Questions involving Viscosity and Buoyancy

The following simple question involving viscous force and force of buoyancy is meant for checking whether you have a good understanding of basic points:

A wooden ball of relative density 0.5 is released from the bottom of a still water reservoir. Its acceleration while moving up will

(a) go on decreasing initially

(b) go on increasing initially

(c) remain constant at g/2 (in magnitude)

(d) remain constant at g (in magnitude)

(e) be zero throughout

The net force acting on the ball at the moment of releasing is its apparent weight which is upwards. [The apparent weight = Real weight – up thrust = Vρg – Vσg. Since the density of wood (ρ) is less than that of water (σ), the apparent weight is negative which means it is directed upwards]

The sphere therefore moves up with an acceleration. But, when the velocity (v) increases from zero, the opposing viscous force (6πrηv) also increases thereby reducing the net upward force. The upward acceleration therefore goes on decreasing. So, the correct option is (a).

The following question which appeared in Kerala Engineering Entrance 2008 question paper also involves viscous force and force of buoyancy; but the latter is negligible:

Eight drops of a liquid of density ρ and each of radius ‘a’ are falling through air with a constant velocity of 3.75 cm s^–1. When the eight drops coalesce to form a single drop, the terminal velocity of the new drop will be

(a) 1.5×10^–2 ms^–1

(b) 2.4×10^–2 ms^–1

(c) 0.75×10^–2 ms^–1

(d) 25×10^–2 ms^–1

(e) 15×10^–2 ms^–1

The viscous force acting on a sphere of radius ‘a’ is 6πahv where h is the coefficient of viscosity of the fluid and ‘v’ is the velocity of the sphere. Terminal velocity is attained when the opposing viscous force is equal in magnitude to the apparent weight of the sphere. Therefore,

6πahv = (4/3)πa³(ρ – σ)g where ρ is the density of the sphere, σ is the density of the fluid and ‘g’ is the acceleration due to gravity.

The above equation shows that the terminal velocity ‘v’ is directly proportional to the square of the radius of the sphere.

In the above problem, eight identical drops coalesce to form a single drop. The new drop thus formed has eight times the volume of each small drop. The radius ‘R’ of the new drop is given by

(4/3)πR³ = 8×(4/3)πa³

Therefore, R =2a

Since the radius of the new drop is twice that of each small drop, the terminal velocity of the new drop must become four times. The correct option therefore is 15×10^–2 ms^–1.

Multiple Choice Questions on Viscosity

As promised in the last post, we will discuss some multiple choice questions on viscosity. Consider the following MCQ):

The rate of steady volume flow of water through a capillary tube of radius ‘r’ and length ‘L’ under a pressure difference P between its ends is V. This tube is connected in series with another tube of radius 2r and length 8L. Then the rate of volume flow through the series combination under the same pressure difference between the ends of the combination is

(a) V (b) 2V/3 (c) 3V (d) V/2 (e) V/4

We have 1/Q_series = 1/Q₁ +1/Q_2.

Here Q₁ = V and Q₂ = 2V since Q α r⁴/L (When the radius is doubled, the rate of flow becomes 16 times; when the length is made 8 fold, the rate of flow becomes one eighths so that the rate of flow with the second tube is twice that with the first tube).

The net rate of flow Q_series through the series combination is therefore given by

1/Q_series = 1/V + 1/2V from which Q_series = V×2V/(V+2V) = 2V/3

[If there is sufficient water column, the acceleration will finally become zero and the sphere will then move with constant (terminal) velocity].

You may be remembering that rain drops arriving near the earth’s suface move down with constant (terminal) velocity. The terminal velocty will depend on the radius of the drop. Here is a question on this:

Two rain drops of radii ‘r’ and √r arriving at the ground will have their terminal velocities in the ratio

(a) 1:1

(b) r:1

(c) 2:1

(d) 1:2

(e) √r:1

Since the rain drops move with terminal velocity v_terminal, we can equate the magnitude of viscous force to the apparent weight of the rain drop so that

6πrηv_terminal = (4/3)πr³(ρ–σ)g with usual notations.

Therefore, v_terminal α r².

