1

Section-A1. Draw a graph to show variation of capacitive-reactance

with frequency in an a.c. circuit.2. What is the function of a ‘Repeater’ used in communication

system?3. The line AB in the ray diagram represents a lens State

whether the lens is convex or concave.

A

B

4. The field lines of a negative point charge are as shown inthe figure. Does the kinetic energy of a small negativecharge increase or decrease in going from B to A?

AB

– q

GENERAL INSTRUCTIONS(i) There are 26 questions in all. All questions are compulsory.(ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E.(iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section

C contains twelve questions of three marks each, Section D contains one value based question of four marks andSection E contains three questions of five marks each.

(iv) There is no overall choice. However, an internal choice has been provided in one question of two marks, one questionof three marks and all the three questions of five marks weightage. You have to atempt only one of the choices insuch questions.

(v) You may use the following values of physical constants wherever necessary:c = 3 × 108 m/sh = 6.63 × 10–34Jse = 1.6 × 10–19Cm0 = 4p × 10–7 TmA–1

9 2 –2

0

1 9 10 Nm C4

= ´pe

me = 9.1 ×10–31 kgmass of neutron = 1.675 × 10–27 kgmass of proton = 1.673 ×10–27 kgAvogadro’s number = 6.023 × 1023 per gram molBoltzmann constant = 1.38 × 10–23 JK–1

5. Distinguish between emf and terminal voltage of a cell.Section-B

6. How does one explain, using de Broglie hypothesis,Bohr’s second postulate of quantization of orbital angularmomentum?

7. In a meter bridge shown in the figure, the balance pointis found to be 40 cm from end A. If a resistance of 10 Wis connected in series with R, balance point is obtained 60cm from A. Calculate the values of R and S.

G

( )

A B

R S

8. What is ground wave communication? Explain why thismode cannot be used for long distance communicationusing high frequencies.

9. The equivalent wavelength of a moving electron has thesame value as that of a photon of energy 6 × 10 –17 J.Calcu late the momentum of the electron.

10. A ray of light passes through an equilateral glass prismsuch that the angle of incidence is equal to the angle ofemergence and each of these angles is equal to 3/4 ofangle of prism. Find the angle of deviation.

2

ORCalculate the speed of light in a medium whose criticalangle is 45°. Does critical angle for a given pair of mediadepend on the wavelength of incident light? Give reason.

Section-C11. Write two important considerations used while fabricating

a Zener diode. Explain, with the help of a circuit diagram,the principle and working of a Zener diode as voltageregulator.

12. Draw the necessary energy band diagrams to distinguishbetween conductors, semiconductors and insulators. Howdoes the change in temperature affect the behaviour ofthese materials ? Explain briefly.

13. (a) What are the three basic units in communicationsystems?Write briefly the function of each of these,

(b) Write any three applications of the Internet used incommunication systems.

14. (a) Write the necessary conditions to obtain sustainedinterference fringes.

(b) In Young’s double slit experiment, plot a graphshowing the variation of fringe width versus thedistance of the screen from the plane of the slitskeeping other parameters same. What informationcan one obtain from the slope of the curve?

(c) What is the effect on the fringe width if the distancebetween the slits is reduced keeping other parameterssame?

15. A series LCR circuit is connected across an a.c. source ofvariable angular frequency ‘w’. Plot a graph showingvariation of current ‘i’ as a function of ‘w’ for tworesistances R1 and R2 (R1 > R2).Answer the following questions using this graph.(a) In which case is the resonance sharper and why?(b) In which case is the power dissipation more and

why?16. (a) Give two reasons to explain why reflecting telescopes

are preferred over refracting type.(b) Use mirror equation to show that convex mirror

always produces a virtual image independent of thelocation of the object.

17. State Lenz’s law. Illustrate, by giving an example, how thislaw helps in predicting the direction of the current in aloop in the presence of a changing magnetic flux.In a given coil of self-inductance of 5 mH, current changesfrom 4 A to 1 A in 30 ms. Calculate the emf induced in thecoil.

ORIn what way is Gauss’s law in magnetism different fromthat used in electrostatics ? Explain briefly.The Earth’s magnetic field at the Equator is approximately

0.4 G. Estimate the Earth’s magnetic dipole moment. Given: Radius of the Earth = 6400 km.

18. How are electromagnetic waves produced? What is thesource of energy of these waves?Draw a schematic sketch of the electromagnetic wavespropagating along the + x–axis. Indicate the directions ofthe electric and magnetic fields. Write the relation betweenthe velocity of propagation of wave and the magnitudesof electric and magnetic fields.

