
The force acting on a window of area $50cm$×$50cm$ of a submarine at a depth of $2000m$ in an ocean, the interior of which is maintained at sea level atmospheric pressure is (density of seawater = ${10^3}kg/{m^3}$ , g = $10m{s^{ - 2}}$ )
A) $5 \times {10^5}N$
B) $25 \times {10^5}N$
C) $5 \times {10^6}N$
D) $25 \times {10^6}N$
Answer
233.1k+ views
Hint: Calculate the net pressure on the window by subtracting pressure on different sides of the window from each other. Take the absolute value of the answer afterwards. Then use the force formula to calculate net force acting on the window which will be the final answer.
Formula Used: In an open container filled with liquid, the net pressure on the walls = $P$ = ${P_0} + \rho g{h_{}}$
$Force = {\text{Pressure}} \times {\text{Area}}$
Complete step by step answer:
Consider the sea as an open container and the wall of submarine will be equivalent to the walls of the container. Therefore, net pressure exerted on the walls from the outside of the submarine will be equal to ${P_0} + \rho g{h_{}}$. Where ${P_0}$ is the sea level atmospheric pressure and $\rho gh$ is the pressure exerted by just the liquid on the walls of submarine, where, $\rho $ is the density of water, and $h$ is the depth of submarine in the ocean.
On the inside, it is given that sea level atmospheric pressure is maintained. Therefore, pressure on the inside = ${P_0}$ .
Now, on subtracting the equations as mentioned above we get,
Net Pressure $(P)= {P_0} + \rho g{h_{}}$−${P_0}$ which gives $P= \rho gh$.
Now, we are given that $\rho = {10^3}kg/{m^3}$ , $g= 10m{s^{ - 2}}$ , $h= 2000m$.
Therefore, $P$= ${10^3} \times 10 \times 2000=2 \times {10^7}Kg{m^{ - 1}}{s^{ - 2}}$
Now we multiply this with the area of the window to get the final answer.
Therefore, net force acting on the window = $2 \times {10^7}Kg{m^{ - 1}}{s^{ - 2}}$ × \[0.5 \times 0.5{m^2}\]
$ \Rightarrow F= 5 \times {10^6}N$
The correct answer is (C) $5 \times {10^6}N$
Note: Don’t forget to convert given values to SI units. In an open container, to calculate the net pressure on the walls, the sea level atmospheric pressure is also added to the pressure exerted by the liquid on the walls. Take the absolute value of pressure obtained by the difference whenever required.
Formula Used: In an open container filled with liquid, the net pressure on the walls = $P$ = ${P_0} + \rho g{h_{}}$
$Force = {\text{Pressure}} \times {\text{Area}}$
Complete step by step answer:
Consider the sea as an open container and the wall of submarine will be equivalent to the walls of the container. Therefore, net pressure exerted on the walls from the outside of the submarine will be equal to ${P_0} + \rho g{h_{}}$. Where ${P_0}$ is the sea level atmospheric pressure and $\rho gh$ is the pressure exerted by just the liquid on the walls of submarine, where, $\rho $ is the density of water, and $h$ is the depth of submarine in the ocean.
On the inside, it is given that sea level atmospheric pressure is maintained. Therefore, pressure on the inside = ${P_0}$ .
Now, on subtracting the equations as mentioned above we get,
Net Pressure $(P)= {P_0} + \rho g{h_{}}$−${P_0}$ which gives $P= \rho gh$.
Now, we are given that $\rho = {10^3}kg/{m^3}$ , $g= 10m{s^{ - 2}}$ , $h= 2000m$.
Therefore, $P$= ${10^3} \times 10 \times 2000=2 \times {10^7}Kg{m^{ - 1}}{s^{ - 2}}$
Now we multiply this with the area of the window to get the final answer.
Therefore, net force acting on the window = $2 \times {10^7}Kg{m^{ - 1}}{s^{ - 2}}$ × \[0.5 \times 0.5{m^2}\]
$ \Rightarrow F= 5 \times {10^6}N$
The correct answer is (C) $5 \times {10^6}N$
Note: Don’t forget to convert given values to SI units. In an open container, to calculate the net pressure on the walls, the sea level atmospheric pressure is also added to the pressure exerted by the liquid on the walls. Take the absolute value of pressure obtained by the difference whenever required.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

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

