
The pressure gauge reading in metre of water column shown in the given figure will be:

(A) $3.20m$
(B) $2.72m$
(C) $2.52m$
(D) $1.52m$
Answer
220.2k+ views
Hint: In the given figure, the air exerts a pressure on the water, which in turn exerts pressure on the mercury, causing it to rise inside the U- tube. The gauge tells the differential pressure in the water column by relating it to the pressure head of the water.
Formula used:
$P = \rho gh$
Where $\rho $ is the density of the fluid.
$g$ is the acceleration due to gravity.
$h$ is the height of the column.
Complete step by step solution:
Here, the height of the column corresponds to the pressure of the different fluids present.
The pressure in the mercury column is equal to the pressure in water column, and is given by-
$P = \rho gh$
Let the pressure in the mercury column be ${P_1}$ and that in the water column be ${P_2}$. Then, assuming no pressure loss we have-
${P_1} = {P_2}$
Which can be written as-
${\rho _1}{h_1}g = {\rho _2}{h_2}g$
Where ${\rho _1}$is the density of mercury at room temperature which is about $13534kg/{m^3}$.
${\rho _2}$is the density of water at room temperature which is about $997.77kg/{m^3}$.
It is given that the height of the mercury column, ${h_1} = 20cm = 0.2m$
On equating both the quantities, we get-
$13534 \times 0.2 = 997.77{h_2}$
Where ${h_2}$ is the height of the water in the column.
${h_2} = \dfrac{{13534}}{{997.77}} \times 0.2$
${h_2} = 2.712m \approx 2.72m$
This reading will be shown by the gauge.
Hence, Option (B) is correct.
Note: Pressure head refers to the representation of pressure in terms of height of fluid raised or fallen. The common pressure measuring term in fluid mechanics, the $mm - Hg$represents pressure head because it tells the height of liquid mercury which would rise in a column provided the same pressure. The gauge here also tells the pressure head, as the meters of water are raised.
Formula used:
$P = \rho gh$
Where $\rho $ is the density of the fluid.
$g$ is the acceleration due to gravity.
$h$ is the height of the column.
Complete step by step solution:
Here, the height of the column corresponds to the pressure of the different fluids present.
The pressure in the mercury column is equal to the pressure in water column, and is given by-
$P = \rho gh$
Let the pressure in the mercury column be ${P_1}$ and that in the water column be ${P_2}$. Then, assuming no pressure loss we have-
${P_1} = {P_2}$
Which can be written as-
${\rho _1}{h_1}g = {\rho _2}{h_2}g$
Where ${\rho _1}$is the density of mercury at room temperature which is about $13534kg/{m^3}$.
${\rho _2}$is the density of water at room temperature which is about $997.77kg/{m^3}$.
It is given that the height of the mercury column, ${h_1} = 20cm = 0.2m$
On equating both the quantities, we get-
$13534 \times 0.2 = 997.77{h_2}$
Where ${h_2}$ is the height of the water in the column.
${h_2} = \dfrac{{13534}}{{997.77}} \times 0.2$
${h_2} = 2.712m \approx 2.72m$
This reading will be shown by the gauge.
Hence, Option (B) is correct.
Note: Pressure head refers to the representation of pressure in terms of height of fluid raised or fallen. The common pressure measuring term in fluid mechanics, the $mm - Hg$represents pressure head because it tells the height of liquid mercury which would rise in a column provided the same pressure. The gauge here also tells the pressure head, as the meters of water are raised.
Recently Updated Pages
Mass vs Weight: Key Differences Explained for Students

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

