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Question: A liquid having coefficient of viscosity \(0.02decapoise\) is filled in a container of cross-section...

A liquid having coefficient of viscosity 0.02decapoise0.02decapoise is filled in a container of cross-section area 20m220{{m}^{2}}. If viscous drag between two adjacent layers in flowing is 1N1N, then the viscous gradient is:
A)2.0s−1A)2.0{{s}^{-1}}
B)2.5s−1B)2.5{{s}^{-1}}
C)3.0s−1C)3.0{{s}^{-1}}
D)3.5s−1D)3.5{{s}^{-1}}

Explanation

Solution

Viscous drag refers to a force which acts opposite to the relative motion of any object moving with respect to a surrounding fluid. Viscous drag between adjacent layers in a fluid refers to the force, which is acting opposite to the relative motion of one layer of the fluid, flowing with respect to the other layer of the fluid. It is proportional to the coefficient of viscosity of the fluid, the cross-sectional area of the container in which the fluid is filled as well as the viscous gradient of the fluid.

Complete answer:
Viscous drag between adjacent layers in a liquid refers to the force which is acting opposite to the relative motion of one layer of the liquid, flowing with respect to its adjacent layer of the liquid. Mathematically, it is given by
F=μA(dvdx)F=\mu A\left( \dfrac{dv}{dx} \right)
where
FF is the viscous drag between adjacent layers of the liquid
μ\mu is the coefficient of viscosity of the liquid
AA is the cross-sectional area of the container in which the liquid is flowing
dvdx\dfrac{dv}{dx} is the viscous gradient of the liquid
Let this be equation 1.

We are supposed to find the viscous gradient of the liquid. Viscous gradient of a liquid measures the rate of shear deformation or shear velocity. On rearranging equation 1, viscous gradient can be expressed as
(dvdx)=FμA\left( \dfrac{dv}{dx} \right)=\dfrac{F}{\mu A}
Let this be equation 2.

From the question, we are provided that
μ=0.2decapoise=0.2Nsm−2\mu =0.2decapoise=0.2Ns{{m}^{-2}}
A=20m2A=20{{m}^{2}}
F=1NF=1N
Substituting these values in equation 2, we have
(dvdx)=FμA=1N0.02Nsm−2×20m2=10.4s−1=2.5s−1\left(\dfrac{dv}{dx}\right)=\dfrac{F}{\mu A}=\dfrac{1N}{0.02Ns{{m}^{-2}}\times20{{m}^{2}}}=\dfrac{1}{0.4}{{s}^{-1}}=2.5{{s}^{-1}}
Therefore, viscous gradient of the liquid is given by
(dvdx)=2.5s−1\left( \dfrac{dv}{dx} \right)=2.5{{s}^{-1}}

So, the correct answer is “Option B”.

Note:
Viscous gradient of a liquid measures the rate of shear deformation or shear velocity. Shear velocity is nothing but shear stress, rewritten in units of velocity. Shear stress refers to the component of stress coplanar with a material cross section. It arises from shear force, which is the component of force parallel to the material cross-section. Hence, shear velocity can be identified as the frictional velocity between the layers of a fluid.
Students need to be aware of the unit decapoise(daP)decapoise(daP). It is the unit of viscosity in CGS System of Units. The conversion of decapoise(daP)decapoise(daP) from CGS to SI is as given below: (CGS)1daP=10P=1Nsm−2(SI)(CGS)1daP=10P=1Ns{{m}^{-2}}(SI)