A metal plate of area 10^{-2} ,m^2 rests on a | vedclass

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(B) The viscous force $F$ acting on the plate is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.Here, $\eta = 1.55 \,Ns\, m^{-2}$ is t

Vedclass · Accepted Answer

$0.2325$

Vedclass · Answer

(B) The viscous force $F$ acting on the plate is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.Here, $\eta = 1.55 \,Ns\, m^{-2}$ is the coefficient of viscosity, $A = 10^{-2} \,m^2$ is the area of the plate, $v = 3 	imes 10^{-2} \,ms^{-1}$ is the velocity, and $h = dx = 2 	imes 10^{-3} \,m$ is the thickness of the oil layer.Substituting the values: $F = 1.55 	imes 10^{-2} 	imes \frac{3 	imes 10^{-2}}{2 	imes 10^{-3}}$.$F = 1.55 	imes 10^{-2} 	imes 1.5 	imes 10^1$.$F = 1.55 	imes 1.5 	imes 10^{-1} = 2.325 	imes 10^{-1} = 0.2325 \,N$.

$A$ metal plate of area $10^{-2} \,m^2$ rests on a layer of castor oil, $2 \times 10^{-3} \,m$ thick, whose coefficient of viscosity is $1.55 \,Ns\, m^{-2}$. The approximate horizontal force required to move the plate with a uniform speed of $3 \times 10^{-2} \,ms^{-1}$ is (in $N$)

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