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Sale of application forms for Secondary & Senior School Certificate Examination, 2009
(FOR DELHI REGION)
PRIVATE CANDIDATE :
The application forms for private candidates for 2009 Exams. (Class X & XII) under All India/Delhi Scheme are available for sale w.e.f. 18.08.2009 from the selected branches of the Syndicate Bank against payment of Rs. 10/-each.
NOTE: Application forms for Private Candidates will be available in the selected branches during business hours.
All the private candidates appearing under All India and Delhi Scheme should submit the examination forms duly completed in all respect along with requisite fee in cash in any abovementioned branches of Syndicate Bank nearest to their area. The candidates need not come to CBSE office for this purpose as no form will be issued/accepted in the Board’s office.
REGULAR CANDIDATE :
In respect of regular candidates, the schools will deposit the examination fee along with list of candidates through Bank Draft in the Board’s Office.
| Candidate | Last date Without late fee | Last date With late fee of Rs. 100/- | Last date With late fee of Rs. 200/- | Last date With late fee of Rs. 300/- |
| Regular Candidate |
10.09.2008 | 20.09.2008 | 06.10.2008 | 31.10.2008 |
| Private Candidates |
05.09.2008 | 15.09.2008 | 30.09.2008 | 25.10.2008 |
SCHEDULE OF LAST DATE
The Bank branches will accept the examination form and fee up to 25.10.2008 with late fee.
No form or fee shall be accepted after the expiry of the scheduled date.
Higher Order Thinking Questions- a Necessity
In the Board’s question paper in Economics of Class XII, there will be higher order thinking questions of 15 to 20 marks. The objective is to assess the students’ understanding, analytical ability and interpretation.
In the course content for Economics, an attempt is made to introduce basic concepts used in the study of this subject. Many of these concepts are also interrelated.
Memorising these concepts and their relationship will not help in understanding the economic laws and principles. These have to be understood in totality by asking questions such as why and how and going a step further by applying them to various situations.
Some of these concepts are:
Just memorising the meaning of concepts does not help in understanding the subject and the use of these concepts. For instance just memorising the meaning of average and marginal costs and their relationship will not help you in answering question, like.
To give a practice of the application of these concepts and to test whether the students have really understood them, the situations from day to day life should be put before
them. They should then be asked to analyse them. Some topics can also be taken up for debate, such as “Are borrowings bad?” or “Is inflation harmful?”
Some examples of higher order thinking questions in Economics for class XII are given below:-
1 When price of a good rises from Rs.5 per unit to Rs.6 per unit, 3
its demand falls from 20 units to 10 units. Compare expenditures
on the good to determine whether demand is elastic or inelastic.
2 What is the relation between good X and good Y in each case, if 3
with fall in the price of X demand for good Y (i) rises and (ii) falls?
Give reason.
3 Giving reasons explain how the following are treated while estimating 2,2,2
national income:
(i) Payment of fees to a lawyer engaged by a firm.
(ii) Rent free house to an employee by an employer.
(iii) Purchases by foreign tourists.
4 Explain what happens to the profits in the long run if the firms are free to 3
enter the industry.
5 Given market equilibrium of a good, what are the effects of simultaneous 6
increase in both demand and supply of that good on its equilibrium price
and quantity?
6 Explain the implications of the following :
(i) The feature ‘differentiated products’ under monopolistic competition.
(ii) The feature’Large number of sellers’ under perfect competition.
7 At a given market price of a good a consumer buys 120 units. When price
falls by 50 percent he buys 150 units. Calculate price elasticity of demand.
8 Explain, by giving examples, how do the following determine price elasticity
of demand:
(i) nature of the good
(ii) availability of substitutes
9 In the following table, identify the different phases of the law
of variable proportions and also explain the causes:
|
Variable input (units) |
1 |
2 |
3 |
4 |
5 |
6 |
|
Total product (units) |
10 |
22 |
32 |
40 |
40 |
35 |
10 Giving reasons, explain how the following are treated in estimating 6
National Income:
(i) Purchase of a truck to carry goods by a production unit.
(ii) Payment of income tax by a production unit.
(iii) Services rendered by family members to each other.
