P_B)$ is : (Given $P_A$ and $P_B$ are liquid pressures at A and B points. Density $\rho=1000$ kg m$^{-3}$ A and B are on the axis of tube

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Fluid Mechanics

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

^{2}

, and B is

25

^{2}

, if the speed of liquid at B is

60

cm/s then

(P_{A} - P_{B})

is :
(Given

P_{A}

and

P_{B}

are liquid pressures at A and B points.
Density

ρ = 1000

kg m

^{- 3}

A and B are on the axis of tube

Answer: (A)

Using the continuity equation

A_{A} v_{A} = A_{B} v_{B}

1.5 \times v_{A} = 0.25 \times 60

v_{A} = \frac{15}{1.5} = 10 cm/s = 0.1 m/s

Applying Bernoulli's equation (horizontal tube, same height):

P_{A} + \frac{1}{2} ρ v_{A}^{2} = P_{B} + \frac{1}{2} ρ v_{B}^{2}

P_{A} - P_{B} = \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) = \frac{1}{2} \times 1000 \times [(0.6)^{2} - (0.1)^{2}]

= 500 \times (0.36 - 0.01) = 500 \times 0.35 = 175 Pa

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option