Question

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 (Acceleration due to gravity is $g$):

• A. $\eta =\frac{\rho agt\sin \theta }{V}$
• B. $\frac{\rho ag{t}^2\sin \theta }{V}$
• C. $\frac{V}{\rho agt\sin \theta }$
• D. none of these

Answer 1

The correct option is A $\eta =\frac{\rho agt\sin \theta }{V}$

Since the block moves with constant velocity, net force F on the block is zero.
The forces acting on the block are:
$mg\sin \theta =a^3\rho g\sin \theta g$ ....(1) downward along the inclined plane
Viscous force $F_{viscous}=\eta A\frac{dV}{dy}=\eta \ast a^2\ast \frac{V}{t}$ .....(2) where
$A=a^2$ is the are of a face of the cube which is in contact with the surface of the inclined plane.
$\eta$ is the coefficient of viscosity
$\frac{V}{t}$ is the velocity gradient
Equating (1) and (2),
$a^3\rho g\sin \theta g=\eta \ast a^2\ast \frac{V}{t}$
$\Rightarrow \eta =\frac{t\rho g\sin \theta a}{V}$