Question

A cubical block (of side $2\text{m}$) of mass $20\text{kg}$ slides on an inclined plane, lubricated with the oil of viscosity $\eta ={10}^{-1}\text{poise}$, with a constant velocity of $10\text{m/s}$. The thickness of layer of liquid will be, $(\text{Take}g=10{\text{m/s}}^2)$

• A. $2\text{mm}$
• B. $3\text{mm}$
• C. $4\text{mm}$
• D. $5\text{mm}$

Answer 1

The correct option is C $4\text{mm}$

We know the viscous force $F$ is given by,
$F=\eta A\frac{dv}{dz}$

$\Rightarrow mg\sin \theta =F$ (since acceleration of block is zero)
$\Rightarrow 20\times 10\times \sin {30}^{\circ }=\eta \times (2{)}^2\times \frac{10}{h}$

Here, $\eta ={10}^{-1}\text{poise}={10}^{-2}{\text{N-s-m}}^{-2}$

$h=\frac{40\times {10}^{-2}}{100}=4\times {10}^{-3}\text{m}=4\text{mm}$


Hence, $\left(C\right)$ is the correct answer.