The ratio of terminal velocities in the present case is (r)²:(√r)² = r:1

Here is another MCQ involving terminal velocity:

A spherical glass bead released at the top of a column of castor oil in a tall jar is found to attain a terminal velocity of 2 cm s^–1. If it is pulled up with a force equal in magnitude to three times its apparent weight in castor oil, its terminal velocity will be

(a) 4 cm s^–1

(b) 6 cm s^–1

(c) 3 cm s^–1

(d) 2 cm s^–1

(e) 1 cm s^–1

We have (from Stokes formula)

6πrηv_terminal = W_app

where W_app is the apparent weight, which tries to move the sphere downwards. When terminal velocity is attained, the opposing viscous force (6πrηv_terminal) becomes equal in magnitude to the apparent weight. Since 6πrη is constant, the terminal velocity v_terminal is directly proportional to the force trying to move the sphere. Therefore,

v_terminal α W_app

When the sphere is pulled up with a force equal in magnitude to 3 times the apparent weight, the net force on it is 2W_app and hence we have

v’_terminal α 2W_app.

From the above we obtain v’_terminal = 2 v_terminal = 2×2 = 4 cm s^–1

Now, consider the following MCQ:

Two horizontal capillary tubes of lengths L and 3L having radii r and 2r are connected in series and a liquid is flowing slowly and steadily through this combination. If P₁ is the pressure difference between the ends of the first tube and P₂ is that between the ends of the second tube, then P₁/P₂ is

(a) 9/5

(b) 1/6

(c) 12/5

(d) 8/3

(e) 16/3

Since the tubes are in series, the rate of flow (πPr⁴/8Lη) through them is the same. Therefore we have,

πP₁r⁴/ 8Lη = πP₂ (2r)⁴/ 8(3L)η, from which P₁/P₂ = 16/3.

Viscosity – Equations to be Remembered

The essential things you must remember in the section, ‘viscosity’ are given below:

1. Viscous force between two layers of a fluid = ηAdv/dx where η is the coefficient of viscosity, A is the common area of the layers and dv/dx is the velocity gradient.

2. Poiseuille’s formula for the volume (V) of a liquid flowing through a capillary tube of radius 'r’ in a time ‘t’ under a pressure difference ‘P’ between the ends of the tube is

V = πPr⁴t / 8Lη

where L is the length of the tube and η is the coefficient of viscosity of the liquid.

If the liquid flows through a horizontal capillary tube under a constant hydrostatic pressure produced by a height ‘h’ of liquid column, V = πhρgr⁴t / 8Lη where ρ is the density of the liquid.

It will be better to remember the rate of flow (which is the volume flowing per second) as

Q = V/t = πPr⁴/ 8Lη

If Q₁ and Q₂ are the rates of flow through two tubes (of radii r₁, r₂ and lengths L₁, L₂) under a given pressure head P, then the rate of flow (Q_series) under the same pressure head P when the tubes are connected in series is given by the reciprocal relation,

1/Q_series = 1/Q₁ +1/Q₂.

[You can easily prove this by combining the equations Q₁ = πPr₁⁴/ 8L₁η, Q₂ = πPr₂⁴/ 8L₂η, P = P₁+P₂ (where P₁ and P₂ are the pressures between the ends of the two tubes when they are in series) and Q = πP₁r₁⁴/ 8L₁η = πP₂ r₂⁴/ 8L₂η. Do this as an exercise]

If there are many tubes in series, the above relation gets modified as

1/Q_series = 1/Q₁ +1/Q₂ + +1/Q₃ +1/Q₄ +…etc.

3. Reynold’s number, R = vρr/ η where ‘v’ is the velocity of the liquid of density ρ and viscosity (coefficient) η through a tube of radius ‘r’.

Note that R is dimensionless and that the flow will be streamlined only if R is less than 2000 (approximately). It therefore follows that stream lined flow is more likely in the case of liquids of small density and large viscosity.

4. Stokes formula for the viscous force (F) on a sphere of radius ‘r’ moving with a velocity ‘v’ through a fluid having coefficient of viscosity η is

F = 6πrηv

If the sphere moves with terminal velocity v_terminal as is the case when it moves down under gravity through a column of viscous medium, we can equate the magnitude of viscous force to the apparent weight of the sphere so that

6πrηv_terminal = (4/3)πr³(ρ–σ)g where ρ is the density of the material of the sphere and σ is the density of the viscous medium.

Note that the terminal velocity is directly proportional to the radius of the sphere.

In the next post we will discus some typical multiple choice questions on viscosity.

Merry Christmas!