A flat plate moves normally with a speed v 1 towards a horizontal jet of water of uniform area of crosssection. The jet discharges water at the rate of volume V per second at a speed of v 2 . The

Question

A flat plate moves normally with a speed $v_{1}$ towards a horizontal jet of water of uniform area of crosssection. The jet discharges water at the rate of volume V per second at a speed of $v_{2}$ . 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 jet of water is :-

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Answer:

B

Solution:

$F = \frac{Δ p}{Δ t} = \frac{m ( v _{1} + v _{2} )}{Δ t} = V_{ρ} (v_{1} + v_{2})$ where $\frac{m}{Δ t} = V =$ volume $/$ second .

TestHub · Accepted Answer

F = Δ t Δ p ​ = Δ t m ( v 1 ​ + v 2 ​ ) ​ = V ρ ​ ( v 1 ​ + v 2 ​ ) where Δ t m ​ = V = volume / second .

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