19. Obtain the relation between the decay constant and halflife of a radioactive sample.The half life of a certain radioactive material against a-decay is 100 days. After how much time, will theundecayed fraction of the material be 6.25%?

20. Find the equivalent capacitance of the network shown inthe figure, when each capacitor is of 1 mF. When the endsX and Y are connected to a 6 V battery, find out (i) thecharge and (ii) the energy stored in the network.

X Y

21. State the underlying principle of a potentiometer. Writetwo factors by which current sensitivity of a potentiometercan be increased. Why is a potentiometer preferred overa voltmeter for measurirg the emf of a cell?

22. (a) Define the term ‘intensity of radiation’ in terms ofphoton picture of light.

(b) Two monochromatic beams, one red and the otherblue, have the same intensity. In which case (i) thenumber of photons per unit area per second is larger,(ii) the maximum kinetic energy of the photoelectronsis more? Justify your answer.

Section-D23. During a thunderstorm the ‘live’ wire of the transmission

line fell down on the ground from the poles in the street.A group of boys, who passed through, noticed it andsome of them wanted to place the wire by the side. Asthey were approaching the wire and trying to lift the cable,Anuj noticed it and immediately pushed them away, thuspreventing them from touching the live wire. Duringpushing some of them got hurt. Anuj took them to adoctor to get them medical aid.Based on the above paragraph, answer the followingquestions :(a) Write the two values which Anuj displayed during

the incident.(b) Why is it that a bird can sit on a suspended live wire

without any harm whereas touching it on the groundcan give a fatal shock ?

(c) The electric power from a power plant is set up to avery high voltage before transmitting it to distantconsumers. Explain, why.

Section-E

3

24 (a) State Kirchhoff’s rules and explain on what basisthey are justified.

(b) Two cells of emfs E1 and E2 and internal resistancesr1 and r2 are connected in parallel. Derive theexpression for the (i) emf and (ii) internal resistanceof a single equivalent cell which can replace thiscombination.

OR(a) “The outward electric flux due to charge + Q is

independent of the shape and size of the surfacewhich encloses it.” Give two reasons to justify thisstatement.

(b) Two identical circular loops ‘1’ and ‘2’ of radius Reach have linear charge densities –l and +l C/mrespectively. The loops are placed coaxially with theircentres 3R distance apart. Find the magnitude anddirection of the net electric field at the centre of loop‘1’.

25. (a) Use Huygens’ principle to show the propagation ofa plane wavefront from a denser medium to a rarermedium. Hence find the ratio of the speeds ofwavefronts in the two media.

(b) (i) Why does an unpolarised light incident on apolaroid get linearly polarised ?

(ii) Derive the expression of Brewster’s law whenunpolarised light passing from a rarer to adenser medium gets polarised on reflection atthe inteface.

ORA biconvex lens with its two faces of equal radius ofcuivature R is made of a transparent medium of refractiveindex m1. It is kept in contact with a medium of refractiveindex m2 as shown in the figure.

R

m1

m2

R

(a) Find the equivalent focal length of the combination.(b) Obtain the condition when this combination acts as

a diverging lens.(c) Draw the ray diagram for the case m1 > (m2 + 1) / 2,

when the object is kept far away from the lens. Pointout the nature of the image formed by the system.

26. Two infinitely long straight parallel wires, ‘1’ and ‘2’,carrying steady currents I1 and I2 in the same direction areseparated by a distance d. Obtain the expression for themagnetic field B

ur due to the wire ‘1’ acting on wire ‘2’.

Hence find out, with the help of a suitable diagram, themagnitude and direction of this force per unit length onwire ‘2’ due to wire ‘1’. How does the nature of this forcechanges if the currents are in opposite direction? Use thisexpression to define the S.I. unit of current.

ORDraw a necessary arrangement for winding of primary andsecondary coils in a step-up transformer. State itsunderlying principle and derive the relation between theprimary and secondary voltages in terms of number ofprimary and secondary turns. Mention the two basicassumptions used in obtaining the above relation.State any two causes of energy loss in actual transformers.

.

4

1. As capacitive reactance

1 12CX

C fC= =

w pHence, it is inversely proportional to frequency f.Graph: XC versus f

f

CX

2. A repeater is an amplifying device used to increase therange of the transmission in communication systems withthe help of a set of receiver and transmitter

3. As the rays diverging from lens so, the lens is concavein nature.

4. Since we know that a negative charge always experiencesa force in the direction opposite to that of the electric fieldpresent, the negative charge will experience the forceaway from the centre. This will cause its motion to retardwhile moving from B to A. Hence, its kinetic energy willdecrease in going from B to A.

5.

