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JEE Main 20268 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2026 · 8 April · Evening Shift · Physics · Mechanical Properties of Fluids

A liquid of density 600 kg/m ^3 flowing steadily in a tube of...

A liquid of density 600600 kg/m3^3 flowing steadily in a tube of varying cross-section. The cross-section at a point AA is 1.01.0 cm2^2 and that at BB is 2020 mm2^2. Both the points AA and BB are in same horizontal plane, the speed of the liquid at AA is 1010 cm/s. The difference in pressures at AA and BB points is ________ Pa.

Options

  1. A1818
  2. B144144
  3. C3636
  4. D7272

Correct answer

D. 7272

Step-by-step solution

Using the equation of continuity:

AAvA=ABvBA_A v_A = A_B v_B

(1.0×10−4)×0.1=(20×10−6)×vB(1.0 \times 10^{-4}) \times 0.1 = (20 \times 10^{-6}) \times v_B

vB=10−52×10−5=0.5v_B = \dfrac{10^{-5}}{2 \times 10^{-5}} = 0.5 m/s

Using Bernoulli's equation for a horizontal tube:

PA+12ρvA2=PB+12ρvB2P_A + \dfrac{1}{2} \rho v_A^2 = P_B + \dfrac{1}{2} \rho v_B^2

PA−PB=12ρ(vB2−vA2)P_A - P_B = \dfrac{1}{2} \rho (v_B^2 - v_A^2)

Substituting the given values:

PA−PB=12×600×(0.52−0.12)P_A - P_B = \dfrac{1}{2} \times 600 \times (0.5^2 - 0.1^2)

PA−PB=300×(0.25−0.01)P_A - P_B = 300 \times (0.25 - 0.01)

PA−PB=300×0.24=72P_A - P_B = 300 \times 0.24 = 72 Pa

Answer: 7272
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