Question

Coefficient of viscosity of a given viscous liquid by measuring terminal velocity of a given spherical body

Answer 1

Answer:

What's the question? I can't understand

Answer 2

Terminal velocity of the given spherical body is $\text{v}=\frac{2}{9} \frac{r^{2}(\rho-\sigma) \text{g}}{\eta}$

Explanation:

To find the terminal velocity of the spherical body with an experiment

Let us take a long glass cylindrical jar is fully filled water.

Place a thermometer inside the jar with the help of a stand holding on it.

Put a small spherical ball inside the jar.

Diameter of the spherical ball in one direction = D (cm)

Diameter of the spherical ball in perpendicular direction = D' (cm)

$\text{Mean diameter (D)} = \frac{\text{D} + \text{D'}}{2}\ \ \ \text {(cm)}$

$\text{Mean radius (r)} = \frac{\text{D}}{2}\ \ \ \text{(cm)}$

Fallen Distance = S (cm)

$\text{Time taken} = \text{t}_1, \ \text{t}_2, \ \text{t}_3 \ \ \ \text{(seconds)}$

$\text{Mean Time (T)} = \frac{\text{t}_1 + \ \text{t}_2 + \ \text{t}_3}{3} \\ \ \ \ \text{(seconds) }$

Specific density of water = ρ

acceleration due to gravity = g

Then the terminal velocity can be written as ,

$\text{Terminal velocity of the spherical body} \ \text{(v)}=$ $\text{v}=\frac{2}{9} \frac{r^{2}(\rho-\sigma) \text{g}}{\eta} \ \ (\text{C.G.S units})$