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 viscous fluid of thickness '$t$' between the plane and the block. Then the coefficient of viscosity of the thin film will be:

• A. $\frac{3\rho agt}{5v}$
• B. $\frac{4\rho agt}{5v}$
• C. $\frac{\rho agt}{5v}$
• D. none of these

Answer 1

Answer: (A)

$\frac{3\rho agt}{5v}$

We have the shear stress as $\tau =\frac{F}{A}$
Thus force F is due to the weight in the direction of sliding.
$F=a^3\rho gsin{37}^o$ and the surface area of the bottom part of the cube is $a^2$. Thus we get $\tau =\frac{3}{5}a\rho g$.
We have the velocity as v and the thickness of the fluid is given as t. Using the relation $\tau =\mu \frac{du}{dy}$ we get $\frac{3}{5}a\rho g=\mu \frac{v}{t}$
or
$\mu =\frac{3\rho agt}{5v}$