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(B) The viscous force $F$ acting on a plate moving in a fluid is given by $F = \eta A \frac{dv}{dx}$.Here,the plate is placed midway between two surfa

Vedclass · Accepted Answer

$2.5 	imes 10^{-2}$

Vedclass · Answer

(B) The viscous force $F$ acting on a plate moving in a fluid is given by $F = \eta A \frac{dv}{dx}$.Here,the plate is placed midway between two surfaces,so the distance from each surface is $dx = \frac{2.5 \ cm}{2} = 1.25 \ cm = 1.25 	imes 10^{-2} \ m$.The velocity gradient is $\frac{dv}{dx} = \frac{v}{dx} = \frac{0.5 \ m/s}{1.25 	imes 10^{-2} \ m} = \frac{0.5}{0.0125} = 40 \ s^{-1}$.Since the plate is moving,it experiences viscous drag from both sides. Thus,the total force is $F = F_{top} + F_{bottom} = 2 	imes (\eta A \frac{dv}{dx})$.Given $F = 1 \ N$,$A = 0.5 \ m^2$,and $\frac{dv}{dx} = 40 \ s^{-1}$,we have:$1 = 2 	imes \eta 	imes 0.5 	imes 40$$1 = 40 \eta$$\eta = \frac{1}{40} = 0.025 \ kg/(m \cdot s) = 2.5 	imes 10^{-2} \ kg/(m \cdot s)$.

$A$ space $2.5 \ cm$ wide between two large plane surfaces is filled with oil. The force required to drag a very thin plate of area $0.5 \ m^2$ just midway between the surfaces at a speed of $0.5 \ m/s$ is $1 \ N$. The coefficient of viscosity in $kg/(m \cdot s)$ is:

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