It is the maximum potential It is the potential differencedifferencethat can be delivered across the terminalby a cell whenno current flowsthrough the circuit.

Emf Terminal voltage

s of the load when current flows throughthe circuit.

It is represented by It is represented by V andE and remains depends on the internalconstant for a cell. resistance of the cell.

6. According to de-Broglie hypothesis, a stationary orbit isthe one that contains an integral number of de-Brogliewaves associated with the revolving electron.For the permissible orbit, we have

2prn = nl,where l is the wavelength, rn - radius of orbitAccording to de-Broglie, wavelength of matter waves isgiven by

n

hmv

l =

wherevn = speed of electron in nth orbit

SOLUTIONS

\ 2 nn

nhrmv

p =

Þ 2n nnhmv r =p

This is Bohr’s second postulate of quantisation of orbitalangular momentum

7. When balance point is at 40 cm, we have

40 40100 – 40 60

RS

= =

Þ23

RS

=

Þ 3 R = 2 S .... (i)When a resistance of 10 W is added in series with R,Then, equivalent reistance of R’ = R + 10Now, balance point is obtained at 60 cm.

\10 60 60

100 – 60 40R

S+

= =

Þ10 3

2R

S+

=

Þ 2 R + 20 = 3S …(ii)Solving eqs (i) and (ii), we getS = 12 W and R = 8 W

8. Ground waves are the radio waves that travel along thesurface of the Earth. The propagation of the wave isguided along the Earth’s surface and follows the curvatureof the Earth.The propagation of high frequency wave is not possiblethrough ground waves for long distance communicationbecause while progressing, ground waves induce currentin the ground and bend round the corner of the objectson the Earth due to which the energy of the ground wavesof high frequency is almost absorbed by the surface of theEarth after travelling a small distance. Thus, ground wavecommunication is not suited for high frequency.

9. According to the de Broglie hypothesis, the momentum(p) of an electron is given by

hp =l

… (i)

Energy of photon, –176 10hcE J= = ´l

(Given)

Þ–34 8

–17(6.625 10 ) (3 10 )

(6 10 )´ ´

l =´

Þ l = 3.31 × 10–9 mFrom eqn (i), we have

5

–34

–96.626 103.31 10

p ´=

´ = 2 × 10– 25 kg m s –1

10. The angle of deviation d for a ray of light in a prism isgiven byd = i + e – Awhere i = Angle of incidence of ray

e = Angle of emergence andA = Angle of prism

Þ d = 2i – A (Q e = i)

Given : i = e = 34

A

\ d = 2 × 34

A – A = 12

A

As the prism is equilateral, ÐA = 60°

\ d = 12

× 60° = 30°

ORAccording to Snell’s Law, we know

1sinC

m =

1sin

cv C

\ = CV

æ öm =ç ÷è øQ

Þ v = sin C × cÞ v = sin 45° × 3 × 108

Þ v = 2.12 × 108

Therefore, speed of light in the medium is 2.12 × 108 ms–1.We know that the critical angle of the medium depends onits refractive index and the refractive index µ of a mediumis inversely proportional to the wavelength of incidentlight. So, the critical angle of the medium also dependsupon the wavelength of incident light.

11. Two important considerations while fabricating a Zenerdiode are as follows:(i) The Zener diode should be fabricated by heavily

doping both p and n junctions.(ii) Proper wafer-handling devices must be used to avoid

contamination, which may affect the quality andstability of the diode.

PrincipleZener diode when reverse biased shows a suddenincrease in current to a high value at a certain voltageknown as the breakdown voltage or the Zener voltage.Working

input voltage Constantoutput voltageRLZener

diode

A Zener diode is connected across the fluctuating voltagesource through a resistance R. A constant voltage supply

is maintained across the load resistance RL.When the input voltage increases, the resistance of theZener diode decreases and hence the current through thediode increases. Thus, a large voltage drop is seen acrossthe resistance R and the output voltage ac’oss RL remainsat a constant value. When the input voltage decreases,the resistance of the Zener diode increases and hence thecurrent through the diode decreases. Only a small voltagedrop takes place now across the resistance R and theoutput voltage at RL remains constant. Thus, we get aconstant voltage in spite of the fluctuating input voltage.In this way, a Zener diode acts as a voltage regulator.

