A metal cube of side 10 ,cm rests on a film o | vedclass

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$(A)$ Given:Side of the cube, $L = 10 \,cm = 0.1 \,m$.Area of the base, $A = L^2 = (0.1 \,m)^2 = 0.01 \,m^2$.Thickness of the liquid film, $dx = 0.2 \

Vedclass · Accepted Answer

$2.5 	imes 10^{-2} \frac{Ns}{m^2}$

Vedclass · Answer

$(A)$ Given:Side of the cube, $L = 10 \,cm = 0.1 \,m$.Area of the base, $A = L^2 = (0.1 \,m)^2 = 0.01 \,m^2$.Thickness of the liquid film, $dx = 0.2 \,mm = 0.2 	imes 10^{-3} \,m$.Applied force, $F = 0.1 \,N$.Constant velocity, $v = 0.08 \,m/s$.Since the cube moves with a constant speed, the net force is zero, meaning the applied force equals the viscous drag force: $F = F_{drag}$.According to Newton's law of viscosity, $F = \eta A \frac{dv}{dx}$.Substituting the values: $0.1 = \eta 	imes 0.01 	imes \frac{0.08}{0.2 	imes 10^{-3}}$.$0.1 = \eta 	imes 0.01 	imes 400$.$0.1 = \eta 	imes 4$.$\eta = \frac{0.1}{4} = 0.025 \,Ns/m^2 = 2.5 	imes 10^{-2} \,Ns/m^2$.

$A$ metal cube of side $10 \,cm$ rests on a film of a liquid of thickness $0.2 \,mm$. If upon applying a horizontal force $\overrightarrow{F}$ of magnitude $0.1 \,N$ the cube slides with a constant speed of $0.08 \,m/s$, then the coefficient of viscosity is

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