
A stream of water flowing horizontally with speed of $15m{s^{ - 1}}$ gushes out of a tube of cross sectional area ${10^{ - 2}}{m^2}$, and hits a vertical wall normally. Assuming that it does not rebound from the wall, the force exerted on the wall by the impact of water is:
A) $1.25 \times {10^3}N$
B) $2.25 \times {10^3}N$
C) $3.25 \times {10^3}N$
D) $4.25 \times {10^3}N$
Answer
217.5k+ views
Hint: To find the solution for the given question, the velocity of stream of water and the area of cross sectional is given. We need to find the force exerted on the wall by the impact of water in which the stream of water hits vertically to the wall.
Complete step by step answer:
When the liquid is at equilibrium, then the force depends on the depth below the surface, the density of the liquid and the acceleration due to gravity.
The given
Velocity $v = 15m{s^{ - 1}}$
The area of cross sectional $A$$ = {10^{ - 2}}{m^2}$
The density of the water $\rho $$ = {10^3}kg{m^{ - 3}}$
The volume of water which hitting the wall per second is,
$ \Rightarrow $ $V = av$
$ \Rightarrow $ $V = 1 \times {10^{ - 2}}{m^2} \times 15m$
$ \Rightarrow $ $V = 15 \times {10^{ - 2}}$${m^3}{s^{ - 1}}$
In order to find the force, first we need to find momentum while impacting which we will need a mass of water impacting the wall.
Mass of the water which is hitting the wall per second is,
$ \Rightarrow $ $m = volume \times density$
$ \Rightarrow $ $m = 15 \times {10^{ - 2}}{m^3}{s^{ - 1}} \times {10^3}kg{m^{ - 3}}$
$ \Rightarrow $ $m = 150$\[kg{s^{ - 1}}\]
\[\]The initial momentum of the water which is hitting on the wall is $p$ $ = m \times V$
$ \Rightarrow $ $ = 150 \times 15$
$ \Rightarrow $ $ = 2250kgm{s^{ - 2}}$
$ \Rightarrow $ $ = 2.25 \times {10^3}$$N$
Hence the force exerted on the wall by the impact of water is $2.25 \times {10^3}$$N$.
Therefore the option (B) is correct.
Note: In mechanics the impact which has high force is applied over a short time period when there are two or more bodies colliding. This force or an acceleration has a greater effect than of a lower applied over a proportional over a longer period. The deformation of the slow down distance is important and key for limiting the force acting on the passenger while in a car crash.
Complete step by step answer:
When the liquid is at equilibrium, then the force depends on the depth below the surface, the density of the liquid and the acceleration due to gravity.
The given
Velocity $v = 15m{s^{ - 1}}$
The area of cross sectional $A$$ = {10^{ - 2}}{m^2}$
The density of the water $\rho $$ = {10^3}kg{m^{ - 3}}$
The volume of water which hitting the wall per second is,
$ \Rightarrow $ $V = av$
$ \Rightarrow $ $V = 1 \times {10^{ - 2}}{m^2} \times 15m$
$ \Rightarrow $ $V = 15 \times {10^{ - 2}}$${m^3}{s^{ - 1}}$
In order to find the force, first we need to find momentum while impacting which we will need a mass of water impacting the wall.
Mass of the water which is hitting the wall per second is,
$ \Rightarrow $ $m = volume \times density$
$ \Rightarrow $ $m = 15 \times {10^{ - 2}}{m^3}{s^{ - 1}} \times {10^3}kg{m^{ - 3}}$
$ \Rightarrow $ $m = 150$\[kg{s^{ - 1}}\]
\[\]The initial momentum of the water which is hitting on the wall is $p$ $ = m \times V$
$ \Rightarrow $ $ = 150 \times 15$
$ \Rightarrow $ $ = 2250kgm{s^{ - 2}}$
$ \Rightarrow $ $ = 2.25 \times {10^3}$$N$
Hence the force exerted on the wall by the impact of water is $2.25 \times {10^3}$$N$.
Therefore the option (B) is correct.
Note: In mechanics the impact which has high force is applied over a short time period when there are two or more bodies colliding. This force or an acceleration has a greater effect than of a lower applied over a proportional over a longer period. The deformation of the slow down distance is important and key for limiting the force acting on the passenger while in a car crash.
Recently Updated Pages
Elastic Collision in Two Dimensions Explained Simply

Elastic Collisions in One Dimension Explained

Electric Field Due to a Uniformly Charged Ring Explained

Electric Field of Infinite Line Charge and Cylinders Explained

Electric Flux and Area Vector Explained Simply

Electric Field of a Charged Spherical Shell Explained

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

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

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

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

Understanding Atomic Structure for Beginners