12. In conductors, the conduction and the valence bandoverlap each other.As the temperature increase, the conductivity of theconductors decreases due to increase in the thermalmotion of the free electrons.Energy band diagram for a conductor

Elec

tron

ener

gy

Overlappingconduction band

Valenceband

Ev

Ec

( 0)gE »

Energy band diagram for an insulator

Elec

tron

ener

gy

Emptyconductionband

Valenceband

Ev

Ec

> 3gE eV

Here, the valence band is completely filled and theconduction band is empty. The energy band gap of theinsulator is quite large. So, on increasing the temperature,the electrons of the valence band are not able to reach theconduction band. Therefore, electrical conduction in thesematerials is almost impossible.Energy band diagram for a semiconductor

Elec

tron

ener

gy

Ev

Ec

< 3gE eV

In semiconductors, the valence band is totally filled andthe conduction band is empty but the energy gap betweenconduction band and valence band is quite small. At 0 K,electrons are not able to cross even this small energy gapand, hence, the conduction band remains totally empty. Atroom temperature, some electrons in the valence bandacquire thermal energy greater than energy band gap,which is less than 3 eV and jump over to the conductionband where they are free to move under the influence ofeven a small change in the temperature. As the temperature

6

increases more and more electrons crosses the band gapand hence conductivity increases.

13. (a) A communication system consists of three basicunits:

(i) Transmitter: This unit is used for transmitting theinformation after modifying it to a suitable form. Itbasically consists of a transducer that convertssignal in any physical form to electrical signal fortransmission. After that, the signal is modulated totransmit over long distances.

(ii) Communication channel: This unit carries themodulated signal from the transmitter to the receiver.E.g free space, transmission lines acts as acommunication channel.

(iii) Receiver: This receives the signal, followed by ademodulator, an amplifier and a transducer. Thedemodulator demodulates the modulated signal, theamplifier boosts up its intensity and the transducerconverts it back again electrical form to the neededphysical form.

(b) The list of uses of Internet is given below.(i) E-banking: It is an electronic payment system that

enables customers of a financial institution (usualy abank) to proceed for financial transactions on awebsite operated by that institution. To bank onlinewith an institution, it is required that the customer isa member of that institution and has the access ofInternet.

(ii) Internet surfing: Navigation over World Wide Web(www) from one webpage/website to another iscalled Internet surfing. It is an interesting way ofsearching and viewing information on any topic ofinterest.

(iii) E-shopping (E-commerce): It is a form of electroniccommerce, by which the consumers can buy theproducts or the services over Internet.Apart from the above internet is used for sendinge-mail, e-ticketing, etc.

14. (a) The necessary conditions to obtain sustainedinterference fringes are:

(i) The two sources of light must be coherent.(ii) The two sources should preferably be monochromatic.(iii) The coherent sources must be very close to each

other.(b) The fringe width in Young’s double slit experiment is

given by

Dd

lb =

where l = wavelength of source D = distance between the slits and screen d = distance between the slits

Þ Db µThe variation of fringe width with distance of screen from

the slits is given by the graph shown below:

b

D

It is a linear graph with slope equal to l/d , So. for thefringe width to vary linearly with distance of screen fromthe slits, the ratio of wavelength to distance between theslits should remain constant. Therefore, it is advised totake wavelengths of incident light nearly equal to thewidth of the slit.(c) Fringe width is given by

Dd

lb =

Hence, if the distance between the slits is reduced thenthe width of the fringes increases.

15. The variation of current with angular frequency for thetwo resistances R1 and R2 shown in the graph below.

w

R1

R2i

wr

Y

X

Here,wr= Resonance frequency(a) From the graph, we can see that resonance for the

resistance R2 is sharper than for R1 because resistanceR2 is less than resistance R1 Therefore, at resonance,the value of peak current will rise more abruptly fora lower value of resistance.

(b) Power associated with the resistance is given byP = Ev Iv

From the graph, we can say that the current in case of R2is more than the current in case of R1. Hence, the powerdissipation in case of the circuit with R2 is more than thatwith R1

16. (a) The reflecting telescopes are preferred over refractingtype because of the following reasons:

(i) There is no chromatic aberration in case of reflecting

7

telescopes as the objective is a mirror.(ii) Spherical aberration is reduced in case of reflecting

telescopes by using mirror objective in the form of aparaboloid.

(b) For convex mirror, f = +veobject distance, u = –ve or u < 0.Using the mirror formula we have,

1 1 1f v u

= + Þ 1 1 1–v f u

=

Since f > 0 and u < 0, then from the above equations, weget that

1 0v

> Þ v < 0

Hence, a image is always formed at the backside of themirror. Therefore, the image formed by the convex mirroris always virtual in nature.

