Liquid Friction!

Consider an elemental ring of radius $r$ in the disk. This elemental ring is moving with a velocity $v = \omega r$ in the liquid. The area of this element is $dA= 2 \pi r dr$ . Now the liquid in contact with this elemental ring would also have the velocity of the same. So now by Newton's formula, the force on this ring is given by : $$ $dF = \large{ 2\frac{\eta A v}{h} = 2\frac{ \eta wr 2\pi r dr}{h} }$ We can obtain the power due to this force $(P=Fv)$ , and on simplifying we get $$ $dP=\large{ \frac{4\pi \eta \omega^2 r^4 dr} {h} }$ Integrating this result form $r=0$ to $R$ , we get the power developed as $$ $P= \large{\frac{\pi \eta \omega^2 R^4 } {h}} = 179.945$

|dF|=\eta (2\pi xdx)(\frac { x\omega }{ h\ } )*2\quad \quad \quad \quad \quad \quad (F=\eta A\frac { dv }{ dx } )\quad \quad (force\quad multiplied\quad by\quad two\quad as\quad there\quad are\quad two\quad surfaces)\ dP=dF.v\quad where\quad v=x\omega \ substituting\quad the\quad values\quad ,we\quad get:\ 4\pi \frac { \eta }{ h } { \omega }^{ 2 }\int _{ 0 }^{ R }{ { x }^{ 3 } } dx\quad =\quad \pi \eta { \omega }^{ 2 }{ R }^{ 4 }/h