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.

KEAM2007 Questions on Electromagnetic Waves

Questions on electromagnetic waves are simple at the 12^th class level and you can easily score full marks in this section. Here are the three questions which appeared in the Kerala Medical (Question No.1) and Engineering (Question Nos.2 & 3) Entrance 2007 question papers:

(1) An electromagnetic radiation has energy of 13.2 keV. Then the radiation belongs to the region of

(a) visible light (b) ultraviolet (c) infrared

(d) X-ray (e) microwave

If an electron accelerated by a potential difference of 13.2 kilo volt strikes a target, the maximum energy of the photon generated will be 13.2 kilo electron volt. You know that an accelerating voltage of 13.2 kV in an X-ray tube can produce X-rays. So, the correct option is (d).

If the energy given were in the electron volt (eV) range, you will be in confusion. To be fool proof, it will be better to calculate the wave length of the radiation, noting that the product of the wave length in angstrom (Ǻ) and the energy in electron volt (in the case of photons) is approximately 12400. The wave length of the photon in the present case is therefore given by

λ = (12400/13200) Ǻ = 0.94 Ǻ, which is in the X-ray region.

(2) The dielectric constant of air is 1.006. The speed of electromagnetic wave traveling in air is a×10⁸ ms^–1 where ‘a’ is about

(a) 3 (b) 3.88 (c) 2.5 (d) 3.2 (e) 2.8

This is an unusually simple question which the question setter might have deliberately included! You know that the speed of electromagnetic waves in air is almost equal to that in free space and hence the answer is 3 [Option (a)].

Generally, in any medium, the speed (v) of electromagnetic waves is given by

v = c/√(μ_rε_r) where c is the speed in free space (which is 3×10⁸ ms^–1), μ_r is the relative permeability and ε_r is the relative permittivity (dielectric constant) of the medium.

(3) a. The wave length of microwaves is greater than that of UV-rays

b. The wave length of IR rays is less than that of UV-rays

c. The wave length of microwaves is less than that of IR rays

d. Gamma ray has the shortest wave length in the electromagnetic spectrum

Of the above statements

(A) a and b are true (B) b and c are true

(C) c and d are true (D) a and c are true

(E) a and d are true

This too is a simple knowledge based question and the correct option is (E).

IIT-JEE Matrix Match Type Question on Electromagnetism

The following Matrix Match type question which appeared in the IIT-JEE 2007 question paper aims at checking your understanding of basic principles of electricity and magnetism. Here is the question:

Column I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4×4 matrix given in the OR

Column I------------------------------ Column II

(A) A charged capacitor is---------------(p) A constant current flows

connected to the ends of---------------- through the wire

the wire

(B) The wire is moved -----------------(q) Thermal energy is generated

perpendicular to its length with -------- in the wire

a constant velocity in a uniform

magnetic field perpendicular to

the plane of motion

(C) The wire is placed in a constant ---(r) A constant potential

electric field that has a direction ------difference develops between

along the length of the wire-----------the ends of the wire

(D) A battery of constant emf is ------(s) Charges of constant

connected to the ends of the----------magnitude appear at the ends

wire ---------------------------------of the wire
(A) When a charged capacitor is connected to the ends of a wire, a discharge current flows through the wire and the wire is heated. This is given in statement (q) under column II so that (A) is to be matched with (q).

(B) When the wire is moved perpendicular to the magnetic field, free electrons in the wire experience a magnetic (Lorentz) force and shift towards one end. Consequently, a potential difference develops between the ends of the wire. Since the wire is moving with constant velocity in a constant magnetic field, the charges developed at the ends as well as the potential difference between the ends are constant. Therefore, (B) is to be matched with (r) and (s).

(C) When the wire is placed in the electric field, free electrons shift towards the end of the wire which is at higher potential. This shifting of free electrons results in the accumulation of charges at the ends of the wire. But there is no potential difference between the ends of the wire since the shifting of free electrons just nullifies the potential difference caused by the external electric field. Therefore, (C) is to be matched with (s) only.

(D) When a battery of constant emf is connected to the ends of the wire, the effects given at (p) (q) and (r) under Column II occur. Therefore, (D) is to be matched with (p), (q) and (r).

