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A thin smooth plate 1 m wide and 2 m long is towed through water at a velocity of 2 m/s. Assuming that boundary remains laminar, then drag on both sides of the plate is (kinematic viscosity = 10-6 m2/s)
  • a)
    5.3 N
  • b)
    26.6 N
  • c)
    53 N
  • d)
    72.5 N
Correct answer is option 'A'. Can you explain this answer?
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A thin smooth plate 1 m wide and 2 m long is towed through water at a ...

The drag force on both sides of the plate,
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A thin smooth plate 1 m wide and 2 m long is towed through water at a ...
Given:

Width of the plate, b = 1 m

Length of the plate, L = 2 m

Velocity of the plate, V = 2 m/s

Kinematic viscosity, v = 10^-6 m^2/s

To find:

Drag force on both sides of the plate

Assumptions:

1. The flow is laminar i.e. Reynolds number (Re) < 2300="" />

2. The plate is thin i.e. thickness < length="" and="" of="" the="" plate="" />

Formula used:

Drag force on both sides of the plate, F = Cd * (ρ*V^2/2) * A

Where,

Cd = Drag coefficient

ρ = Density of water

V = Velocity of the plate

A = Surface area of the plate

Surface area of both sides of the plate, A = 2*b*L = 2*1*2 = 4 m^2

Density of water, ρ = 1000 kg/m^3

Calculation:

Reynolds number, Re = (ρ*V*L)/v = (1000*2*2)/10^-6 = 4*10^6

Since Re > 2300, the flow is turbulent. But as per the given question, the flow is laminar.

Therefore, we need to assume that the flow is laminar i.e. Re < 2300.="" />

For laminar flow, the drag coefficient is given as:

Cd = 1.328/√Re

Cd = 1.328/√(1000*2*2/10^-6) = 0.053

Substituting the given values in the formula:

F = Cd * (ρ*V^2/2) * A

F = 0.053 * (1000*2^2/2) * 4

F = 5.3 N

Therefore, the drag force on both sides of the plate is 5.3 N.

Hence, option (a) is the correct answer.
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A thin smooth plate 1 m wide and 2 m long is towed through water at a velocity of 2 m/s. Assuming that boundary remains laminar, then drag on both sides of the plate is (kinematic viscosity = 10-6 m2/s)a)5.3 Nb)26.6 Nc)53 Nd)72.5 NCorrect answer is option 'A'. Can you explain this answer?
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