2 phy cbse_2012-13_12th_05-03-12

155/1/1 P.T.O.

Series : SKS/1

Roll No.

Code No. 55/1/1Candidates must write the Code onthe title page of the answer-book.

Please check that this question paper contains 15 printed pages.

Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate.

Please check that this question paper contains 29 questions.

Please write down the Serial Number of the questions before attempting it.

15 minutes time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m.From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not write any answer on theanswer script during this period.

PHYSICS (Theory)[Time allowed : 3 hours] [Maximum marks : 70]

General Instructuions:

(i) All questions are compulsory.

(ii) There are 29 questions in total. Questions numbered 1 to 8 are very short-answer questions and carry 1 markeach.

(iii) Questions numbered 9 to 16 carry two marks each, Question Nos. 17 to 25 carries three marks each and QuestionNos. 27 to 29 carry five marks each.

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

(v) Question No. 26 is value based question carries four marks.

(vi) Use of calculators is not permitted. However, you may use log tables if necessary.

(vi) You may use the following values of physical constants wherever necessary:

c = 3 × 108 m/s

h = 6.63 × 10–34 Js

e = 1.6 × 10–19 C

0 = 4 × 10–7 T mA–1

0

14 = 9 × 109 Nm2C–2

me = 9.1 × 10–31 kg

Mass of the Neutron = 1.675 × 10–27 kg

Mass of the Proton = 1.673 × 10–27 kg

Studymate Solutions to CBSE Board Examination 2012-2013

255/1/1 P.T.O.

1. What are permanent magnets? Give one example.

Ans. Magnets which possess higher retentivity and coercivity and retain the magnetic field forlonger period are called permanent magnets. Example. Alnico, Nipermag, Steel.

2. What is the geometrical shape of equipotential surfaces due to a single isolated charge?

Ans. Concentric spheres with the gap between them not being uniform as 1

Vr

3. Which of the following waves can be polarized

(a) Heat waves (b) Sound waves?

Ans. Heat waves can be polarised as they are transverse by nature.

4. A capacitor has been charged by a dc source. What are the magnitudes of conduction anddisplacement currents, when it is fully charged?

Ans. On full charging the source will maintain the potential across the plates. Displacement currentand conduction current will be zero.

5. Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimumdeviation for a triangular prism.

Ans. Angle of incidence mA + δi =

2

6. The given graph shows the variation of photo-electric current (I) versus applied voltage (V) fortwo different photosensitive materials and for two different intensities of the incident radiation.Identify the pairs of curves that correspond to different materials but same intensity of incidentradiation.

1

2

3

4

I

VAns. The pairs (2,4) and (1,3) are having the same intensity and different metal.

7. A 10 V battery of negligible internal resistance is connected across a 200 V battery and aresistance of 38 as shown in the figure. Find the value of the current in circuit.

10V

38 200VAns. Applying Kirchoff’s rule, we get

200 – I(38) – 10 = 0

I = 190 95

5A38 19

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8. The emf of a cell is always greater than its terminal voltage. Why? Give reasons.

Ans. Emf is the p.d. when no current is drawn. When current is drawn, there will be potential dropacross the internal resistance of the cell. So, terminal voltage will be less than the emf.

9. (a) Write the necessary conditions for the phenomenon of total internal refletion of occur.

(b) Write the relation between the refractive index and critical angle for a given pair ofoptical media.

Ans. (a) (i) The light should traverse from denser to rarer media.

(ii) The angle of incidence should be more than the critical angle.

(b) d

r c

1

sin i

where d and

r are the refractive index of the denser and the rarer media.

10. State Lenz’s Law.

A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Willthere be an emf induced at its ends? Justify your answer.

Ans. According to Lenz’s Law, “the emf produced due to the change in magnetic flux will oppose thecause of the emf.”

Yes, there will be an emf as the horizontal component of field of earth, velocity of the motionof the rod and the length of the rod are all perpendicular to each other.

11. A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focallength 20 cm. Determine the power of the combination. Will the system be converging ordiverging in nature?