These are to be marked in the 4×4 matrix as shown:

IIT-JEE 2008 Application from 23-11-2007

The Joint Entrance Examination 2008 (JEE-2008) for admission to the seven IITs at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee as well as to Institute of Technology, Banaras Hindu University, Varanasi and the Indian School of Mines, Dhanbad, will be held on April 13, 2008 (Sunday) as per the following schedule:
09:00 – 12:00 hrs Paper - 1
14:00 – 17:00 hrs Paper – 2

Application form and Prospectus of the Joint Entrance Examination 2008 (IIT-JEE-2008) will be issued with effect from November 23, 2007 (Friday). On-line submission of Application will also commence on the same day.

Application form along with the Information Brochure can be purchased from any one of the selected branches of Canara Bank/State Bank of India/ Union Bank/ Punjab National Bank (visit the site: http://www.iitkgp.ernet.in/jee/advt.html for the list of branches) or from any of the IITs between 23.11.2007 (Friday) and 4.1.2008 (Friday) by paying Rs. 500/- in the case of SC/ST/Female candidates and Rs. 1000/- in the case of all other candidates by cash. SC/ST/Female candidates will get the materials in a WHITE envelope while the other candidates will get the materials in a BLUE envelope.

Application Material by Post from IITs:

The request for Application Form and Prospectus by Post from any of the IITs will also be accepted from November 23, 2007. Application Form and Prospectus can be obtained by post from any of the IITs by sending a request along with two self- addressed slips and a Demand Draft for Rs.500/- (in case of SC/ST/Female applicants) and for RS.1000/- in case of other applicants, payable to the “CHAIRMAN, JEE” of the respective IIT, at the corresponding city.[For example, those applying to IIT, Madras, should take the DD in favour of “Chairman, JEE, IIT Madras” and payable at Chennai]. Such requests will be accepted from 23.11.2007(Friday) to 21.12.2007 (Friday).

Submission of filled up Application Forms: Duly completed Application Form, refolded only along the original fold should be inserted in the envelope supplied, along with the attested copy of the 10th Class Pass, or Equivalent Examination Certificate, and the Acknowledgement Card. These items should not be stapled or pasted with the Application form. Irrespective of the Bank/Institute from where the Application has been obtained, they should be re-submitted along with the contents by Registered Post/Speed Post only to the IIT located in the Zone where the centre of the examination chosen by the applicant is located. They may also submitted in person at the JEE office of the IIT concerned.

The last date for receipt of the completed application at the IITs is 17:00 hours on January 4, 2008 (Friday).

Online Submission of Application: Facility for online submission of applications will be available between 23.11.2007 (Friday) and 5 pm on 28.12.2007 (Friday) through the JEE websites of the different IITs. The JEE websites of the IITs are given below:
IIT Bombay: http://www.jee.iitb.ac.in
IIT Delhi: http://www.jee.iitd.ac.in
IIT Guwahati: http://www.iitg.ac.in/jee
IIT Kanpur: http://www.iitk.ac.in/jee
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Madras: http://jee.iitm.ac.in
IIT Roorkee: http://www.iitr.ac.in/jee

Visit the website http://www.iitkgp.ernet.in/jee/ for more details.

Two Questions (MCQ) on Elasticity

Have you ever thought why a tree does not grow beyond a limiting height? It is the breaking stress of the tree that limits its height. Here is a question that high lights this point:

A tree has a breaking stress of 4.5×10⁵ Nm^–2. If the density of the tree is 900 kg m^–3, the maximum height up to which it can grow is (assuming that the tree is cylindrical in shape)

(a) 100 m (b) 75 m (c) 50 m (d) 30 m (e) 25 m

The weight of the tree is AHρg where A is the area of cross section, H is the height, ρ is the density and ‘g’ is the acceleration due to gravity.

Maximum stress will be developed at the base of the tree and will be equal to AHρg/A = Hρg.

Since the breaking stress is 4.5×10⁵ Nm^–2, we can write

Hρg = 4.5×10⁵ Nm^–2 so that H = 4.5×10⁵ Nm^–2/ρg = 4.5×10⁵ Nm^–2/9000 nearly.

This works out to 50 m.