Maths
1 Relations and Functions.
2 Inverse Trigonometric Functions.
3 Matrices.
4 Determinants.
5 Continuity and Differentiability.
6 Application of Derivatives.
Appendix 1: Proofs in Mathematics.
Appendix 2: Mathematical Modelling.
Answers.
Physics
1 Electric Charges and Fields.
2 Electrostatic Potential and Capacitance.
3 Current Electricity.
4 Moving Charges and Magnetism.
5 Magnetism and Matter.
6 Electromagnetic Induction.
7 Alternating Current.
8 Electromagnetic Waves.
Answers.
Maths.
1 Real Numbers.
2 Polynomials.
3 Pair of Linear Equations in Two Variables.
4 Quadratic Equations.
5 Arithmetic Progressions.
6 Triangles.
7 Coordinate Geometry.
8 Introduction to Trigonometry.
9 Some Applications of Trigonometry.
10 Circles.
11 Construction.
12 Areas Related to Circles.
13 Surface Areas and Volumes.
14 Statistics.
15 Probability.
Appendix 1: Proofs in Mathematics.
Appendix 2: Mathematical Modelling.
Answers/Hints.
Science.
1 Chemical Reactions and Equations.
2 Acids, Bases and Salts.
3 Metals and Non-Metals.
4 Carbon and its Compounds.
5 Periodic Classification of Elements.
6 Life Processes.
7 Control and Coordination.
8 How do Organisms Reproduce.
9 Heredity and Evolution.
10 Light – Reflection and Refraction.
11 Human Eye and Colourful World.
12 Electricity.
13 Magnetic Effects of Electric Current.
14 Sources of Energy.
15 Our Environment.
16 Management of Natural Resources.
Answers.
Electrostatics deals with the study of electric charges at rest.
When certain substances are rubbed with other suitable substances, they acquire the property of attracting small pieces of straw, feather etc, towards it. Eg: A glass rod rubbed with silk, an ebonite rod rubbed with fur etc. The attractive property is due to the presence of electric charges on them. The glass rod and the ebonite rod are said to be electrified or electrically charged.
When a glass rod rubbed with silk is brought near another glass rod similarly rubbed with silk, the two rods are found to repel each other. Hence there must be a force of repulsion between the charges developed on the two rods. But if an ebonite rod rubbed with fur is brought near a glass rod rubbed with silk, the two rods attract each other. So there must be a force of attraction between the charges on the two rods. From these experiments it is clear that there are two kinds of electric charges. The charge acquired by the glass rod rubbed with silk is called positive electric charge and that acquired by the ebonite rod rubbed with fur is called negative electric charge. The above experiments show that like charges repel and unlike charges attract each other.
Conductors, Insulators and Semiconductors
Substances which allow electric charges to flow through them are called conductors. Eg. All metals, Carbon etc.
Substances which are not allowed electric charges to flow through them are called insulators. Eg. Paper, leather, wood, glass, ebonite, wax etc.
Substances which have an electrical conductivity intermediate between those of conductors and insulators are called semiconductors. Eg. Germanium, Silicon, Selenium, Silver Sulphide etc.
Electrostatic Induction
It is possible to charge a conductor by bringing a charged body near it. When a positively charged body A is brought near a conductor BC, negative charges appear at the near end B and positive charges at the farther end C. Thus unlike charges are induced at the near end and like charges at the farther end of the conductor. This process is called electrostatic induction or electrification by induction. The charge on the body A is called the inducing charge and the charges appearing on BC are called inducted charges. The induced charges (each type) will be equal in magnitude to the inducing charge. When the body A is removed, the charges on BC will disappear. This shows that the induced charges will exist only as long as the inducing charge is present.
Coulomb’s Law (Inverse Square Law) of electrostatic force
The force of attraction or repulsion between two stationary charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let r be the distance between two stationary charges q1 and q2 in free space. Then the force of attraction or repulsion between them can be given by
F α q1q2 and F α r2
Ie. F α q1q2/r2
Or F = K q1q2/r2
Electric Field
The region around any electric charge where its effects (attractive force or repulsive force) can be experienced is called the electric field of the charge. The electric fields produced by a stationary electric charges is called an electrostatic field.