17. According to Lenz’s law, the polarity of emf induced in thecoil always opposes the change in the magnetic fluxresponsible for production of induced emf.E.g., When a north pole of a bar magnet is brought closerto a coil, as shown in figure, the amount of magnetic fluxlinked with the coil increases. So, the emf induced in thecoil is directed such that it opposes this change inmagnetic flux responsible for production of induced emf,which is possible only when the current induced in thecoil is in anticlockwise direction.Similarly, the direction of induced current can be foundwhen north pole of bar magnet is moved away from thecoil. As the magnetic flux in this case decreases, theinduced current is in the clockwise direction.

NS

Inducedcurrent

Mag. field frominduced current

N

Approaching

N S

magnet

N S

Inducedcurrent

Mag. field frominduced current

Approaching

N S

magnet

Given:Self inductance, L = 5 mH = 5 × 10–3 HChange in current, dl = (1 – 4) = –3 AChange in time, dt = 30 ms = 30 × 10–3 sThe emf induced in the coil is given by

– dIe Ldt

=

Þ–3

–3(–5 10 ) (–3)

30 10e ´ ´

=´

Þ e = 0.5 VOR

Gauss’s law for magnetism states that net magnetic flux

Bf through any closed surface is zero.

. 0B B dsf = =òuv uuv

Ñ

On the other hand, Gauss’s law for electrostatics statesthat electric flux through a closed surface S is given by

0.E

qE dsf = =eò

uv uuv

Ñ

So, if an electric dipole is enclosed by surface, electric fluxwill be zero. But in magnetism there is no counterpart ofisolated charge as in electricity. So, we can say thatisolated magnetic poles or monopoles do not exist.

As we know 034

MB

dm

=p

Hered = R = Radius of the EarthGiven :B = 0.4G = 0.4 × 10–4 T

04m

p = 10–7 Wb AA–1 m–1

d = 6400 km = 6. 4 × 106 m

\3

0

4 BdM p=

m

Þ–4 6 3

–7(0.4 10 )(6.4 10 )

10M ´ ´

=

Þ M = 1.05 × 1023 Am2

18. Electromagnetic waves are produced by a moving chargewith non-zero acceleration. When the charge moves withacceleration, both the magnetic and electric fields changecontinuously. This change produces electromagneticwaves. An accelerated charge is the source of energy ofthese waves.Schematic sketch of the electromagnetic waves:

Magnetic induction vectorE BB

XO

Z

YElectric intensity vector

E EB

The relation between the velocity of propagation and themagnitudes of electric and magnetic fields is E = CB.

8

19. Relation between decay constant and half life of aradioactive substance:The number of atoms at any instant in a radioactivesample is given by

N = N0 e–lt

N = total number of atoms at any instantN0 = number of atoms in radioactive substance att = 0When t = T (Where T is the half life of the sample)

02

NN =

Þ –002

TNN e l=

Þ elT = 2Taking log on both the sides, we getlT = loge 2 = 2.303 log10 2

Þ 102.303log 2=

lT

Þ0.693

=l

T

Let t be the required time after which the undecayedfraction of the material will be 6.25%.

\6.25 1100 16

=

\ N = 016N

But N = 012

nN æ ö

ç ÷è ø

where tnT

=

Þ 00

116 2

nNN æ ö= ç ÷

è øÞ n = 4\ Time’ t = n × TÞ t = 4 × 100

t = 400 days20. The equivalent circuit is given below.

1

X Y

Fm

Fm Fm

Fm1

1 1

There are two capacitors in one branch in series. So, theequivalent capacitance of one branch will be

1 1 21 1

F+ = m

The arrangement will be further reduced as

2 Fm

2 Fm

X Y

Now, both the capacitors are in parallel, so the equivalentcapacitance will be2 + 2 = 4mF(i) Voltage, V = 6 VThe charge in the network is given byq = CVNow,q= 4 × l0–6 × 6= 24 × 10–6 C = 24 mC(ii) The energy stored in the network is given by

212

E CV=

–6 21 4 10 (6)2

= ´ ´ ´

–61 4 36 102

= ´ ´ ´

= 72 × 10–6 J = 72 mJ21. (a) Principle of potentiometer: When a steady current

flows through the potentiometer wire then, thepotential difference across the uniform wire is directlyproportional to the length of the part across whichthe potential is measured.

(b) Current sensitivity can be increased by(i) increasing the length of the wire(ii) decreasing the current in the wire using a rheostat(c) Potentiometer is preferred over voltmeter because it

measures accurate emf of the cell. It uses nullmethod, so no current is drawn by the galvanometerfrom the cell in balanced condition of potentiometerand a voltmeter measures the voltage across theterminals of a cell when the cell is in closed circuit.This voltage is called terminal voltage of a cell notemf.

22. (a) Intensity of radiation is defined as the number ofphotons falling per unit area in unit time.

(b) (i) Number of photons per unit per second in boththe beams is same as they have the sameintensity.

