The perpendicular force to flow exerted by a jet on stationary inclined flat plate = _____ , when velocity of jet = 20m/s, angle between jet and plate = 60° and area of cross section of jet = 0.01 m2

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SSC JE CE Previous Year Paper 10 (Held on 30 Oct 2020 Evening)
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  1. 1732.05 N
  2. 1682.03 N
  3. 1542.05 N
  4. 1563.03 N

Answer (Detailed Solution Below)

Option 1 : 1732.05 N
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Detailed Solution

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Concept:

Vr = relative velocity of the jet with respect to the plate

  • When the plate is moving (u) in the direction of the jet (V) i.e. away from the jet: Vr = V - U
  • When the plate is moving (U) in the opposite direction of the jet (V) i.e. towards the jet: Vr = V + U

Force exerted by the jet on the plate in the direction normal to the plate:

Force exerted by the jet on the plate in the direction of the jet (Fx) and is perpendicular to the direction of flow (Fy)

Calculation:

Given, velocity of jet = 20 m/s 

Angle between jet and plate = 60° and area of cross-section of jet = 0.01 m2

Flate plate is stationary, Then

Relative velocity of the jet with respect to the plate(Vr) = 20 - 0 = 20 m/s

The perpendicular force to flow exerted by a jet on a stationary inclined flat plate,

Fy = ρAVr2 sin θ cos θ

Fy = 1000 × 0.01 × 202 × sin 60 × cos 60

Fy = 1732.05 N.

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