Intensity of Electric Field (E)
The intensity of electric field (or electric field strength) at a point in an electric field is the force exerted on a unit positive charge placed at that point.
Ie E = F/q
Therefore the Unit of E = Unit of Force/Unit of q
= Newton/Coulomb (N/C)
It may be shown that Volt per metre is also the unit for electric field.
Dimensions of E = Dimensions of Force/Dimensions of Charge
= MLT-2/IT
=M1L1I-1T-3
Lines of Force
An electric filed may be represented by a number of imaginary lines of force. A line of force may be defined as a smooth, continuous curve along which unit positive charges free to move would move. The tangent at any point on the curve gives the direction of the electric field at that point.
Properties
1) Since the intensity of the field due to a positive charge is directed away from it, the lines of force proceed from the positive charge. Since the intensity of the field due to a negative charge is directed towards it, the lines of force proceed towards the negative charge and end at it. Thus lines of force start from positive charges and go to infinity or end at negative charges.
2) The number of lines of force passing through unit area taken around a point perpendicular to the direction of the electric filed gives the magnitude of the field at the point.
3) The direction of the tangent drawn to the line of force at any point gives the direction of the electric field at that point.
4) Only one line of force can pass through a given point, since the intensity of the electric field has a unique direction at each point. Thus two lines of force can never intersect.
Lines of force in a uniform electric filed are parallel, equidistant and in the same direction. Lines of force due to an isolated positive charge, an isolated negative charge, two positive charges, and due to a positive and a negative charge are shown in figures.
Electric Flux
Consider an arbitrary closed surface A in an electric filed E. Let the surface be divvied into a large number of elements, each of area dA. Each elementary area dA is so small that it is practically flat and the electric field intensity E within the surface is practically constant. Let the direction of E make angle θ with PN, the normal to dA. The normal PN is drawn so as to be directed away from dA. It is called the outward drawn normal to dA at P. The component of the elementary area dA, perpendicular to E is dAcosθ . Since the intensity of the electric field is E, the number of lines of force passing normally through unit are E. Therefore the number of lines of force passing normally through the area dAcosθ is EdAcosθ. This quantity is called the normal electric elementary areas and the sum taken, it gives the total outward normal flux of E over the area A.
It is given by Φ = ΣEdA Cosθ = EA Cos θ
Using Vector notation, it may be expressed as Φ = E.A
Gauss’s Theorem
It has been shown that electric flux is associated with electric field. But electric field arises from electric charge. So there is close relationship between electric charge and electric flux. Gauss’s theorem gives a relation between the flux through any closed surface and the net charge enclosed within the surface. The theorem may be stated as follows: The total flux of the electric field (also called total normal induction) over a closed surface is 1/e0 times the net charge enclosed by the surface. That is, the total electric flux over a closed surface enclosing charge q in free space is given by,
Φ = q/Є0
The closed surface taken in the electric field is called Gaussian surface.
Electric dipole and dipole moment
Two equal and opposite point charges separated by a small distance is called an electric dipole.
The product of one of the charges and the distance between the charges is called the electric dipole moment.
Let two charges +q and –q be separated by a small distances 2a. Then the electric dipole moment, p = q.2a
= 2a.q
The electric dipole moment is a vector directed along the axis of the dipole from the negative4 to the positive charge.
Electric Potential.
Consider a region free of any electric charge. Let a small charge +q be placed at any point in this region. It can be made to move about anywhere in the region without doing any work because this charge does not experience any electrical force.
To consider another case. Let a charge +q be placed in a region. Thus an electric field is produced around +q. Let a charge +q be introduced into this region. Now to move the charge +q from one point to another, work has to be done since this charge +q experiences a force of repulsion due to the presence of +q. Thus the presence of charge +q in the medium confers a property to each point of the medium by virtue of which work becomes necessary to move an electric charge up to that point. This property is called electrostatic or electric potential.
Usually the earth and the bodies connected to the earth by a conductor (earthed) are considered to be at zero potential. The potentials of other bodies are measured relative to the potential of the earth.
Electric potential at appoint is measured by the work done in taking unit positive charge from infinity (or zero potential) to that point.
