Answer :

Given:

Speed of water stream, v = 15 ms-1


Cross-sectional area of the tube, A = 10-2 m2


Density of water, ρ = 1000 Kgm-3


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


V = A× v …(1)


V = 10-2 m2× 15 ms-1


V = 0.15 m3s-1


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


M = ρ× V …(2)


M = 1000 Kgm-3× 0.15 m3s-1


M = 150 Kgs-1


Since the water doesn’t rebound after striking the Wall, hence according to Newton’s 2nd 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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