Question

A cubical block (side $2\text{m}$) of mass $20\text{kg}$ slides on inclined plane lubricated with the oil of viscosity $\eta ={10}^{-1}\text{Poise}$ with constant velocity of $10\text{m/s}$. Find out the thickness of the layer of liquid. $(g=10{\text{m/s}}^2)$

• A. $4\times {10}^{-3}\text{m}$
• B. $3\times {10}^{-3}\text{m}$
• C. $2\times {10}^{-3}\text{m}$
• D. $1\times {10}^{-3}\text{m}$

Answer 1

The correct option is A $4\times {10}^{-3}\text{m}$

We know that , $F_v=\eta A\frac{dv}{dz}=\eta A\frac{v}{h}$


Balancing forces along inlined plane,
$mg\sin \theta =F_v=\eta A\frac{v}{h}$
$\Rightarrow 20\times 10\times \sin {30}^{\circ }={10}^{-2}\times 4\times \frac{10}{h}$
$\Rightarrow h=4\times {10}^{-3}\text{m}=4\text{mm}$