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# 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) $4.25 \times {10^3}N$

Last updated date: 23rd Jul 2024
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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.

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}}$
$\Rightarrow$ $V = av$
$\Rightarrow$ $V = 1 \times {10^{ - 2}}{m^2} \times 15m$
$\Rightarrow$ $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$.