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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.

**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.

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