Now, consider the following MCQ involving compressibility:

A solid sphere made of copper having compressibility K is placed in an air chamber and the pressure of air is reduced by P. The fractional change in the radius of the sphere will be

(a) PK (b) 3PK (c) P/3K (d) K/3P (e) PK/3

Compressibility (K) is the reciprocal of bulk modulus (B) and we have

B = –P/(dV/V) where dV is the change in volume V due to a pressure change P. The negative sign indicates that a decrease in pressure will produce a increase in volume.

In terms of compressibility, the above equation can be written as

B = 1/K = –P/(dV/V) so that the fractional change in volume is given by

dV/V = PK.

But V = (4/3) πR³ where R is the radius of the sphere so that dV/V = 3 dR/R. The fractional change in radius is therefore given by dR/R = (dV/V)/3 = PK/3.

You may find all posts on elasticity on this site by clicking on the label ‘elasticity’ below this post or on the side of this page.

CBSE announces dates of All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008

Central Board of Secondary Education (CBSE), Delhi has announced the dates of All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments and Municipal or other local authorities in India except in the States of Andhra pradesh and Jammu&Kashmir. The dates of the Examination are:

(1) Preliminary Examination : 6^th April 2008 (Sunday)

(2) Final Examination : 11^th May 2008 (Sunday)

Candidates can apply for the All India Pre-Medical/Pre-Dental Entrance Examination in the following two ways:-

(i) Online submission

Online submission of application may be made by accessing the website www.cbse.nic.in from 16.10.2007 (10.00 AM) to 26.11.2007 (5.00 PM). Candidates are required to take a print out of the Online Application after successful submission of data. The print out of the computer generated application, complete in all respect as applicable for Offline submission should be sent to the Deputy Secretary (AIPMT), Central Board of Secondary Education, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post only. Candidate should pay the examination fee of Rupees 400/- for General Category and Rupees 200/- for SC/ST Category through a Demand Draft in favour of the Secretary, Central Board of Secondary Education, Delhi drawn on any nationalized bank payable at Delhi. Instructions for Online submission of Application Form will be made available on the website www.cbse.nic.in.

(ii) Offline submission

Offline submission of Application Form may be made on the prescribed Application Form. The Information Bulletin and Application Form costing Rs.400/- for General Category candidates and Rs.200/- for SC/ST candidates inclusive of examination fee can be obtained against Cash Payment from designated branches of Canara Bank/ Regional Offices of the CBSE from 16-10-2007 to 26-11-2007. The details of Banks are given in the Admission Notice which is available on CBSE website www.cbse.nic.in.

Designated branches of Canara Bank in Kerala are:

KOTTAYAM	P.B. No.122, K.K.Road, Kottayam-686 001
QUILANDY	Fasila Buidling, Main Road, Quilandy-673 305
TRIVANDRUM	Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023
TRIVANDRUM	Plot no.2, PTP Nagar Trivandrum-695 038
TRIVANDRUM	TC No.25/1647, Devaswom Board Bldg. M.G. Road, Trivandrum-695 001
ERNAKULAM	Shenoy’s Chamber, Shanmugam Road, Ernakulam, Cochin-682 031
CALICUT	9/367-A, Cherooty Road, Calicut-673 001
TRICHUR	Trichur Main Ramaray Building, Round South, Trichur-680001
QUILON	Maheshwari Mansion, Tamarakulam, Quilon-691 001
PALGHAT	Market Road, Big Bazar, 20/68, Ist floor,Palghat-678 014.

The Information Bulletin and Application Form can also be obtained by Speed Post/Registered Post by sending a written request with a Bank Draft/Demand Draft for Rs.450/- for General Category and Rs.250/- for SC/ST Category payable to the Secretary, Central Board of Secondary Education, Delhi along with a Self Addressed Envelope of size 12” x 10”. The request must reach the Deputy Secretary (AIPMT), CBSE, 2, Community Centre, Preet Vihar, Delhi-110 301 on or before 15-11-2007. The request should be super scribed as “Request for Information Bulletin and Application Form for AIPMT, 2008”.

Completed Application Form is to be dispatched by Registered Post/Speed Post only. Last Date for receipt of completed Application Forms for both Offline and Online in CBSE is 28-11-2007.