The unit of electric potential is the volt.
The volt
The potential at a point is one volt if the work done in taking one coulomb of charge from infinity up to that point is one joule.
Ie, volt = joule/coulomb
Dimensions of potential
Potential = Work/Charge
Dimensions of Potential = Dim of work/Dim. of charge
= M1L2T-2/IT (Charge = Current x Time )
Therefore Dimensions of potential are M1L2I-1T-3
Potential Difference ( Pd )
The potential difference between two points in an electric field is measured by the work done in taking unit positive chare between those points.
Pd is also measured in volt.
Relation between intensity of electric field and potential – Intensity is Negative potential gradient.
Let E be the electric intensity at a point B due the presence of an electric charge +q at any point A. This field E acts along AB.
The force acting on the unit positive charge place at the point B is E. Then the work done in moving the unit positive charge from B to C, towards A, through a small distance dx can be given by
W = – E dx
The – sings shows that the work done is against the direction of the field. But this work done gives the potential difference dV between B and C
DV = -Edx
Ie E = -dv/dx
Dv/dx is the change of potential with respect to the distance. It is called the potential gradient. Thus from equn it amy he said that the intensity of electric field at any point is the negative potntail gradient at aht point. Or electric field may be represented as numerically equal to the p space rate of change of lelectrric potential.
The Electron Volt
Let V be the potential difference between two point sin an electric field, ie, the work done in moving unit positive charge between these points is V. Threfore the work required in moving a charge q between these points in the electric fild is qV. This work done will be stored in the charge as its energy. Hence the energy acquired by a charge q in moving between twopoints in an electric field under a potential difference V volts is qV Joules.
Thus, the energy acquired by an electron of charge e coulomb in moving between two points under a potential difference V volts = ev Joules. If V = 1 volt, the energy acquired by the electrons becomes e joules. This quantity I staken as a unit to measure the energy of charged particles and is called the electron volt. ( ev)
An electron volt is the energy acquired by an electron when it moves under a potential diference of oen volt.
1 ev = e Joules = 1.6021 x 10-19 J.
Electric Field due to a point charge
Let a point charge q be placed at a point in free space. This charge produces an electric field.
Let B be a point at a distance r from A. The intensity E of the electric field at B is to be determined.
Let a charge Q be placed at B. By Coulomb’s law, the force on this charge is given by
From this equation it is clear that the electric field at a point due toa point charge is iversily proportional to r2.
If the charge is in a medium of relative permittivity er, the intensity of the field is given by
The field acts along the line joining the point charge q and the point B. If the charge q is positie, the direction of the field will e away from the charge and if it is negative the field will e towards the charge.
CAPACITANCE
When electric charge is given to a conductor, the potential of the conductor increases. The change in potential is found to be directly proportional to the charge given to the conductor. Thus, if a charge Q raises the potential of a conductor by V,
V α Q
Or Q α V
Q = CV
Where C is a constant known as the capacitance of the conductor.
From the above equation, if V = 1, C = Q
Therefore, the capacitance of a conductor may be defined, as the charge required raising its potential by unity.
The unit of capacitance is the Farad
From equation (1)
C = Q/V
Ie, Capacitance = Charge / Potential
Therefore 1 farad = 1coulomb/1 volt
Hence the capacitance of a conductor is one farad, if a charge of one coulomb raises its potential by one volt.
Farad is too large a unit for practical purposes. Hence capacitance is usually expresses in macro farad (mf or μF) or Pico farad (micro-micro farad, μμf, or pf )
1μf = 10-6F
1pf = 10-6μF=10-12F
Dimensions of Capacitance
Capacitance = Charge/Potential
Dimensions of capacitance = Dimensions of charge/ Dimensions of potential
IT/M1L2I-1T-3 = M-1L-2I2T4
Ie Capacitance has dimensions M-1L-2I2T4
Capacitor or condenser
Any arrangement by which the capacity of a conductor is increased it called a capacitor or electrical condenser. Thus a capacitor is an electrical device for storing electric charges. In its simplest practical form, a capacitor consists of two conducting plates arranged parallel and close to each other with a dielectric medium ( may be air) between them.