Ans. f1 = 25 cm f

2 = –20 cm

Peq

= P1 + P

2 =

1 2

1 1100

f f

Peq

= 1 1 20 25

100 10025 20 500

= –1 Dioptre

The system will be a diverging lens as if has negative power.

12. An ammeter of resistance 0.80 can measure current upto 1.0 A.

(a) What must be the value of shunt resistance to enable the ammeter to measure currentupto 5.0 A?

(b) What is the combined resistance of the ammeter and the shunt?

Ans. (a) Rg = 0.80 , Ig = 1.0 A, I = 5A

We know that, Shunt resistance (S) = g g

g

I R 1 0.8

I I 4

= 0.2

(b) The combined resistance will be R = g

g

R S 0.80 0.2

R S 1.00

= 0.160

13. In the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would (a) thebrightness of the lamp and (b) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’is decreased? Justify your answer.

455/1/1 P.T.O.

R

V L+

9V–

Ans. As R is decreased, Ib increases. So, I

c increases as =

c

b

I

I. With increase in I

c, power dissipated

by bulb is more. So, voltmeter reading and brighness of the bulb, both increase.

14. (a) An em wave is travelling in a medium with a velocity ˆv vi . Draw a sketch showing

the propagation of the em wave, indicating the direction of the oscillating electric andmagnetic fields.

(b) How are the magnitudes of the electric and magnetic fields related to the velocity ofthe em wave?

Ans. (a) Propagation direction is X.

Ey

Bz

X

E

B

Electric field Along y

Magnetic field Along z

(b) If Em and B

m are the magnitude of the electric and magnetic fields, then, the velocity of

e.m. wave is given by C = m

m

E

B

15. Block diagram of a receiver is shown in the figure:

Received Signal

Receiving Antenna

Amplifier X Detector YOutput

(a) Identify ‘X’ and ‘Y’.

(b) Write their functions.

Ans. (a) X – Intermediate frequency amplifier

Y – Audio frquency amplifier

(b) The intermediate frquency amplifier brings down the frequency to a lower value. Audioamplifier amplifies the final demodulated output.

16. Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how itis used to detect the optical signals.

555/1/1 P.T.O.

OR

Mention the important considerations required while fabricating a p-n junction diode to beused as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it isrequired to emit light in the visible range?

Ans. It is a p-n junction fabricated with a transparent window to allow light photons to fall on it.These photons generate electron hole pairs upon absorption. If the juction is reverse biasedusing an electrical circuit, these electron hole pair move in opposite directions so as to producecurrent in the circuit. This current is very small and is detected by the microammeter placedin the circuit. The magnitude of the photocurrent depends on the intensity of incident light(photocurrent is proportional to incident light intensity). It is easier to observe the change inthe current with change in the light intensity. This property is used in directing opticalsignal.

np A

R. B.

I

I ( A)

(m A)

V

I >I >I >I4 3 2 1

I1I2I3I4

OR

Ans. The important considerations required while fabricating a p-n junction diode to be used as aLight Emitting Diode (LED) are

(i) the light emitting efficiency is maximum.

(ii) care should be taken that high reverse voltages do not appear across them.

The semiconductor used for fabrication of visible LEDs must at least have a band gap of1.8 eV (spectral range of visible light is from about 0.4 m to 0.7 m, i.e., from about 3 eV to1.8 eV).

17. Write three important factors which justify the need of modulating a message signal. Showdiagrammatically how an amplitude modulated wave is obtained when a modulating signalis superimposed on a carrier wave.

Ans. Need for Modulation

(a) Size of antenna: For transmitting a signal, the antenna must have a size h = /4. For

e.m wave of frequency 20 kHz, l = 15 km. Such a long antenna is not practical. So,there is a need of translating the low frequency signal into high radio frequencies.

(b) Power radiated: Is proportional to 2

l. l – antenna length to have good power radiation,

high frequency (low wavelength) transmission is needed.

(c) If many transmitters are transmitting message signals simultaneously, all thesesignals will get mixed up and it will be difficult to distinguish them. Possible solution isusing communication at high frequencies and allotting a band of frequencies to eachmessage signal for its transmission.

