A flat plate moves normally with a speed v₁ 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 v₂. 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 of water is

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$ρ V v_{1}$

$ρ [\frac{V}{v_{2}}] {(v_{1} + v_{2})}^{2}$

$ρ V (v_{1} + v_{2})$

$\frac{ρ V}{v_{1} + v_{2}} v_{1}^{2}$

answer is D.

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Detailed Solution

$Force acting on plate, F = \frac{d p}{d t} = v (\frac{d m}{d t})$

Mass of water reaching the plate per second,

$\frac{d m}{d t} = A v ρ = A (v_{1} + v_{2}) ρ = \frac{V}{v_{2}} (v_{1} + v_{2}) ρ$

$(v = v_{1} + v_{2} = velocity of water coming out of jet w.r.t. plate)$

$(A = Area of cross-section of jet = \frac{V}{v_{2}})$

$\begin{array}{l} ∴ F = \frac{d m}{d t} v = \frac{V}{v_{2}} (v_{1} + v_{2}) ρ \times (v_{1} + v_{2}) \\ = ρ [\frac{V}{v_{2}}] {(v_{1} + v_{2})}^{2} \end{array}$

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