Principle of the Capacitor
Consider an insulated metal plate A. Let it be charged with a charge +Q so that its potential becomes V. Now the capacitance of the plate, C = Q/V
Let another insulated metal plate be placed near A. The plate B gets charged by induction. The side of B nearer to A gets negatively charged and the farther side gets positively charged. The negative charge on B tends to lower the potential of A, while the positive charge on B tends to raise the potential of A. But as the negative charge is nearer to A its effect will be more predominant on the positive charge on B. Hence the potential of A gets slightly lowered. Therefore the capacitance of A is slightly increased.
Now, let the plate B be earthed. The positive charge on the outer surface of B flows to the earth. The negative charge on B remains due to the electrostatic attraction with the positive charge on A. Therefore the potential of A is very much lowered. Hence the capacitance of A increases considerably. This is the principle of the capacitor.
A capacitor has a capacitance of one farad if a charge of one coulomb flows from one conductor to the other when the pd between the conductors is one volt.
Capacitance of a Parallel Plate capacitor
The parallel plate capacitor consists of two parallel metal plates kept separated by a small distance. The capacitance of a parallel plate capacitor is the ratio of the charge on any one plate to the potential difference between the plates.
Let A and B be two metal plates each of area A. Let them be kept separated by a small distance d. The plate B is earthed.
Let a charge +Q be given to the plate A. The surface charge density, or charge per unit area of the plate,
σ = Q/A
or Q = Aσ.
The charge +Q on A induces a charge –Q on the inner surface of B, and a charge +Q on the outer surface of B. The charge +Q on B leaks to the earth. The electric field between the plates is uniform.
The intensity of the electric field at any point between the plates is given by
E =σ /ε0 where ε0 is the permittivity of free space.
Therefore work done in moving unit positive charge from A to B = Ed = σ / ε0 d.
This is equal to the potential difference V between the plates.
V = σ d/ ε0
Therefore Capacitance of the parallel plate capacitor,
C = Charge on any one plate/Potential Difference = Q / σ d/ ε0
= A σ ε0/ σ d ( Q = A σ )
= A ε0/ d Farad.
If the space between the plates is filled with a medium of relative permittivity εr,
C = A εr ε0/ d Farad.
From the above equation it is clear that the capacitance of a parallel plate capacitor may be increased by
1 increasing the area of the plates.
2.decreasing the separation between the plates, and
3. Using a medium having a high value for the dielectric constant.
If the parallel plate capacitor has n identical plates arranged at equal distance d apart and alternate plates connected together, its capacitance is given by C = εr ε0(n-1)A/d farad.
Here A represents the area of each plate.
Effect of Dielectric between the plates
A dielectric medium between the plates increases the capacitance. This is quite clear from the expression for the capacitance of the parallel plate capacitor. Also it is clear that thinner the dielectric, greater is the value fo the capacitance.
It keeps the plates very close together without actual contact.
It also prevents the electric discharge between the plates.
Different Types of Capacitors
a) Paper capacitor
In this type two long strips of thin tin foil or aluminum foil act as the conducting plates. Paraffin paper acts as the dielectric medium. It is placed between the metal foil and then rolled up into a cylindrical form. These capacitors are very cheap and convenient to handle. They are widely used in many electronic devices.
b) Electrolytic capacitor
It has been mentioned earlier that the capacitance of a capacitor can be increased by decreasing he thickness of the dielectric medium between the plates. In electrolytic capacitors, the thickness of the dielectric is extremely small (of the order of 10-6cm) Therefore the capacitance is very large. Hence these capacitors are used in electrical circuits where large capacitance is required.
There are two types of electrolytic capacitors – the wet type and the dry type. The wet type of capacitor consists of an aluminium cylinder A, kept dipped in a solution of ammonium borate taken in another cylindrical aluminium vessel C. A is the anode and C is the cathode. Electrolysis takes place when a D C potential of about 450 volts is applied between the anode and cathode. Then a very thin coating of aluminium oxide is formed on the anode. This film offers a very high resistance to the flow of current in one direction and a very low resistance to the current in the opposite direction. Thus the foil acts as a dielectric for currents in one direction. This behaves as a parallel plate capacitor. The thickness of the dielectric being very small, electrolytic capacitors have large capacitance.