655/1/1 P.T.O.

Diagrammatic representation of Amplitude Modulation

In A.M., the amplitude of carrier wave is changed in accordance with information signal.

+ cA

0

– cA

Am

0

0

– mA

t

m(t)

c(t)

c (t)m

t

Carrier wave = sin tc cc(t) A

Message signal = sin tm mm(t) A

A

t

Amplitude modulated wave

18. A capacitor of unknown capacitance is connected across a battery of V volts. The chargestored in it is 360 C. When potential across the capacitor is reduced by 120 V, the chargestored in it becomes 120 C.

Calculate:

(a) The potential V and the unknown capacitance C.

(b) What will be the charge stored in the capacitor, if the voltage applied had increased by120 V?

OR

A hollow cylindrical box of length 1 m and area of cross-section 25 cm2 is placed in a threedimensional coordinate system as shown in the figure. The electric field in the region is

given by ˆ50E xi , where E is in NC–1 and x is in metres. Find

(a) Net flux through the cylinder.

(b) Charge enclosed by the cylinder.

Y

Z

XO1 m

Ans. (a) Let the capacitance be C

so charge Q1 = cV

or 360 C = cV ...(i)

V

Q1 CIn second case

Q2 = c(V – 120)

or 120 C = c(V – 120) ...(ii)

755/1/1 P.T.O.

From equation (i) and (ii),

360 C cV

120 C c V 120

(V – 120)

Q2 C3 = V

V 120or 3V – 360 = V

or 2V = 360

V = 180 volt

From equation (i),

360 C = C × 180

C = 360 C

180V

= 2 f

C = 2 f

(b) Energy stored V = 1

2cV2

U = 1

2 × 2 × 10–6 × (180 + 120)2

= 61

2 10 300 3002

U = 9 × 10–2 J

OR

Ans. (a) Flux acoss curved surface 2 = E.A 0

( = 90°)

Flux acoss first surface 1 = E.A

= EA cos 0°

1 = (50 × 2) × 25 × 10–4

= 100 × 25 × 10–4

1 = 25 × 10–2 Vm

Flux across third surface 3 = E.A

= EA cos 180° ( = 180°)

= (50 × 1) × 25 × 10–4 (–1)

= –1250 × 10–4

= –12.50 × 10–2 Vm

Y

Z

XO1 m

III

II

ISo, net flux =

1 +

2 +

3

= 0 + 25 × 10–2 – 12.50 × 10–2

= 12.50 × 10–2 Vm

(b) From Gauss Law = en

0

q

or 12.50 × 10–2 = en

0

q

qen

= 0 × 12.50 × 10–2

= 8.85 × 10–12 × 12.50 × 10–2

qen

= 110.625 × 10–14 C

855/1/1 P.T.O.

19. (a) In a typical nuclear reaction, e.g.2

1H + 2

1H 3

2He + n + 3.27 MeV

although number of nucleons is conserved, yet energy is released. How? Explain.

(b) Show that nuclear density in a given nucleus is independent of mass number A.

Ans. (a) In all types of nuclear reaction, the law of conservation of number of nucleons is followed.

But during reaction, the mass of final product is found to be slightly less than the sumof the masses of reactant components. This difference in masses is called mass defect.So, as per mass energy relation E = (m)c2, energy is released.

In the given reaction, the sum of the masses of two deutrons is more than the mass ofhelium and neutron. Energy equivalent of mass defect is released.

(b) Let m = mass of one nucleon

A = mass number

R = radius of nucleus

Density (d) = mass of nucleus

volume

= 1

33

30

m.A mA4 4R R A3 3

= 3m A

304 R A

d = 30

3m

4 R

We can see that density is independent of mass number.

20. (a) Why photoelectric effect can not be explained on the basis of wave nature of light? Givereasons.

(b) Write the basic features of photon picture of electromagnetic radiation on whichEinstein’s photoelectric equation is based.

