Viscosity is a measure of a fluid's resistance to flow, and understanding it is crucial in various fields ranging from engineering to biology. By determining the coefficient of viscosity of a given liquid, we gain insights into its flow behaviour, which can be essential for applications such as lubricant formulation, blood flow analysis, and industrial processes involving fluid transport. One practical method to measure viscosity involves observing the terminal velocity of a spherical body as it moves through the liquid. For example, consider a small ball bearing falling through a thick syrup; the speed it eventually reaches, where the forces of gravity and viscous drag balance out, provides valuable data to calculate the liquid's viscosity. This principle not only aids in academic experiments but also helps design efficient fluid systems in real-world applications like oil pipelines and medical diagnostics.
To determine the coefficient of viscosity of a given viscous liquid by measuring the terminal velocity of a given spherical body.
A half-metre high, 5 cm broad glass cylindrical jar with millimetre graduations along its height, transparent viscous liquid, one steel ball, screw gauge, stop clock/watch, thermometer, and clamp withstand.
Theory
$\eta=\frac{2}{9}(\rho-\sigma) g \frac{r^2}{v}$
$
\begin{aligned}
& \eta=\text { coefficient of viscosity } \\
& \rho=\text { density of ball } \\
& \sigma=\text { density of liquid } \\
& \mathrm{r}=\text { radius of ball } \\
& \mathrm{v}=\text { terminal velocity of ball }
\end{aligned}
$
Knowing $\mathrm{r}, \rho$ and $\sigma_t$ and measuring $\mathrm{v}, \eta$ can be calculated.
Procedure
1. Take some viscous liquid which must be homogeneous and transparent.
2. Clean the glass cylinder and fill it with this liquid.
3. Find the least count of the vertical scale (graduation) of the glass cylinder.
4. Find the least count and zero error of the stopwatch.
5. Find the least count and zero error of the screw gauge.
6. Mark the given balls as 1, 2, 3, 4, 5. Giving number 1 to the ball of smallest diameter and number 5 to the ball of maximum diameter.
7. Find the diameters of the balls as explained in Experiment 2. (Section A)
8. Make two convenient marks on the cylinder about 40 cm apart, one at a distance of 40 cm from the top and the other at a distance of 20 cm from the bottom.
9. Drop the ball 1 gently in the liquid. For about 1/3rd of the liquid column's height, the ball falls with the accelerated velocity. After that, this ball falls down with the uniform velocity called terminal velocity.
10. When the ball covers about 40 cm from the top, start the stopwatch.
11. Stop the stopwatch when the ball reaches the lower mark which is at 20 cm from the bottom.
12. Note the distance covered by the ball and the time taken to cover this distance.
13. Repeat the experiment one more time with the steel ball of the same diameter.
14. Repeat steps 9 to 13 with the other four balls.
Mean diameter
$
D=\frac{D_1+D_2}{2} \mathrm{~mm}
$
Mean radius
$
r=\frac{D}{2} \mathrm{~mm}=
$
$\qquad$ cm
Meantime
$
t=\frac{t_1+t_2+t_3}{3}=
$
$\qquad$
Mean terminal velocity, $v=\frac{S}{t}=\ldots \mathrm{cms}^{-1}$
From formula,
$\eta=\frac{2 r^2(\rho-\sigma) g}{9 v}=$ $\qquad$ C.G.S. units.
1. Liquid should be transparent to watch the motion of the ball.
2. The ball should be perfectly spherical.
3. Velocity should be noted only when it becomes constant.
Example 1: A uniform rod of length L and density $\rho$ is being pulled along A smooth floor with a horizontal acceleration $\alpha$ what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod?
1) $\frac{1}{3} \mathrm{~L} \rho \alpha$
2) $\frac{2}{3} \mathrm{~L} \rho \alpha$
3) $\mathrm{L} \rho \alpha$
4) $\frac{1}{2} \mathrm{~L} \rho \alpha$
Solution:
Let the cross-sectional area of the rod be A.
So, the mass of the rod $=\rho(A L)$
Now, we have to find the tension at the midpoint of the rod.
$\begin{aligned} & T=\frac{\rho A L}{2}(\alpha) \\ & \text { stress }=\frac{T}{\text { area }}=\frac{\rho \alpha L}{2}\end{aligned}$
The correct option is (4).
Example 2: Which of the following is responsible for the viscous force acting on a spherical body?
1) The radius of the spherical body
2) Coefficient of viscosity of the liquid
3) Terminal velocity of the spherical body
4) All of the above
Solution:
wherein
$\begin{aligned} & \eta=\frac{2}{9}(\rho-\sigma) g \frac{r^2}{v} \\ & \eta=\text { coefficient of viscoscity } \\ & \rho=\text { density of ball } \\ & \sigma=\text { density of liquid } \\ & \mathrm{r}=\text { radius of ball } \\ & \mathrm{v}=\text { terminal velocity of ball } \\ & F=6 \pi \eta r v \\ & \Rightarrow F \alpha \eta \\ & \quad \alpha r \\ & \alpha v\end{aligned}$
Example 3: A cylindrical glass tube with an inner radius of 0.20 mm is dipped into a beaker of mercury. Given the surface tension of mercury is 0.075 N/m and the density is 1000 kg/m3, calculate the height to which the mercury rises in the tube. Assume standard gravity as(g=10 m/s2 ).