Electrolytic capacitors are widely used tin power supply units for smoothening of varying currents. They can be used only with D C since a reversal of polarity causes the break down of the dielectric film.
Combination of Capacitors.
Capacitors may be connected either in series or in parallel.
a) Capacitors in series
Let three capacitors of capacities C1, C2 and C3 be connected in series as in figure between two points A and B. Let a charge +Q be given to a plate of capacitor C1. This induces a charge –Q on the other plate. Since the capacitors are connected in series, the charge on all the capacitors will be the same, but the potential across each capacitor will be different. Let V1, V2 and V3 be the potentials across the plates of C1, C2 and C3 respectively.
Then V1 = Q/C1, V2 = Q/C2, and V3 = Q/C3.
Let V be the pd between A and B.
Then V = V1 + V2 + V3
Ie V = Q/C1 + Q/C2 + Q/C3.
Let the three capacitors be replaced by a single capacitor or capacitance C such that when the pd between A and B is V, the charge on it is the same as before. Then C is called the effective of equivalent capacitance of the combination.
P.d between A and B, V = Q/C
From equn (4) and (5)
Q/C = Q/C1 + Q/C2 + Q/C3
1/C = 1/C1 + 1/C2 +1/C3
Thus, when a number of capacitors are connected in series, the reciprocal of the effective capacitance is the sum of the reciprocals of the individual capacitances.
It may be noted that when several capacitors are connected in series,
1. The charge on each capacitor is the same, and
2. The effective capacitance is less than the smallest individual capacitance.
The series combination is used to reduce the capacitance.
Capacitors in Parallel
Let three capacitors of capacitances C1,C2 and C3 be connected in parallel between A and B. Let V be the p.d applied between A and B. Therefore the Pd across the plate of each capacitor is V. Let Q1, Q2 and Q3 be the charges acquired by C1, C2 and C3 respectively.
Total Charge, Q = Q1 + Q2 + Q3
But Q1 = C1V, Q2 = C2V, Q3 = C3V
Total charge on the three capacitors
Q = Q1 + Q2 + Q3
= C1V + C2V + C3V
= (C1 + C2 + C3) V
Let the three capacitors be replaced by a single capacitor of capacitance C such that it acquires the same charge Q when the p.d between its plates is V. Then this capacitor is called the equivalent capacitor of the combination.
Then Q = CV
CV = (C1 + C2 + C3 ) V
Or C = C1 + C2 + C3
Thus, when a number of capacitors are connected in parallel, the effective capacitance is the sum of the capacities of the individual capacitors.
It may be noted that when several capacitors are connected in parallel,
The p.d across the plates of each capacitor is the same.
The effective capacity is larger than the largest individual capacitance.
Parallel combination of condensers is used to increase the capacity.
Energy of a charged capacitor.
Let a capacitor be charged with a charge Q. When it is charged, its potential increases uniformly from its initial value zero to the final potential V. Therefore the average potential of the capacitor in the processes of charging = 0 + V/2 = V/2. Let C be the capacitance of the capacitor.
The potential at a point is the work done in taking unit positive charge from zero potential to that point. Therefore the work done in giving a charge Q to the capacitor,
W = Charge x Average potential
= Q.V/2 = ½ QV
This work done will be stored in the capacitor as its electrical potential energy. Thus, energy of the charged capacitor, W = ½ QV
Or W = ½ CV2, since Q =CV
W = ½ Q2/C, since V = Q/C.
PROBLEMS
A parallel plate capacitor of area 1 square metre and relative permittivity 7 is charged to a potential of 300 V. If the distance between the plates is 10-4m, find the capacity and the energy stored in the capacitor.
A parallel plate capacitor of capacity 0.5 mf has its plates separated by 1mm. If the space between the plates is filled with a medium of relative permittivity 1.5, find the plate area of the capacitor.
The plates of a parallel plate capacitor of capacitance 0.01mF are 1m2 each in area. The dielectric constant of the medium between the plates is 2.5. Calculate the distance between the plates.