(ii) The maximum kinetic energy of photoelectronsis given byE = hc/l – f

We know that the wavelength of blue beam is less thanthat of red beam. So, we can say that the maximum kineticenergy of the photoelectrons of the blue beam will bemore.

23 (a) During this incident, Anuj reflected the followingvalues:

(i) He was concerned about the lives of the individuals.

9

(ii) He was well aware about what would be the aftereffect of touching a live wire with naked hands andhe immediately acted upon this situation and savedlives of the boys.

(b) When the bird perches on a live wire, its bodybecomes charged for the moment and has samevoltage as the live wire; however, no current flowsinto its body. Body is a poor conductor of electricityas compared to the copper wire, so there is no reasonfor electrons to take a detour through the bird’sbody. Therefore, no current flows from the body ofthe bird.On the other hand, if the bird touches the groundwhile being in contact with the high voltage live wire,then the electric circuit gets complete and highcurrent flows through the body of the bird, giving ita fatal shock.

(c) The electric power from a power plant is set up at avery high voltage before transmitting it to distantconsumers to reduce the loss of power duringtransmission.

The power loss during the transmission = I2 R. So thepower loss can be minimised by reducing the value of thecurrent. Thus, to reduce the value of current, value of thevoltage of the electric power should be kept high beforetransmitting.Power = Voltage × CurrentFrom the above equation, we can see that the current canbe reduced by increasing the voltage of the electric power.

24. (a) Kirchhoff’s first rule (Junction rule): The algebraicsum of the currents meeting at a point in an electricalcircuit is always zero.

0I =åThis law is justified on the basis of law of conservationof charge.Kirchhoff’s second law (Loop rule): In a closed loop,the algebraic sum of the emfs is equal to the algebraicsum of the products of the resistances and thecurrent flowing through them.

0I Re + =å åThis law is justified on the basis of law of conservationof energy.(b)

r1

r2

I2E2

E1

II BA

I1

Terminal potential difference across the first cell isV = E1– l1r1

Þ1

11

–E VI

r=

For the second cell, terminal potential difference will beequal to that across the first cell. So, V = E2 – l2 r2

Þ2

22

–E VIr

=uv

Let E be effective emf and r is effective internal resistance.Let I be the current flowing through the cell.I = I1 + I2

Þ 1 2

1 2

– –E V E VI

r r= +

Þ 2 1 1 2

1 2

( – ) ( – )r E V r E VI

r r+

=

Þ Ir1r2 = E1r2 + E2r1– (r1 + r2)V

Þ 1 2 2 1 1 2

1 2 1 2–

E r E r Ir rV

r r r r+

=+ +

Comparing the equation withV = E – Ir, we get

Emf, 1 2 2 1

1 2

E r E rE

r r+

=+

Internal resistance, 1 2

1 2

r rr

r r=

+

OR(a) The outward electric flux due to charge +Q is

independent of the shape and size of the surface,which encloses it because :

(i) Number of electric field lines coming out from aclosed surface enclosing the charge depends on thecharge enclosed by the surface,

(ii) Number of electric field lines coming out from aclosed surface enclosing the charge is independentof the position of the charge inside the closedsurface.

(b) Magnitude of electric field at any point on the axisof a uniformly charged loop is given by,

30 2 2 2

2( )

rRE

r R

l=

Î+

(i)

10

O O’

R R’

E’

Loop 1 Loop 2

-l +l3R

Electric field at the centre of loop 1 due to charge presenton it is zero. [From (i), when r = 0]Hence, electric field at the centre of the loop 1 due tocharge present on the loop 2 is,

2 2 3/20

( 3)2 [( 3) ]

R RER R

+l=

Î +

2

30

32 8

RR

l=

e

Þ0

316

ERl

=Î

The direction of this net field is from loop 2 to loop 1 asshown.

25. (a) Let XY be the surface separating the denser mediumand the rarer medium. Let:v1 = Speed of lightwave in the denser mediumv2 = Speed of light wave in the rarer medium

X

Incidentwave frontA¢

Densermedium,

Refractedwavefront

rr CA

Y

B

1v i

i

D

Rarer medium

v1

v2

Let us consider a plane wavefront AB. Let this wavefrontincident on the interface at an angle of incidence i.Let t be the time taken by the wavefront to travel thedistance BC in denser medium.Þ BC = v1tTo determine the shape of refracted wavefront, we willdraw a sphere of radius v2t from point A in the rarermedium. Let CD represent a tangent plane drawn frompoint C onto the sphere.Now, AD = v2tHere, CD would represent the refracted wavefront.Considering the triangles ABC and ADC, we get

1sin v tBCiAC AC

= =

2sinv tADr

AC AC= =

Þ2

1

sinsin

vri v

= ....(1)

Since r > i, the speed of light in the rarer medium (v2) willbe greater than the speed of light in the denser medium(v1).

