Question

A rectangular metal plate has dimensions of $10\text{cm}\times 20\text{cm}$. A thin film of oil separates the plate from a fixed horizontal surface. The separation between the lower surface of the rectangular plate and the upper portion of the horizontal surface is $0.2\text{mm}$. An ideal string is attached to the plate and passes over an ideal pulley to a mass $m$. When $m=125\text{gm}$, the metal plate moves at a constant speed of $5\text{cm/s}$ across the horizontal surface. Then, the coefficient of viscosity of oil in ${\text{dyne-s/cm}}^2$ is $(\text{Use}g=1000{\text{cm/s}}^2)$

• A. $5$
• B. $25$
• C. $2.5$
• D. $50$

Answer 1

The correct option is C $2.5$

The coefficient of viscosity is the ratio of shear stress (tangential stress ) at the top surface of the film (exerted by the block) to that of the velocity gradient of the film.

Since mass $m$ moves with constant velocity i.e $a=0$
or, $T=mg$, for equilibrium condition
The string will exert a force equal to $mg$ on the plate towards right.
Hence, oil shall exert an equal but opposite tangential force on the plate towards left.
$\therefore F_v=mg$
Applying equation of viscosity,
$\Rightarrow \frac{F_v}{A}=\eta \frac{dv}{dz}...\left(i\right)$
where $\frac{dv}{dz}=\frac{v-0}{\Delta z}$
$\Delta z=0.2\text{mm}=0.02\text{cm}$, represents the thickness of oil film between the lower surface of the plate and the upper portion of the horizontal surface.

Substituting in Eq.$\left(i\right)$,
$\eta =\frac{\frac{F_v}{A}}{\frac{(v-0)}{\Delta z}}=\frac{\frac{125\times 1000}{10\times 20}}{\frac{(5-0)}{0.02}}$
$\because m=125\text{gm},g=1000{\text{cm/s}}^2,A=20\times 10{\text{cm}}^2$
$\Rightarrow \eta =2.5{\text{dyne -s/cm}}^2$