Answer :

Given:

Speed of water stream, v = 15 ms^{-1}

Cross-sectional area of the tube, A = 10^{-2} m^{2}

Density of water, ρ = 1000 Kgm^{-3}

Volume of water leaving the tube per second = A× v

V = A× v …(1)

V = 10^{-2} m^{2}× 15 ms^{-1}

V = 0.15 m^{3}s^{-1}

Mass of water flowing out of the pipe = density of fluid× volume

M = ρ× V …(2)

M = 1000 Kgm^{-3}× 0.15 m^{3}s^{-1}

M = 150 Kgs^{-1}

Since the water doesn’t rebound after striking the Wall, hence according to Newton’s 2^{nd} law , we can write,

F = Rate of Change of momentum

F = change in momentum / time

F = |mv-mu|/ t …(3)

Where,

M = mass of water

v = final velocity = 0

u = initial velocity

t = time = 1sec

the equation (3) becomes,

F = mu/t

By putting values in the above equation,

F = 150 Kgs^{-1}× 15 ms^{-1}

F = 2250 N.

The water exerts a force of 2250 N on the wall.

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