
The height of a mercury barometer is 75cm at sea level and 50cm at the top of a hill. Ratio of density of mercury to that of air is 104. The hill is:
A) 1.25Km
B) 2.5Km
C) 250m
D) 750m
Answer
233.1k+ views
Hint: The above problem is based on the concept of pressure exerted by liquid when the liquid is at rest which is equal to the height of the liquid density of the liquid and gravitational acceleration.
$P = \rho gh$ ($\rho $is the density of the liquid, h is the height of the liquid and g is the acceleration).
Using the above mentioned relation we will find the height of the hill.
Complete step by step solution:
Let us discuss some of the properties of the pressure exerted by the liquid kept in a vessel.
When a liquid of some density is kept in a vessel with height h. The weight of the liquid column exerts a downward force and thus downward pressure and hence we have the formula:
$P = \rho gh$
Let us do the calculation now:
Pressure difference caused due to sea level and top of the hill;
$ \Rightarrow \Delta P = ({h_1} - {h_2}){\rho _m}g$ (Height at sea level and hill top is subtracted and ${\rho _m}$ is density of mercury)
$
\Rightarrow \Delta P = (75 - 50){\rho _m}g \\
\Rightarrow \Delta P = 25{\rho _m}g \\
$(We substituted the numerical value of height)..................(1)
Pressure difference due to air with height h
$ \Rightarrow \Delta P = {\rho _a}hg$ ..................(2)
Equating equation 1 and 2
$
\Rightarrow {\rho _a}hg = 25{\rho _m}hg \\
\Rightarrow h = 25 \times {10^4} \\
$(Ratio of density of mercury to air is 104)
The above calculated quantity is in cm to convert it in m we will multiply by $10^{-2}$
$
\Rightarrow h = 25 \times {10^4} \times {10^{ - 2}} \\
\Rightarrow h = 2.5Km \\
$
Note: We will have many other applications where pressure has different effects for example a sharp knife cuts easily while blunt knife take much force in cutting, when we stand on our feet pressure is more than we lay down, sharp needle is easily able to pierce the skin rather than a dull needle, camel can easily walk on sand than human beings because area of feet of camel is larger than human being.
$P = \rho gh$ ($\rho $is the density of the liquid, h is the height of the liquid and g is the acceleration).
Using the above mentioned relation we will find the height of the hill.
Complete step by step solution:
Let us discuss some of the properties of the pressure exerted by the liquid kept in a vessel.
When a liquid of some density is kept in a vessel with height h. The weight of the liquid column exerts a downward force and thus downward pressure and hence we have the formula:
$P = \rho gh$
Let us do the calculation now:
Pressure difference caused due to sea level and top of the hill;
$ \Rightarrow \Delta P = ({h_1} - {h_2}){\rho _m}g$ (Height at sea level and hill top is subtracted and ${\rho _m}$ is density of mercury)
$
\Rightarrow \Delta P = (75 - 50){\rho _m}g \\
\Rightarrow \Delta P = 25{\rho _m}g \\
$(We substituted the numerical value of height)..................(1)
Pressure difference due to air with height h
$ \Rightarrow \Delta P = {\rho _a}hg$ ..................(2)
Equating equation 1 and 2
$
\Rightarrow {\rho _a}hg = 25{\rho _m}hg \\
\Rightarrow h = 25 \times {10^4} \\
$(Ratio of density of mercury to air is 104)
The above calculated quantity is in cm to convert it in m we will multiply by $10^{-2}$
$
\Rightarrow h = 25 \times {10^4} \times {10^{ - 2}} \\
\Rightarrow h = 2.5Km \\
$
Note: We will have many other applications where pressure has different effects for example a sharp knife cuts easily while blunt knife take much force in cutting, when we stand on our feet pressure is more than we lay down, sharp needle is easily able to pierce the skin rather than a dull needle, camel can easily walk on sand than human beings because area of feet of camel is larger than human being.
Recently Updated Pages
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

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

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

