2 glass plates each of length 16cm and breadth 8cm move parallel to each other in WATER with a relative velocity of 6m/s. if the viscous force is 175×10^-5, what is their distance of separation? (given viscosity of water = 0.001PL)
Answers
Distance of separation between two plates is 43.89 mm.
Explanation:
=> Here, It is given that
Viscosity of water, η = 0.001 PL = 10⁻³ PL
Velocity, V = 6 m/s
Viscous force, F = 175×10⁻⁵
Length of glass plates, L = 16 cm = 16 * 10⁻² m
Breadth of glass plates, B = 8 cm = 8 * 10⁻² m
Distance of separation between two plates, l = ?
=> Thus, Area of glass plates , A:
A = L * B
= 16 * 10⁻² * 8 * 10⁻²
= 128 * 10⁻⁴
=> According to the formula of viscosity, η:
η = F*l / V*A
∴ l = η*V*A/F
= 10⁻³ * 6 * 128 * 10⁻⁴ / 175 * 10⁻⁵
= 43.89 mm
Thus, distance of separation between two plates is 43.89 mm.
Learn more:
Q:1 The viscous force 'F' acting on a small sphere of radius 'r' moving with velocity 'v' through a liquid is given by F = 6πɳrv. Calculate dimensions of ɳ.
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Q:2 To determine the coefficient of viscosity of a liquid by stokes method
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Explanation:
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