A cubical block of side a and density rho sli | vedclass

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(A) Since the block moves with constant velocity,the net force acting on it is zero.The component of the gravitational force acting down the incline i

Vedclass · Accepted Answer

$\frac{ho agt \sin 	heta}{v}$

Vedclass · Answer

(A) Since the block moves with constant velocity,the net force acting on it is zero.The component of the gravitational force acting down the incline is $F = mg \sin 	heta$.Here,$m = ho V = ho a^3$,so $F = ho a^3 g \sin 	heta$.According to Newton's law of viscosity,the viscous force is $F = \eta A \left(\frac{dv}{dx}ight)$,where $A = a^2$ is the contact area and $\frac{dv}{dx} = \frac{v}{t}$ is the velocity gradient.Equating the forces: $ho a^3 g \sin 	heta = \eta a^2 \left(\frac{v}{t}ight)$.Solving for the coefficient of viscosity $\eta$: $\eta = \frac{ho a^3 g \sin 	heta \cdot t}{a^2 v} = \frac{ho agt \sin 	heta}{v}$.

$A$ cubical block of side $a$ and density $\rho$ slides over a fixed inclined plane with constant velocity $v$. There is a thin film of viscous fluid of thickness $t$ between the plane and the block. Then the coefficient of viscosity of the thin film will be

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