Physics Q33 – JEE-MAIN 2026

Home Physics Mechanical Properties of FluidsQuestion #11

A liquid of density

600 {kg/m}^{3}

flowing steadily in a tube of varying cross-section. The cross-section at a point A is

1.0 {cm}^{2}

and that at B is

20 {mm}^{2}

. 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.

Clue 1 of 3: Fluid Dynamics Principles

This problem involves the steady flow of an incompressible fluid through a tube of varying cross-section, requiring the application of fundamental fluid dynamics principles.

🧠 Full Solution Path

Step 1: Convert Units and Calculate Velocity at B✦ Active

First, convert all given values to SI units:

A_{A} = 1.0 {cm}^{2} = 1.0 \times 10^{- 4} m^{2}

A_{B} = 20 {mm}^{2} = 20 \times (10^{- 3})^{2} m^{2} = 2.0 \times 10^{- 5} m^{2}

v_{A} = 10 cm/s = 0.1 m/s

ρ = 600 {kg/m}^{3}

Apply the Equation of Continuity ( $A_{A} v_{A} = A_{B} v_{B}$ ) to find the velocity at point B ( $v_{B}$ ):

(1.0 \times 10^{- 4} m^{2}) \times (0.1 m/s) = (2.0 \times 10^{- 5} m^{2}) \times v_{B}

v_{B} = \frac{1.0 \times 10^{- 5}}{2.0 \times 10^{- 5}} = 0.5 m/s

Step 2: Apply Bernoulli's Principle○ Expand

Since points A and B are in the same horizontal plane, their heights are equal ( $h_{A} = h_{B}$ ). Bernoulli's equation ( $P + \frac{1}{2} ρ v^{2} + ρ g h = constant$ ) simplifies to:

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

Rearrange the equation to find the pressure difference $P_{A} - P_{B}$ :

P_{A} - P_{B} = \frac{1}{2} ρ v_{B}^{2} - \frac{1}{2} ρ v_{A}^{2} = \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2})

💡 Teacher's Secret Hint

Remember to cancel out the potential energy term if the points are at the same height.

Step 3: Calculate the Pressure Difference○ Expand

Substitute the known values into the simplified Bernoulli's equation:

P_{A} - P_{B} = \frac{1}{2} \times 600 {kg/m}^{3} \times ((0.5 m/s)^{2} - (0.1 m/s)^{2})

P_{A} - P_{B} = 300 \times (0.25 - 0.01)

P_{A} - P_{B} = 300 \times 0.24

P_{A} - P_{B} = 72 Pa

💡 Teacher's Secret Hint

Ensure all calculations are performed with consistent units to avoid errors.

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Physics Question 33 – JEE-MAIN 2026

Clue 1 of 3: Fluid Dynamics Principles

🧠 Full Solution Path