- Home
- Standard 11
- Physics
9-1.Fluid Mechanics
hard
A plastic circular disc of radius $R$ is placed on a thin oil film, spread over a flat horizontal surface. The torque required to spin the disc about its central vertical axis with a constant angular velocity is proportional to
A
$R^2$
B
$R^3$
C
$R^4$
D
$R^6$
Solution
Let coefficient of viscosity $\&$ thickness of film
$\rightarrow \eta \&$ x respectively
Viscous force on element ring is
$\mathrm{dF}=\eta(2 \pi \mathrm{rdr})\left[\frac{\mathrm{wr}-0}{\mathrm{x}}\right]$
$\mathrm{d} \tau=(\mathrm{dF}) r$
$\therefore \tau=\frac{\eta 2 \pi \mathrm{w}}{\mathrm{t}} \int_{0}^{\mathrm{R}} \mathrm{r}^{3} \cdot \mathrm{dr}=\left(\frac{\eta(2 \pi) \mathrm{w}}{4 \mathrm{x}}\right) \mathrm{R}^{4}$
$\therefore \tau \propto R^{4}$
Standard 11
Physics
