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Fluid Mechanics

Section: Physics

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Question : 2 of 18

Marks: +1, -0

Water flowing through a pipe of area of cross-section

2 \times 10^{-3} \text{m}^2

hits a vertical wall horizontally with a velocity of

12 \text{m} \text{s}^{-1}

. If the water does not rebound after hitting the wall, then the force acting on the wall due to water is

[AP EAPCET 23 May 2025 S2]

$24 \text{N}$
$144 \text{N}$
$288 \text{N}$
$72 \text{N}$

Solution:

Step 1: Find the volume of water hitting the wall each second.

The formula for flow rate (volume per second) is:

Volume per second

=

Area × Velocity

Here, Area

=2 \times 10^{-3} \text{m}^2

and Velocity

=12 \text{ms}^{-1}

Volume per second

=2 \times 10^{-3} \times 12=24 \times 10^{-3} \text{m}^3/\text{s}

Step 2: Find the mass of water hitting the wall each second.

To get the mass per second, multiply the volume flow by the density of water (

\rho=1000 \text{kg}/\text{m}^3

):

Mass per second

=

Density × Volume per second

=1000 \times 24 \times 10^{-3}=24 \text{kg}/\text{s}

Step 3: Calculate the force on the wall.

Force is the change in momentum per second. The water hits the wall at velocity

12 \text{ms}^{-1}

and stops instantly.

Force

=

Mass per second × Velocity

=24 \times 12=288 \text{N}

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