Question

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

• A. $2.5mm$
• B. $6mm$
• C. $4mm$
• D. $5.5mm$

Answer 1

The correct option is C $4mm$

We know that
$F=\eta A\left(\frac{dv}{dz}\right)$
Given, side of the block, $a=2m$
Mass of the block, $m=20kg$
Viscosity of the oil,
$\eta ={10}^{-1}poise={10}^{-2}Nm/s^2$
Velocity of the block, $v=10m/s$
Let $h$ be the thickness of the layer of the liquid.
the block moves with constant velocity,
From $FBD$ of the block,
$F=F^{\prime }=mg\sin \theta$
$\eta A\frac{dv}{dz}=mg\sin \theta$
Here, velocity gradient, $\frac{dv}{dz}=\frac{v}{h}$
$\eta A\frac{v}{h}=mg\sin \theta$
$\frac{{10}^{-2}\times (2{)}^2\times 10}{h}=20\times 10\times \sin {30}^{\circ }$
$h=4\times {10}^{-3}m=4mm$
Final answer: (b)