![JEE Main 2023 Physics, Mechanical Properties of Fluids — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Mechanical_Properties_of_Fluids/648b5a71417cc3fb48d6865d/question_1__q_648b5a71417cc3fb48d6865d__cdn-question-pool.getmarks.app__a528e366-fdce-4841-b6ea-3e4d9f760505__84f58b0b2c_final_ppt_sync.png) The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at $A$ is $1.5{\mathrm{cm}}^{2},$ and $B$ is $25{\mathrm{mm}}^{2},$ if the speed of liquid at $B$ is $60\mathrm{cm}{s}^{-1}$ then $({P}_{A}–{P}_{B})$ is (Given ${P}_{A}$ and ${P}_{B}$ are liquid pressures at $A$ and $B$ points. Density $\rho =1000\mathrm{kg}{m}^{-3}$ $A$ and $B$ are on the axis of tube)

JEE Main 2023 Physics, Mechanical Properties… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

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JEE Main 2023 Physics, Mechanical Properties of Fluids — question figure

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at $A$ is $1.5{\mathrm{cm}}^{2},$ and $B$ is $25{\mathrm{mm}}^{2},$ if the speed of liquid at $B$ is $60\mathrm{cm}{s}^{-1}$ then $({P}_{A}–{P}_{B})$ is

(Given ${P}_{A}$ and ${P}_{B}$ are liquid pressures at $A$ and $B$ points.

Density $\rho =1000\mathrm{kg}{m}^{-3}$

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

Official previous-year question

Held on 13 Apr 2023 · Verified 6 Jul 2026.

Options

A
$135Pa$
B
$27Pa$
C
$175Pa$
D
$36Pa$

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Solution

Using Bernoulli's theorem,

${P}_{A}+\frac{1}{2}\rho {V}_{A}^{2}={P}_{B}+\frac{1}{2}\rho {V}_{B}^{2}...(i)$

Using equation of continuity,

${A}_{A}{V}_{A}={A}_{B}{V}_{B}$

So, $1.5\times {10}^{-4}\times {V}_{A}=25\times {10}^{-6}\times {V}_{B}$

${V}_{A}=\frac{25\times {10}^{-6}}{1.5\times {10}^{-4}}\times 0.6$

$=\frac{1}{10}m{s}^{-1}$

Substituting in equation (i)

$\Rightarrow {P}_{A}-{P}_{B}=\frac{1000}{2}[0.{6}^{2}-0.{1}^{2}]=\frac{1000}{2}\times 0.7\times 0.5$

$=175Pa$

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Solution

Using Bernoulli's theorem,

${P}_{A}+\frac{1}{2}\rho {V}_{A}^{2}={P}_{B}+\frac{1}{2}\rho {V}_{B}^{2}...(i)$

Using equation of continuity,

${A}_{A}{V}_{A}={A}_{B}{V}_{B}$

So, $1.5\times {10}^{-4}\times {V}_{A}=25\times {10}^{-6}\times {V}_{B}$

${V}_{A}=\frac{25\times {10}^{-6}}{1.5\times {10}^{-4}}\times 0.6$

$=\frac{1}{10}m{s}^{-1}$

Substituting in equation (i)

$\Rightarrow {P}_{A}-{P}_{B}=\frac{1000}{2}[0.{6}^{2}-0.{1}^{2}]=\frac{1000}{2}\times 0.7\times 0.5$

$=175Pa$

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics