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Question: A layer of glycerine of thickness 1mm is enclosed between a big plate and another plane of area \({1...

A layer of glycerine of thickness 1mm is enclosed between a big plate and another plane of area 10−2m2{10^{ - 2}}{m^2}. If coefficient of viscosity of glycerine is 1kg/(m−s)1kg/(m-s), then the force in newton required to move plate with a velocity of 0.07m/s0.07m/s will be:
A) 77
B) 0.70.7
C) 7070
D) 0.070.07

Explanation

Solution

The viscosity of a fluid represents the obstruction offered to the flow of the fluid by the subsequent layers. The layers of the fluid undergo shear stress when the force is applied to move the fluid. The relationship is given by:
Newton’s law of viscosity i.e. τ=μdudy\tau = \mu \dfrac{{du}}{{dy}}
μ\mu = coefficient of viscosity of the fluid and dudy\dfrac{{du}}{{dy}}= velocity gradient.

Complete step by step by solution:
Given that, layer thickness
dy=1mm=10−3mdy = 1mm = {10^{ - 3}}m
Area A=10−2m2A = {10^{ - 2}}{m^2}
Coefficient of viscosity μ=1kg/ms\mu = 1kg/ms
Velocity du=0.07m/sdu = 0.07m/s
Using the equation of Newton’s law of viscosity i.e. τ=μdudy\tau = \mu \dfrac{{du}}{{dy}} where
μ\mu = dynamic viscosity of the fluid, and dudy\dfrac{{du}}{{dy}}= velocity gradient.
The shear stress is given as the ratio of the force applied to the area on which the force is applied.
Shear stress,
τ=FA\tau = \dfrac{F}{A}
where F= force and A= Area
Substituting,
⇒\Rightarrow F=μdudyAF = \mu \dfrac{{du}}{{dy}}A
Put all the values in the above equation.
⇒\Rightarrow F=(1)×0.0710−3×10−2=0⋅7NF=(1)\times \dfrac{0.07}{{{10}^{-3}}}\times {{10}^{-2}}=0\cdot 7N

Which is required Force; Hence option (B) is the correct option.

Additional information: Dynamic viscosity is the force required to maintain the fluid flow at a particular rate. Dynamic viscosity is usually used when the fluid is subjected to any external force. The unit of dynamic viscosity is Pa-s, where Pa is N/m2N/{m^2}- SI unit of pressure.
Kinematic viscosity is defined as dynamic viscosity to the density of the fluid. It shows the rate of momentum transfer (how fast the fluid is moving) when the fluid is subjected to deformation. The unit of Kinematic viscosity is m2/s{m^2}/s in SI units.

Note: Not all fluids follow Newton's law of viscosity. There is a special category of fluids called non-Newtonian fluids where the Newton’s law of viscosity is not applicable. Example of such a fluid is quicksand.