1) 0.0160 Pa·s
2) 0.0075 Pa·s
3) 10.44 Pa·s
4) 0.2320 Pa·s
Solution:
Given:
$T=0.075 \mathrm{~N} / \mathrm{mtext}\left(\right.$ surface tension of mercury) $\theta$ is (contact angle) is not provided, typically assumed as ( 0 ) for mercury in glass. $r=0.20 \times 10^{-3} \mathrm{~m} \rho=1000 \mathrm{~kg} / \mathrm{m}^3 g=10 \mathrm{~m} / \mathrm{s}^2$
Substituting the given values into the formula:
$
\begin{aligned}
& h=\frac{2 T \cdot \cos (\theta)}{r \cdot \rho \cdot g} \\
& h=\frac{2 \times 0.075 \mathrm{~N} / \mathrm{m} \times 1}{0.20 \times 10^{-3} \mathrm{~m} \times 1000 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2} \\
& h=\frac{0.15}{200} \mathrm{~m} h=0.00075 \mathrm{~m}=0.0075 \text { Pa.s }
\end{aligned}
$
Hence, the answer is the option (2).
Example 4: A spherical metal ball has a diameter (d) of 0.04 meters, and its coefficient of viscosity in a liquid is 0.8 Ns/m². The ball experiences a terminal acceleration (v) of 0.1 m/s² while moving through the liquid. Use this information to measure the coefficient of viscosity in standard units.
1) 1.60Ns/m2
2) 8.70Ns/m2
3) 0.8Ns/m2
4) 20Ns/m2
Solution:
Given:
Diameter of the ball (d) = 0.04 meters
Coefficient of viscosity (η) = 0.8 Ns/m2
Terminal acceleration (v) = 0.1 m/s2
To measure the coefficient of viscosity, we can use Stokes’ law equation:
$F_{\text {drag }}=6 \pi \eta r v$ ....... (1)
Where:
Fdrag is the drag force on the ball
η is the coefficient of viscosity
r is the radius of the ball
v is the terminal acceleration
The radius is half the diameter:
$
r=\frac{d}{2}=\frac{0.04}{2}=0.02 \mathrm{~m}
$
Using equation (1), we rearrange to solve for $\eta$ :
$
\eta=\frac{F_{\text {drag }}}{6 \pi r v}
$
Plugging in values:
$
\eta=\frac{F_{\mathrm{drag}}}{6 \pi \times 0.02 \mathrm{~m} \times 0.1 \mathrm{~m} / \mathrm{s}^2}=0.8 \mathrm{Ns} / \mathrm{m}^2
$
The coefficient of viscosity is found to be 0.8 Ns/m2 based on the terminal acceleration of the spherical metal ball moving through the liquid.
Hence, the answer is the option (3).
Example 5: A student wants to determine the coefficient of viscosity (η) of a given viscous liquid using the method of measuring the terminal velocity of a spherical body falling through the liquid. The sphere has a radius (r) of 0.02 m and a density (ρ) of 1000 kg/m³. The density of the liquid (ρ0) is 1200 kg/m³, and the acceleration due to gravity (g) is 9.81 m/s². The measured terminal velocity of the sphere in the liquid is 0.1 m/s. Calculate the coefficient of viscosity (η) of the given viscous liquid.
1) 1.60Ns/m2
2) 8.70Ns/m2
3) 0.8Ns/m2
4) 4.36Ns/m2
Solution:
The terminal velocity (vt) of a spherical body falling through a viscous liquid is given by the following equation:
$v_t=\frac{2 \cdot\left(\rho-\rho_0\right) \cdot g \cdot r^2}{9 \cdot \eta}$
Where:
vt is the terminal velocity of the sphere
ρ is the density of the sphere
ρ0 is the density of the liquid g is the acceleration due to gravity
r is the radius of the sphere
η is the coefficient of viscosity of the liquid
We are given:
Sphere radius (r) = 0.02 m
Sphere density (ρ) = 1000 kg/m³
Liquid density (ρ0) = 1200 kg/m³
Acceleration due to gravity (g) = 9.81 m/s²
Terminal velocity (vt) = 0.1 m/s
Let’s rearrange the equation to solve for the coefficient of viscosity (η):
$
\eta=\frac{2 \cdot\left(\rho-\rho_0\right) \cdot g \cdot r^2}{9 \cdot v_t}
$
Now, substitute the given values:
$
\begin{gathered}
\eta=\frac{2 \cdot\left(1000 \mathrm{~kg} / \mathrm{m}^3-1200 \mathrm{~kg} / \mathrm{m}^3\right) \cdot 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot(0.02 \mathrm{~m})^2}{9 \cdot 0.1 \mathrm{~m} / \mathrm{s}} \\
\eta=\frac{2 \cdot\left(-200 \mathrm{~kg} / \mathrm{m}^3\right) \cdot 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot 0.0004 \mathrm{~m}^2}{0.9 \mathrm{~m} / \mathrm{s}} \\
\eta=\frac{-3.9248 \mathrm{Ns} / \mathrm{m}^2}{0.9 \mathrm{~kg} /(\mathrm{ms})} \\
\eta \approx-4.36 \mathrm{Ns} / \mathrm{m}^2
\end{gathered}
$
Since viscosity cannot be negative, it seems there might be an error or discrepancy in the given values or measurements. Please double-check the data provided for accuracy.
Hence, the answer is the option (4).
Measuring the terminal velocity of a spherical object moving through the fluid is a direct way of calculating the coefficient of viscosity of the fluid. At terminal velocity, the effects of gravitational force on the body are balanced by viscosity drag force for an equilibrium state. Thus, from an analysis of this equilibrium condition in accordance with Stokes’ law, it becomes possible to determine the fluidity co-efficient involved.
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