Question
What is viscosity? What are the factors affecting viscous force in a liquid flowing in a tube? Derive the relation for the velocity upto which the liquid can have streamlined flow.
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Answer
Viscosity: The opposing force that exists between the layers of a liquid and the inner walls of the tube in which it flows is called viscous drag or viscous force and the property is called viscosity. The viscous force directly depends on the area of the layer and the velocity gradient.$\text{F}=-\eta\text{A}\frac{\text{dv}}{\text{dx}}$
-ve sign shows the opposing nature n refers to coefficient of viscosity. Factors affecting viscosity:
When terminal velocity is attained, acceleration should be zero and the ner force should be zero.$\therefore\text{mg}-\text{U}-\text{F}_\text{v}=0$
$\Rightarrow\frac43\pi\text{r}^3\rho\text{g}-\frac43\pi\text{r}^3\rho_\text{l}\text{g}-6\pi\eta\text{ rv}=0$
$\therefore\text{v}=\frac{\frac43\pi\text{r}^3\text{g}(\rho-\rho_\text{l})}{6\pi\eta\text{r}}=\frac29\frac{\text{r}^2\text{g}(\rho-\rho_\text{l})}{\eta}$
-ve sign shows the opposing nature n refers to coefficient of viscosity. Factors affecting viscosity:
- Increase in temperature decreases viscosity.
- Increase in pressure increases viscosity in liquids. In water, it decreases while in gases it remains same.
- Weight $=\text{mg}=\frac{4}{3}\pi\text{r}^3\rho\text{g}$
- Upthrust, $\text{U}=\frac43\pi\text{r}^3\rho_\text{l}\text{g}$
- Viscous force, $\text{F}_\text{v}=6\pi\eta\text{ rv}$
When terminal velocity is attained, acceleration should be zero and the ner force should be zero.$\therefore\text{mg}-\text{U}-\text{F}_\text{v}=0$
$\Rightarrow\frac43\pi\text{r}^3\rho\text{g}-\frac43\pi\text{r}^3\rho_\text{l}\text{g}-6\pi\eta\text{ rv}=0$
$\therefore\text{v}=\frac{\frac43\pi\text{r}^3\text{g}(\rho-\rho_\text{l})}{6\pi\eta\text{r}}=\frac29\frac{\text{r}^2\text{g}(\rho-\rho_\text{l})}{\eta}$
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