Now, 11

cv

m = ; 22

cv

m =

wherem1= Refractive index of denser mediumm2= Refractive index of rarer mediumFurther, (1) can be written as m1 sin i = m2 sin rThis is the Snell’s law of refraction.(b) (i) When unpolarised light is passed through a

polaroid, only those vibrations of light passthrough the crystal that are parallel to the axisof the crystal. All other vibrations will beabsorbed by the crystal. In this way, theunpolarised light gets linearly polarised.

Polaroid

Plane of polarisation

Plane of vibration

Unpolarisedlight Plane polarised light

(ii) Let us consider that an unpolarised light is incidentalong XO at an angle ip (angle of polarisation) on theinterface AB, separating air, a rarer medium, from adenser medium of refractive index m.

A

X

N

Z

Y

Densermedium

O

Rarermedium

pr i=

pi

'r

N¢

B

11

It has been experimentally observed that when unpolarisedlight is incident at polarising angle, the reflectedcomponents along OZ and OY are mutually perpendicularto each other. So, we haveÐZOB + ÐBOY = 90°Þ (90° –ip) + (90° – r’) = 90°Þ 90° –ip = r’

By Snell’s law, sinsin

pir

m =¢

\sin

sin(90 – )p

p

ii

m =° = tan ip

This gives us the required expression for Brewster’s law.OR

From the lens maker formula, we have

1 2

1 1 1( –1) –f R R

æ ö= m ç ÷

è øLet fl anf f2 be the focal lengths of the two mediums. Then,

11

1 1 1( –1) – –f R R

é ùæ ö= m ç ÷ê úè øë û

Þ 11

1 2( –1)f R

æ ö= m ç ÷è ø

22

1 1 1( –1) – –f R

é ùæ ö= m ç ÷ê ú¥è øë û

Þ 22

1 1( –1) –f R

æ ö= m ç ÷è ø

(a) If feq is the equivalent focal length of the combination,then

1 2

1 1 1

eqf f f= +

Þ 1 22( –1) ( –1)1 –eqf R R

m m=

Þ1 22 – –11

eqf Rm m

=

Þ1 22 – –1eq

Rf =m m

(b) For the combination to behave as a diverging lens,feq < 0

Þ1 22 – –1

Rm m < 0

Þ 2m1 – m2 – 1 < 0

Þ 21

( 1)2

m +m <

which is the required condition(c) For m1 > (m2 + 1)/2. the combination will behave as

the converging lens. So, an object placed far awayfrom the lens will form image at the focus of the lens.

The image so formed will be real and diminished in nature.26. Consider a straight conductor XY lying in the plane of

paper. Consider a point P at a perpendicular distance ‘a’from straight conductor.

Y

XI

ldl

C f2f1q

f P

O

Now Magnetic field induction at a point P due to currentI passing through conductor XY is given by

[ ]01 2sin sin

4I

Ba

m= f + f

p

At the centre of the infinite long wire,f1 = f2 = 90°

\ [ ]0 sin 90 sin 904

IB

am

= ° + °p

Þ 0 24m

=p

IBa ....(1)

1 2

I1 I2

d

P2F

1B

12

Consider two infinite straight conductors 1 and 2. Let l1and l2 current flowing through the conductor 1 and 2 andthey are d distance apart from each other.The magnetic field induction (B) at a point P on conductor2 due to current l1 passing through conductor 1 is given

by 0 11

24

IB

dm

=p

According to right hand rule, the direction of thismagnetic field is perpendicular to the plane of the paperinward.Now force experienced (F2) by unit length of conductor 2will be F2 = B1I2 × 1 = B1 I2

\ 0 1 22

24

I IFd

m=

pConductor 1 also experiences the same amount of force,directed towards the conductor 2. Hence, conductor 1 andconductor 2 attract each other. Thus, two linear parallelconductors carrying currents in the same direction attractand repel each other when the current flows in theopposite direction.Let I1 = I1 = 1A; r = 1 mThen,

Fl = F2= F = 10–7 2 1 1

1´ ´

Þ F = 2 × 10–7 N/mDefinition of one ampere : One ampere is that value ofconstant current which when flowing through each of thetwo parallel uniform long linear conductors placed in freespace at a distance of 1 m from each other will attract orrepel each other with a force of 2 × 10–7 N per metre oftheir length.