You can obtain complete information at www.cbse.nic.in

Visit the site regularly for information updates.

Three AIEEE 2003 Questions on Waves

The following MCQ which appeared in AIEEE 2003 question paper is worth noting:

A metal wire of linear mass density 9.8 g/m is stretched with a tension of 10 kg wt between two rigid supports 1m apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency ‘n’. The frequency ‘n’ of the alternating current is

(a) 25 Hz (b) 50 Hz (c) 100 Hz (d) 200 Hz

The wire vibrates because of the magnetic force on it. When the alternating current completes one cycle, the wire completes one oscillation and hence the frequency of oscillation of the wir is the same as the frequency of the alternating current (n). Therefore we have

n = (1/2L)√(T/m) where L is the length of the segment of the wire betweenetween the supports, T is the tension and ‘m’ is the linear density (mass per unit length) of the wire. Substituting for L, T and m we have

n = (½)√[(10×9.8)/(9.8×10^–3 )] = 50 Hz.

The following MCQ is a conventional type:

The displacement ‘y’ of a wave traveling in the X-direction is given by

y =10^–4 sin(600t – 2x + π/3) metre,

where x is expressed in metre and t in seconds. The speed of the wave motion in ms^–1 is

(a) 200 (b) 300 (c) 600 (d) 1200

It will be useful to remember that the velocity of the wave, v = Coefficient of t /Coefficient of x. So, the correct option is (b).

Now consider the following MCQ on beats, which is of the type popular among question setters:

A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

(a) 256 + 5 Hz (b) 256 + 2 Hz (c) 256 – 2 Hz (d) 256 – 5 Hz

Since the beat frequency before increasing the tension in the piano wire was 5 Hz, the frequency of vibration of the piano wire was (256 ± 5) Hz. When the tension in the piano wire is increased, its frequency increases. If the original frequency of the piano wire was (256 + 5) Hz, the beat frequency would have increased on increasing the tension in the piano wire. Therefore, the original frequency of the piano wire was (256 – 5) Hz .

You will find more multiple choice questions (with solution) on waves at physicsplus: Multiple Choice Questions on Waves

KEAM (Engineering) 2007 Questions on Elasticity

You will often find questions involving elastic potential energy in entrance examinations for admission to professional and other degree courses. Here is a question which appeared in Kerala Government Engineering Entrance 2007 test paper:

A wire of natural length L_, Young’s modulus Y and area of cross section A is extended by x. Then the energy stored in the wire is given by

(a) (YA/2L)x² (b) (YA/3L)x² (c) (YL/2A)x² (d) (YA/2L²)x² (e) (A/2YL)x²

The elastic potential energy per unit volume is ½ stress×strain. Since the Young’s modulus Y = stress/strain, we can modify the above expression as (½)Y×strain². The elastic potential energy stored in the entire wire is (½)Y×strain² × volume of the wire = (½)Y×strain² × AL = (½)Y×(x/L)² × AL = (YA/2L)x².

Here is another question which appeared in Kerala Government Engineering Entrane 2007 test paper:

The length of a rubber cord is l₁ metre when the tension is 4N and l₂ metre when the tension is 6N. The length when the tension is 9N is

(a) (2.5 l₂ – 1.5 l₁) m (b) (6 l₂ – 1.5 l₁) m (c) (3 l₁ – 2 l₂) m

(d) (3.5 l₂ – 2.5 l₁) m (e) (2.5 l₂ + 1.5 l₁) m

You can use Hooke’s law to solve this question. Since the increase in length is directly proportional to the force (tension) applied, in accordance with Hooke’s law, we have

l₂ – l₁ = K(6 – 4) where K is the constant of proportionality.

[Note that the increase in length from l₁ to l₂ is produced by the increase in tension from 4N to 6 N].

If the length of the rubber cord is l₃ when the tension is 9 N, we have

l₃ – l₁ = K (9 – 4)

Dividing the first equation by the second, we obtain

(l₂ – l₁)/(l₃ – l₁) = 2/5, from which l₃ = (2.5 l₂ – 1.5 l₁) metre.

You will find more questions on elasticity on clicking on the label ‘elasticity’ below this post or on the on the side of this page.