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.