ORTransformer

zInputAC

P1S1

P2S2

OutputR

Laminated core

Principle: It works on the principle of mutual induction.Whenever magnetic flux linked with coil changes, an emfis induced in the neighbouring coil.Working: When an alternating emf is supplied to theprimary coil then magnetic flux linked with the secondarycoil changes. As a result, an emf is induced in thesecondary coil. Voltage at the primary and the secondarycoils depends on the number of turns. Magnetic flux thatis linked with the primary coil is also linked with thesecondary coil. The induced emf in each turn (Eturn) of thesecondary coil is equal to that of the primary coil.Let Ep be the alternating emf applied to primaiy coil and

np be the number of turns in it. If f be the electric fluxassociated with it, then

p pdE ndtf

= - ....(1)

Let Es be the emf across the secondary coil and ns be theno of turns in it. Then,

s sdE ndtf

= - ....(2)

Dividing (2) by (1),

s s

p p

E nk

E n= =

For step up transformer, k > 1\ ES > EPFor step down transformer, k < 1\ ES < EPFor ideal transformer, we assume that1. The resistance of the primary and secondary windings

are negligible.2. The energy loss due to magnetic hysterisis in the

iron core is negligible.Following are the two cause of energy loss:(1) Iron loss: Loss in the bulk of iron core due to the

induced eddy currents. It is minimised by using thinlaminated core.

(2) Hysteresis loss: Alternately magnetising anddemagnetising, the iron core causes loss of energy.It is minimised using a special alloy of iron core withsilicon.

1

(A) Electrostatics

1. Define electric flux.[All India 2008, 2009, Delhi 2008C, 2012, 2012C,

2013C]2. Define electric dipole moment of an electric dipole. Write its SI unit.

[All India 2008C, 2011, Foreign 2009, 2012]3. Define the dielectric constant of a medium. [Delhi 2011C]4. Define quantization of electric charge.5. Define following terms :

(i) Electric field(ii) Electric field intensity

6. Define electric potential energy. [Delhi 2008C]7. Define electrostatic potential.8. Define capacitance of a capacitor. Give its S.I. unit.(B) Current Electricity

9. Define resistivity of a conductor. [Delhi 2008, All India 2011]10. Define Ohm's law.11. Define specific resistance.12. Define mobility of a charge carrier. [Foreign 2008,

All India 2013C, Delhi 2014]13. Define relaxation time of the free electrons drifting in a conductor.

[All India 2012]14. What is the emf of a cell ?15. Define the term 'drift velocity' of charge carriers in a conductor.

[All India 2013C, Delhi 2014]16. Define the term 'electrical conductivity' of a metallic wire.

[Delhi 2014](C) Magnetism

17. Using the concept of force between two infinitely long parallel currentcarrying conductors, define one ampere of current.

WHAT, WHY, WHEREWhat is the defination of......?

(CLASS XII PHYSICS)

2

Buy books : http://www.dishapublication.com/school-foundation-books/books-for-class-11-12.html

[Delhi 2009, All India 2014]18. What is ‘Bohr magnetion’? [All India 2010]19. Define the current sensitivity of a galvanometer.

[All India 2013, 2013C]20. What are permanent magnets? [Delhi 2013]21. Define force on a moving charge.22. Define magnetic field strength.23. Define magnetic susceptibility of a material. [Delhi 2008]24. Define magnetic dipole moment. [Delhi 2008C]25. Define potential energy in magnetic field.26. What is hysteresis ?27. Define magnetic permeability :(D) Alternating current and EM waves

28. Define mutual induction between two long coaxial solenoids.[All India 2008, 2014, Foreign 2008, 2009, Delhi 2009, 2012]

29. Define self inductance of a coil. [Foreign 2009, Delhi 2009,All India 2010, 2013C, 2014]

30. Define magnetic flux (f)31. Define eddy currents.32. What is working principle of a.c. generator.33. Define the quality factor in an a.c. circuit. [Delhi 2009]34. Define power factor.35. What do you mean by root mean square value of an A.C.36. What do you mean by electromagnetic spectrum ?(E) Optics

37. Define power of a lens. [Foreign 2010, 2012]38. Define magnifying power of a telescope.

[Delhi 2010C, 2012, All India 2013]39. Define following terms :

(i) Reflection (ii) Refraction40. Define the phenomenon of total internal reflection.41. Define critical angle:42. How is a wavefront defined ? [Delhi 2008]43. What is plane polarised light? [Delhi 2008]44. What are coherent sources? [All India 2010C]45. Define the diffraction in wave optics.

CBSE Practise Set With Solution ForPhysics Class 12

Publisher : Faculty Notes Author : Panel Of Experts

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