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(B) The viscous drag force is given by $F = \eta A \frac{dv}{dx}$.Since the box moves with a constant speed,the applied force $F$ must be equal to the

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$1.1 	imes 10^{-1} \ Pa \cdot s$

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(B) The viscous drag force is given by $F = \eta A \frac{dv}{dx}$.Since the box moves with a constant speed,the applied force $F$ must be equal to the viscous drag force.Here,$F = \frac{1}{3} \ N$,$A = 0.01 \ m^2$,$dv = 0.09 \ m/s$,and $dx = 0.3 \ mm = 0.3 	imes 10^{-3} \ m$.Substituting these values into the formula:$\eta = \frac{F \cdot dx}{A \cdot dv}$$\eta = \frac{(1/3) 	imes (0.3 	imes 10^{-3})}{0.01 	imes 0.09}$$\eta = \frac{0.1 	imes 10^{-3}}{0.0009} = \frac{10^{-4}}{9 	imes 10^{-4}} = \frac{1}{9} \approx 0.11 \ Pa \cdot s$.Thus,$\eta \approx 1.1 	imes 10^{-1} \ Pa \cdot s$.

$A$ cylindrical metal box whose flat surface has an area of $0.01 \ m^2$ rests on a liquid layer of $0.3 \ mm$ thickness. If upon applying a horizontal force of magnitude $\frac{1}{3} \ N$,the box slides with a constant speed of $0.09 \ m/s$,the coefficient of viscosity of the liquid is nearly:

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