![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
What is the pressure on the swimmer who is 10m below the surface of the lake?
Answer
117.3k+ views
Hint: Pressure on the swimmer will be due to the atmospheric pressure and the column of water of height 10m above him. Pressure is given by the product of height which in this case in depth, density and acceleration due to gravity. Substitute the values in the formula \[P = h\rho g\] and simplify to get the pressure on the swimmer.
Complete step-by-step solution
When an object is below any fluid, it experiences some pressure from the top. This pressure is due to the column of fluid above that body and it is given by:
\[P = h\rho g\]
Where P is the pressure exerted on the body
H is the height of the fluid column
\[\rho \] is the density of the fluid
g is the acceleration due to gravity.
When a swimmer is swimming 10m below the surface of water, he experiences pressure due to the water column above him in addition to the atmospheric pressure. The atmospheric pressure is caused due to the envelope of air above the earth and its value is 1 atm or \[1.01 \times {10^5}\] Pa. So the total pressure on the swimmer will be
\[
P = 1.01 \times {10^5} + h\rho g \\
P = 1.01 \times {10^5} + 10 \times 1000 \times 9.81 = 1.01 \times {10^5} + 0.98 \times {10^5} \\
P = 1.99 \times {10^5} \\
\]
Therefore, the correct answer is \[1.99 \times {10^5}\] pascal
Note
The value of atmospheric pressure can be appreciated by imagining it using the following analogy.
We know that the atmospheric pressure \[P = 1.01 \times {10^5}Pa\] . Let’s say that a human standing on ground takes the area \[A = 1{m^2}\] . So to equate the pressure on this area, the weight required is w=101000N (assuming g=10 \[m/{s^2}\] ). So effectively this is equivalent to 100 motorcycles. Which means, on a daily basis we carry ~100 motorcycles on our heads.
Complete step-by-step solution
When an object is below any fluid, it experiences some pressure from the top. This pressure is due to the column of fluid above that body and it is given by:
\[P = h\rho g\]
Where P is the pressure exerted on the body
H is the height of the fluid column
\[\rho \] is the density of the fluid
g is the acceleration due to gravity.
When a swimmer is swimming 10m below the surface of water, he experiences pressure due to the water column above him in addition to the atmospheric pressure. The atmospheric pressure is caused due to the envelope of air above the earth and its value is 1 atm or \[1.01 \times {10^5}\] Pa. So the total pressure on the swimmer will be
\[
P = 1.01 \times {10^5} + h\rho g \\
P = 1.01 \times {10^5} + 10 \times 1000 \times 9.81 = 1.01 \times {10^5} + 0.98 \times {10^5} \\
P = 1.99 \times {10^5} \\
\]
Therefore, the correct answer is \[1.99 \times {10^5}\] pascal
Note
The value of atmospheric pressure can be appreciated by imagining it using the following analogy.
We know that the atmospheric pressure \[P = 1.01 \times {10^5}Pa\] . Let’s say that a human standing on ground takes the area \[A = 1{m^2}\] . So to equate the pressure on this area, the weight required is w=101000N (assuming g=10 \[m/{s^2}\] ). So effectively this is equivalent to 100 motorcycles. Which means, on a daily basis we carry ~100 motorcycles on our heads.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How to find Oxidation Number - Important Concepts for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How Electromagnetic Waves are Formed - Important Concepts for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electrical Resistance - Important Concepts and Tips for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Average Atomic Mass - Important Concepts and Tips for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Chemical Equation - Important Concepts and Tips for JEE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Login 2045: Step-by-Step Instructions and Details
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Class 11 JEE Main Physics Mock Test 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main Chemistry Question Paper with Answer Keys and Solutions
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids
![arrow-right](/cdn/images/seo-templates/arrow-right.png)