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Q.

A flat plate moves normally with a speed  ν1 towards a horizontal jet of water of uniform area of cross section. The jet discharges water at the rate of volume V per second at a speed of  ν2. The density of water is ρ  . Assume that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the force acting on the plate due to the jet in newton is...

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a

F=ν1ν2ν1+ν22ρ

b

F=ν2ν1ν1ν22ρ

c

F=νν2ν1+ν22ρ

d

F=ν2νν2ν12ρ

answer is C.

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Detailed Solution

Force acting on the plate  F=dpdt=udmdt

Since   Aν2=Vdmdt=Aν1+ν2ρ=Vν2ν1+ν2ρ

(ur=ν1+ν2= velocity of water coming out of jet w.r.t plate)

F=ν1+ν2.νν2ν1+ν2ρ=νν2ν1+ν22ρ

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