Find the area of the paper used in a capacitor of capacitance 0.5mF if the dielectric constant of paper is 2,5 and its thickness is 0.05 mm.
Find the equivalent capacitance of three capacitors of capacitance 2mF, 3mF and 5mF when connected (1) in series, and (2) in parallel.
Three capacitors of 2,3 and 4 mF are connected in series and then in parallel. Compare the effective capacitance in the two cases.
Calculate the capacitance of a capacitor which will acquire a charge of 0.05 micro coulomb at 150 volts
Calculate the current required to charge a capacitor of capacitance 25mF to a potential of 400V in 2 minutes.
Find the work done in charging a capacitor of capacitance 25mF to a potential of 300 volts.
Three capacitors fo 2mF,3mF and 4 mF are connected in series and charged by a potential of 260 volts. Calculate the p.d across each capacitor.
A capacitor of capacitance 1mF is charged with a potential 75 volts and another of 0.5 mF is charged with 80 volts. The two are then connected in parallel. Find the common potential and charges on each capacitor.
Important instructions from Dr Arnab Bhattacharya, Chairperson, IRIS panel of Judges
Congratulations on being selected for the IRIS-2007 National Fair! To help you with your project
presentation, the Scientific Review Committee would like to share with you some information which we
hope will ensure that you highlight the important aspects of your project, and also make the evaluation
process easier for both you as well as the judging team at IRIS 2007.
Do remember that judges will focus on 1) what YOU, the student or team did in the project; 2) how well
you followed the scientific methodologies; 3) the detail and accuracy of research as documented in the data
book; and 4) whether experimental procedures were used in the best possible way. We are always looking
out for innovative ideas, original work, well thought-out research and scientific/engineering skill.
At the IRIS national fair, your project display must be complete, and comprehensive such that, as far as
possible, it is possible for someone visiting your project to get a the total picture of the work done
WITHOUT YOUR PRESENCE. Please note that a part of the judging will be carried out in the absence of
participants, when the judges go through your project display as well as your logbook/original data etc. This
means that it is VITAL that your ORIGINAL DATA, WITHOUT MODIFICATIONS is available to be
viewed. Your results alone without the data to substantiate your work are not enough. Original laboratory
readings are never neat, and we do not expect them to be! Please do NOT “copy into fair” your data, or print
it neatly, it does not impress judges; on the contrary, it is counted against you!
Do make sure that your project display is such that it is readable from a comfortable distance (typically
1meter away). Make your title clear and easy to read. Avoid type styles that may be hard to read. Fonts that
have shadows or outlines may seem like a great idea but they are harder to read than simple lettering. The
title lettering should be at least 5cm high.(For those using computer based word-processors/presentation
software, a minimum font size of 20 is recommended). Do NOT display pages of text, but summarize the
important aspects of your project. Plotting relevant graphs, charts etc. can often be much more instructive
than showing huge tables of data.
An example of a good display and here’s a bad one!
From Janice VanCleave’s Guide to the Best Science Fair Projects, (John Wiley & Sons, Inc., 1997)
While organizing your display, please ensure that there is a sensible and easy progression through the
display so that the average person can easily understand it. While there is no one correct way to set up a
display, it must, however, make sense and be easy to follow. Remember that most people read from left to
right and from top to bottom. Design what the “center” of your display will be. This is where everyone will
look first. Group topics that go together like question, research, and hypothesis; materials and procedures;
analysis and conclusion. Make a small sketch of where everything will go and lay it out before you glue
anything down to make sure it looks good. (If needed, please number your panels for easy readability)
Do make sure that all items that are displayed are permissible as per the display guidelines in the IRIS
handbook. Basically, anything that is or could be hazardous to other participants or the public is prohibited
and cannot be displayed.
If your work involves working with human subjects, consent forms must have been obtained prior to
experimentation, and should be available in case needed. Similarly, any work involving living organisms,
human/animal tissues, DNA etc. must have the appropriate documentation if required.
Please especially ensure that the data presented has the correct number of significant digits and the proper
units.
