Question

A cubical block of side ${}^{\prime }{a}^{\prime }$ and density ${}^{\prime }{\rho }^{\prime }$ slides over a fixed inclined plane with constant velocity ${}^{\prime }{v}^{\prime }$ . There is a thin film of viscous fluid of thickness ${}^{\prime }{t}^{\prime }$ between the plane and the block. Then, the coefficient of viscosity of the thin film will be:
(Acceleration due to gravity is $g$)

• A. $\frac{\rho ag\sin \theta t}{v}$
• B. $\frac{\rho {agt}^2\text{sin}\theta }{v}$
• C. $\frac{v}{\rho agt\sin \theta }$
• D. None of these

Answer 1

The correct option is A $\frac{\rho ag\sin \theta t}{v}$

The viscous force will act at the lower surface of the cubical block in a direction opposite to its velocity.
$v=\text{constant}$ $\Rightarrow a=0$
Applying the equilibrium condition along inclined plane on the block as per the FBD:


$F_v=mg\sin \theta ...\left(i\right)$
From the equation of viscosity,
$F_v=\eta A\frac{dv}{dz}$
The velocity gradient for the liquid film will be,
$\frac{dv}{dz}=\frac{v-0}{t}=\frac{v}{t}$
$\Rightarrow F_v=\eta A\frac{v}{t}...\left(ii\right)$
Equating both equations, we get
$\eta A\frac{v}{t}=mg\sin \theta$
$\Rightarrow \eta a^2\left(\frac{v}{t}\right)=\left(a^3\rho \right)\times g\sin \theta$
$\because A=a^2,V=a^3,\&m=V\times \rho$
$\therefore \eta =\frac{a\rho g\sin \theta t}{v}$