There is a 1, mm thick layer of glycerine bet | vedclass

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(A) To move the plate with a constant velocity,the applied force must be equal to the viscous drag force.The formula for viscous force is $F = \eta A

Vedclass · Accepted Answer

$0.7$

Vedclass · Answer

(A) To move the plate with a constant velocity,the applied force must be equal to the viscous drag force.The formula for viscous force is $F = \eta A \frac{\Delta v}{\Delta z}$.Given values:Coefficient of viscosity $\eta = 1.0\, kg/(m\cdot s)$.Area $A = 100\, cm^2 = 100 	imes 10^{-4}\, m^2 = 10^{-2}\, m^2$.Velocity gradient $\frac{\Delta v}{\Delta z} = \frac{7\, cm/s}{1\, mm} = \frac{7 	imes 10^{-2}\, m/s}{10^{-3}\, m} = 70\, s^{-1}$.Substituting these values into the formula:$F = 1.0 	imes 10^{-2} 	imes 70 = 0.7\, N$.Therefore,the force required is $0.7\, N$.

There is a $1\, mm$ thick layer of glycerine between a flat plate of area $100\, cm^2$ and a big plate. If the coefficient of viscosity of glycerine is $1.0\, kg/(m\cdot s)$,then the force required to move the plate with a velocity of $7\, cm/s$ is $.......\, N$.

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