PRACTICE your presentation! Remember that most judges will spend only about 5-10 minutes with you,
where they would like to focus mostly on what you have done, rather than the background of the problem,
and have enough time to ask you questions and have a discussion. So do not come with a “memorized
speech”, with a long introduction, but be prepared to explain what you have done in your project in as short
or long as a time that may be available.
Some things that judges look for while going through a project:
1. Clearly defined objectives – what did you want to do, and how did you go about doing it? (Presenting
original ideas, Stating the problem clearly, Defining the variables and using controls, Relating background
reading to the problem)
2. Skill in performing the project – (knowledgeable about equipment used, Performing the experiments
with little or no assistance except as required for safety/special equipment/data collection, Demonstrating
the skills required to do all the work necessary to obtain the data reported)
3. Appropriate data collection – (using a log-book and journal to collect data and document research,
Repeating experiment to verify the results, Spending an appropriate amount of time to complete the project,
Having measurable results, appropriate control experiments etc.)
4. Correct data interpretation – (collecting enough data to make a conclusion, Using research to interpret
data collected, Using only data collected to make a conclusion, Using tables, graphs etc. to visualize the
data easily)
5. Project Presentation (Written Materials, Interviews, Displays)
(Having a complete and comprehensive report, Answering questions accurately, Using the display during
oral presentation, Justifying conclusions on the basis of experimental data, Summarizing what was learned,
Presenting an attractive and interesting display)
We hope some of these tips will help you in making a good presentation at the National Fair. There are also
a lot of excellent on-line resources that discuss various aspects of science project data analysis and
presentation.
We’re sure you’ll find the National Fair not only an intense scientific experience but also a lot of fun. See
you all in Delhi, and all the best!
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Initiative for Research & Innovation in Science (IRIS) |
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Intel has joined hands with the Department of Science and Technology (DST) and Confederation of Indian Industry (CII) to jointly promote inventions and innovations among students and youth in India. This is aimed at accelerating Intel India’s ongoing effort towards encouraging young inventors in India.
Launched in February 2006; ‘IRIS – Initiative for Research & Innovation in Science’, an Intel-DST-CII initiative, is the merger of two programs the Intel Science Talent Discovery Fair and Steer the Big Idea. It is India’s largest science initiative for students involving Government and private partnership. The merger creates THE Science Fair of India and achieves a higher level of reach across schools, colleges, labs and research institutions in the country.
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On February 22, 2006 Mr. N. Srinivasan, director general, CII, Prof. V.S. Ramamurthy, secretary, DST and Mr. Amar Babu, Director Intel SE-Asia formally signed an expression of interest leading to the Memorandum of Understanding (MOU). Hon Minister for Science & Technology, Mr Kapil Sibal was the Chief guest for the Launch.
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Intel over the past six years has been organizing an annual nationwide initiative, the Intel Science Talent Discovery Fair (Intel STDF) that aims to infuse a spirit of discovery in school children and increase their interest in science and technology.
The Intel STDF national winners represent India and showcase their work to the international scientific community at the largest pre-college science fair in the world – the Intel International Science and Engineering Fair (Intel ISEF) held every year in USA.
Under the CII-DST initiative – Steer the Big Idea, Indian participation was organized at the International Exhibition for Young Inventors (IEYI) at Tokyo in 2004 and at Kuala Lumpur in 2005. Different countries in Asia host IEYI every year and India has got the opportunity to host IEYI in 2006.
IRIS aims to help build scientific temperament and an innovative culture amongst the youth of the country, as well as help popularise science and technology in the schools and amongst students. It expected to touch over 20 lakh students from over 25000 schools and institutes in India and infuse the spirit of science, research and innovation among them.
IRIS would recognize and reward innovative & outstanding projects and provide a platform for these young innovators to interact with Indian industry as well as get recognized at International events like the ISEF and the IEYI.
IRIS has been initiated in April 2006 and will start by inviting project synopses from participants in four categories from age 06 – 35 years. These entries will go through 2 screening rounds to shortlist 300 finalists from all four categories for participating at the National Fair in December 2006. Apart from getting awards and scholarships; the winners at the National fair will go on to represent India at the International Fairs like the Intel ISEF and IEYI. This year since the IEYI is being hosted by India, the Indian students qualifying for the National Fair will be directly competing with international participants.
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