A square plate of 0.1 ; m side moves parallel | vedclass

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(D) The viscous force $F$ is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.Here, the area of the square plate $A = (0.1 \; m)^2 = 0.0

Vedclass · Accepted Answer

$0.0005$

Vedclass · Answer

(D) The viscous force $F$ is given by Newton's law of viscosity: $F = \eta A \frac{dv}{dx}$.Here, the area of the square plate $A = (0.1 \; m)^2 = 0.01 \; m^2$.The coefficient of viscosity $\eta = 0.01 \; 	ext{poise} = 0.001 \; 	ext{Pa} \cdot s$ (since $1 \; 	ext{poise} = 0.1 \; 	ext{Pa} \cdot s$).The velocity gradient is $\frac{dv}{dx} = \frac{v}{dx}$, where $v = 0.1 \; m/s$ and $dx$ is the distance between the plates.Substituting the values into the formula: $0.002 = 0.001 	imes 0.01 	imes \frac{0.1}{dx}$.Rearranging for $dx$: $dx = \frac{0.001 	imes 0.01 	imes 0.1}{0.002} = \frac{0.000001}{0.002} = 0.0005 \; m$.

$A$ square plate of $0.1 \; m$ side moves parallel to a second plate with a velocity of $0.1 \; m/s$, both plates being immersed in water. If the viscous force is $0.002 \; N$ and the coefficient of viscosity is $0.01 \; \text{poise}$, the distance between the plates in $m$ is:

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