Question

a layer of glycerine of thickness 0.5mm is present between a large fixed surface area A and a surface area B( area of B= 0.1m^2). the small surface is being pulled with a constant velocity of 1m/s as shown in the figure. find the viscous force acting on surface B(in N). (Given that coefficient of viscosity =0.07kgm^-1s^-1 and large surface remains at rest)​

Answer 1

Answer:

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Explanation:

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Answer 2

The viscous force acting on surface B is 70,000 N.

The viscous force acting on surface B can be calculated using the equation for shear stress, which is given by:

τ = η ×  v / h

where:

τ = shear stress (N/m^2)

η = coefficient of viscosity (kgm^-1s^-1)

v = velocity gradient (m/s)

h = height of the glycerine layer (m)

Since the velocity gradient is given by v/h, we can write:

$v/h = v / 0.0005 m = 2000 m/s$

So the shear stress can be calculated as:

τ = η v / h

= $0.07 kg/m/s * 2000 m/s / 0.0005 m$

= 700,000 N/m^2

Finally, the viscous force acting on surface B can be calculated by multiplying the shear stress by the surface area:

F = τ × A

= $700,000 N/m^2 * 0.1 m^2 = 70,000 N$

So the viscous force acting on surface B is 70,000 N.

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