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!

Two Questions on Relative Velocity

The relative velocity of a body A with respect to a body B is given by

v = v_A – v_B

To make the symbol of the relative velocity ‘v’ more informative, it is usually written as v_AB so that the above equation becomes

v_AB = v_A – v_B

Since the velocity is a vector, you have to find the vector difference v_A – v_B to get the relative velocity.

Now, consider the following MCQ:

A river is flowing with a velocity 1.5 î – ĵ with respect to the ground. A boat is moving with a velocity 2.5 î + 2 ĵ with respect to the ground where î and ĵ are unit vectors in the X and Y directions respectively. The relative velocity of the boat with respect to water in the river is

(a) 4 î + 3 ĵ (b) 4 î + ĵ (c) – î + 3 ĵ (d) î – 3 ĵ (e) î + 3 ĵ

To obtain the relative velocity of the boat with respect to water, you have to subtract (vectorially) the velocity of flow of the river from the velocity of the boat. Therefore, the relative velocity is (2.5 î + 2 ĵ) – (1.5 î – ĵ) = î + 3 ĵ

Here is a simple question which appeared in the Kerala Engineering entrance 2007 question paper:

Two trains are moving with equal speed in opposite directions along two parallel tracks. If the wind is blowing with speed u along the track so that the relative velocities of the trains with respect to the wind are in the ratio 1:2, then the speed of each train must be

(a) 3 u (b) 2 u (c) 5 u (d) 4 u (e) u

If ‘v’ is the speed of each train, the relative velocities of the trains with respect to the wind are v – u and v + u. (One train is moving along the direction of the wind and the other is moving opposite to the wind).

Therefore we have (v – u)/( v + u) = ½, so that v = 3u.

Two Questions on Relative velocity

The relative velocity of a body A with respect to a body B is given by

v = v_A – v_B

To make the symbol of the relative velocity ‘v’ more informative, it is usually written as v_AB so that the above equation becomes

v_AB = v_A – v_B

Since the velocity is a vector, you have to find the vector difference v_A – v_B to get the relative velocity.

Now, consider the following MCQ:

A river is flowing with a velocity 1.5 î – ĵ with respect to the ground. A boat is moving with a velocity 2.5 î + 2 ĵ with respect to the ground where î and ĵ are unit vectors in the X and Y directions respectively. The relative velocity of the boat with respect to water in the river is

(a) 4 î + 3 ĵ (b) 4 î + ĵ (c) – î + 3 ĵ (d) î – 3 ĵ (e) î + 3 ĵ

To obtain the relative velocity of the boat with respect to water, you have to subtract (vectorially) the velocity of flow of the river from the velocity of the boat. Therefore, the relative velocity is (2.5 î + 2 ĵ) – (1.5 î – ĵ) = î + 3 ĵ

Here is a simple question which appeared in the Kerala Engineering entrance 2007 question paper:

Two trains are moving with equal speed in opposite directions along two parallel tracks. If the wind is blowing with speed u along the track so that the relative velocities of the trains with respect to the wind are in the ratio 1:2, then the speed of each train must be

(a) 3 u (b) 2 u (c) 5 u (d) 4 u (e) u

If ‘v’ is the speed of each train, the relative velocities of the trains with respect to the wind are v – u and v + u. (One train is moving along the direction of the wind and the other is moving opposite to the wind).

Therefore we have (v – u)/( v + u) = ½, so that v = 3u.

Two Questions (MCQ) on A.C. Circuits

Questions on alternating current circuits were discussed on 31^st October, 2006 and on 20^th March 2007. You will find those questions by clicking on the label ‘AC circuit’ below this post. Here are two more questions on AC circuits:

(1) In a simple AC generator, a plane rectangular coil PQRS rotates in a magnetic field B. During the rotation, the emf induced in the coil will be maximum when the plane of the coil is

(a) parallel to the magnetic field

(b) perpendicular to the magnetic field

(c) inclined at 30º with the magnetic field

(d) inclined at 45º with the magnetic field

(e) inclined at 60º with the magnetic field

It is the magnetic (Lorentz) force on the electrons in the sides QR ans PS of the coil, which is responsible for the shifting of charges and the consequent motional emf in these sides. The force (and therefore, the induced emf) is maximum when these sides move perpendicular to the magnetic field and this happens when the plane of the coil is parallel to the magnetic field. To put this in a different manner, the sides QR and PS will cut the magnetic field lines normally when the plane of the coil is parallel to the magnetic field (fig).

The correct option therefore is (a).

(2) In the AC circuit shown, the inductive reactance is X_L and the capacitive reactance is X_C. The readings of the voltmeter and ammeter are respectively

(a) 110 V, 0.25 A

(b) 55 V, 0.25 A

(c) 55 V, 0.5 A

(d) 110 V, 0.5 A

(e) 0 V, 0.5 A

The voltage across the inductance leads the current in the series circuit by 90º and the voltage across the capacitor lags behind the current by 90º. The voltages across the inductance and capacitance therefore are in opposition (phase difference of 180º). Since the inductive and capacitive reactances have the same value, the voltages across these elements have the same value, but they are in opposition. They get canceled and the voltmeter reading is zero. The entire supply voltage appears across the resistance which alone controls the current. The current is 110/220 = 0.5 A, which is indicated by the ammeter.

[If you remember that the inductive reactance and the capacitive reactance have equal magnitudes at resonance and the circuit behaves as a pure resistance, you will be able to answer this question in no time].

All India Engineering/Architecture Entrance Examination 2008 (AIEEE- 2008)

Application Form and the Information Bulletin of the All India Engineering/Architecture Entrance Examination 2008 (AIEEE-2008) to be conducted by Central Board of Secondary Education (CBSE) on 27^th April 2008 (Sunday) are being distributed from 30.11.2007 (Friday) and will continue till 5.1.2008 (Saturday) through selected Branches of Syndicate Bank, Regional Offices of CBSE and designated institutions.

Here are the important dates in this regard:

1.a.	Date of Examination	27.04.2008
b.	Sale of AIEEE Information Bulletin containing Application Form	30.11.2007 to 05.01.2008
c.	Online submission of application on website http://www.aieee.nic.in	30.11.2007 to 05.01.2008
2.	Last date for
a.	Receipt of request for Information Bulletin and Application Form by Post atAIEEE Unit,CBSE,PS1-2,Institutional Area,IP Extension,Patparganj,Delhi-110092	15.12.2007
b.	Sale of Information Bulletin at designated branches of Syndicate Bank, Regional Offices of the CBSE and designated institutions	05.01.2008
c.	Online submission of applications	05.01.2008
d.	Receipt of complete applications “by post” including Registration Forms with Bank Draft at AIEEE Unit, CBSE, PS1-2, Institutional Area, IP Extension,Patparganj, Delhi-110092	10.01.2008
3	Date of dispatch of Admit Card	10.03.2008 to 31.03.2008
4	Issue/dispatch of duplicate admit card(or request only with fee of Rs. 50/- + postal charges of Rs. 30/- extra for out station candidate.	11.04.2008 to 27.04.2008 (By Hand) 11.04.2008 to 21.04.2008 (By Post)
5	Dates of Examination B.E./B.Tech: B.Arch / B.Planning:	PAPER – 1 27.04.2008 (0930-1230 hrs) PAPER – 2 27.04.2008 (1400-1700 hrs)

Paper-1 is for B.E./B.Tech and Paper-2 is for B.Arch/B.Planning

Visit the site www.aieee.nic.in for complete information in this connection.