Question

26. Find the viscous drag between the two liquid layers each of area 100 cm, and having relative
velocity 8 cm s-1. The viscosity of the liquid is 0.004 Pl and the layers are separated by a
distance 4 cm.​

Answer 1

Answer:

the viscous force between the two liquid layers is $8 \times 10^{-5} N$

Explanation:

As we know that the drag force or viscous force between two layers of liquid is given as

$F = \eta A \frac{dv}{dx}$

here we know that

$A = 100 cm^2$ = Area of contact

$\eta = 0.004 Pl$ = coefficient of viscosity

$dv = 8 cm/s$ = relative speed of two layers

$dx = 4 cm$ = thickness of layers

now by above formula we will have

$F = (0.004)(100\times 10^{-4})(\frac{0.08}{0.04})$

$F = 8 \times 10^{-5} N$

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Topic : Viscous force between two layers

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