
A flat plate of area 10cm2 is separated from a large plate by a layer of glycerine 1mm thick. If the coefficient of viscosity of glycerine is 20 poise, the force required to keep the plate moving with a velocity of 1cm per sec is:
A) 80 dyne
B) 200 dyne
C) 800 dyne
D) 2000 dyne
Answer
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Hint:When the surface is rough and an object is sliding on that surface some heat will be generated due to the interaction between the block and the surface. This is called friction between them and it is a contact force. Not only in this case and in case of liquids also there will be this hindrance force and we call it as a viscosity.
Complete step by step answer:
Normally fluids will be of two types. They are Newtonian fluids and non-Newtonian fluids. The fluid which flows by obeying the newton laws are called Newtonian fluids. The fluids which don't obey Newton's law are called non Newtonian fluids. There exists friction between the layers of the fluid during the flow and it is called viscosity. There will be shear stress induced between the layers due to this friction and due to this stress, the shear force will be generated and that force will depend upon the area of cross section of the surface and coefficient of viscosity and velocity gradient i.e. rate of change of velocity with respect to the displacement.
That force F is given as:
$F = \mu A\dfrac{{dv}}{{dx}}$
Here µ is the viscosity, A is the surface area, dv is the change in velocity and dx is the thickness of the film.
Substitute the known values given above in the above equation. It comes out as:
$F = \dfrac{{20\,poise \times \dfrac{{{{10}^{ - 1}}\,Pa}}{{1\,poise}} \times 10\,c{m^2} \times \dfrac{{{{10}^{ - 4}}\,{m^2}}}{{1\,c{m^2}}} \times 1\,cm/s \times \dfrac{{{{10}^{ - 2}}\,m}}{{1\,cm}}}}{{1\,mm \times \dfrac{{{{10}^{ - 3}}\,m}}{{1\,mm}}}}$
$ \Rightarrow F = 0.02\,N \times \dfrac{{{{10}^5}\;dyne}}{{1\;N}}$
$\Rightarrow F= 2000\;dyne$
Hence, the correct option is D.
Note:Here when we go away from the axis the velocity keeps on decreasing and by the time the liquid reaches the surface liquid velocity will be very less due the more friction between the container surface and the liquid layer.
Complete step by step answer:
Normally fluids will be of two types. They are Newtonian fluids and non-Newtonian fluids. The fluid which flows by obeying the newton laws are called Newtonian fluids. The fluids which don't obey Newton's law are called non Newtonian fluids. There exists friction between the layers of the fluid during the flow and it is called viscosity. There will be shear stress induced between the layers due to this friction and due to this stress, the shear force will be generated and that force will depend upon the area of cross section of the surface and coefficient of viscosity and velocity gradient i.e. rate of change of velocity with respect to the displacement.
That force F is given as:
$F = \mu A\dfrac{{dv}}{{dx}}$
Here µ is the viscosity, A is the surface area, dv is the change in velocity and dx is the thickness of the film.
Substitute the known values given above in the above equation. It comes out as:
$F = \dfrac{{20\,poise \times \dfrac{{{{10}^{ - 1}}\,Pa}}{{1\,poise}} \times 10\,c{m^2} \times \dfrac{{{{10}^{ - 4}}\,{m^2}}}{{1\,c{m^2}}} \times 1\,cm/s \times \dfrac{{{{10}^{ - 2}}\,m}}{{1\,cm}}}}{{1\,mm \times \dfrac{{{{10}^{ - 3}}\,m}}{{1\,mm}}}}$
$ \Rightarrow F = 0.02\,N \times \dfrac{{{{10}^5}\;dyne}}{{1\;N}}$
$\Rightarrow F= 2000\;dyne$
Hence, the correct option is D.
Note:Here when we go away from the axis the velocity keeps on decreasing and by the time the liquid reaches the surface liquid velocity will be very less due the more friction between the container surface and the liquid layer.
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