Ans. (a) (i) According to wave nature, the energy carried by light is measured in terms ofintensity. So, when light of higher intensity falls on metal surfce, it will impartmore energy to electron. Hence, electrons will eject out with more kinetic energy.But experimentally, kinetic energy of photoelectrons is independent of intensityof light.

(ii) According to wave theory, threshold frequency should not exist. Light of allfrequencies should emit electron provided intensity of light is sufficient forelectrons to eject.

(iii) According to wave theory, photoelectric effect should not be instantaneous. Energyof wave cannot be transferred to a particular electron but will be distributed to allthe electrons present in the illuminated portion. Hence, there has to be a timelag between incident of radiation and emission of electrons.

(b) Basic features of photon picture of electromagnetic radiation:

(i) Radiation behaves as if it is made of particles like photons.

(ii) Each photon has energy E (= h) and momentum p h

.

955/1/1 P.T.O.

(iii) Intensity of radiation can be understood in terms of number of photons falling persecond on the surface. Photon energy depends only on frequency and isindependent of intensity.

(iv) Photoelectric effect can be understood as the result of the one to one collisionbetween an electron and a photon.

(v) When a photon of frequency () is incident on a metal surface, a part of its energyis used in overcoming the work function and other part is used in impartingkinetic energy so,

20 max

1h mV

2

21. A metallic rod of light ‘l’ is rotated with a freqneucy with one end hinged at the centre andthe other end at the circumference of a circular metallic ring of radius r, about an axispassing through the centre and perpendicular to the plane of the ring. A constant uniformmagnetic field B parallel to the axis is present every where. Using Lorentz force, explain howemf is induced between the centre and the metallic ring and hence obtain the expression forit.

Ans. As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentzforce and get distributed over the ring. Thus, the resulting separation of charges produces anemf across the ends of the rod. At a certain value of emf, there is no more flow of electronsand a steady state is reached. The magnitude of the emf generated across a length dr of therod as it moves at right angles to the magnetic field is given by

d = Bv dr

Hence,

r r 2 22

0 0

B r B 2 rd Bvdr B r dr Br

2 2

22. Output characteristics of an n-p-n transistor in CE configuration is shown in the figure.

6

5

4

3

2

1

02 4 6 8 10 12 14 16 18

3.5

50µA

40µA

30µA

20µA

10µA

I = 0B

Determine:

(a) dynamic output resistance

(b) dc current gain and

(c) ac current gain at an operating point VCE

= 10 V, when IB = 30 A.

1055/1/1 P.T.O.

Ans. (a) Dynamic output resistance

r0 =

B

CE

C I

V

I

r0 = 3

30 A

12 8

3.6 3.4 10

r0 = 20 k

(b) For IB = 30 A, the I

C = 3.5 A

dc

= 3

6

3.5 10

30 10

= 0.12 × 103

dc

= 120

(c) ac

= CE

C

B V

I

I

For IB = 30 A, I

C = 3.5 mA and for I

B = 40 A, I

C = 4.6 mA

ac

=

CE

3

6V 10V

4.6 3.5 10

40 30 10

= 31.1

1010

ac

= 110

23. Using Bohr’s postulates, obtain the expression for the total energy of the electron in thestationary states of the hydrogen atom. Hence draw the energy level diagram showing howthe line spectra corresponding to Balmer series occur due to transition between energylevels.

Ans. The basic postulates of Bohr’s theory are :

(a) The central core of every atom is called nucleus and it carries only positive charge andalmost the entire mass of the atom. Negatively charged electrons revolve around it witha centripetal force which is provided by the electrostatic force of attraction betweenelectron and nucleus.

F = 2mv

r

where m and v are mass and velocity of electron moving in orbit of radius r. .....(1)

(b) Electron revolve in discrete non-radiating orbits (stationary orbits) with angular

momentum equal to integral multiple of h

2 (h is Planck’s constant)

i.e., mvr = nh

2.....(2)

where mvr = angular momentum of e–

(c) Radiation of energy takes place when electron jumps from one orbit to another and theamount of energy radiated is equal to the difference in energies of the two permittedorbits i.e.,

hv = E2 – E1 .....(3)

where v = frequency of radiation emitted

We know, mvr = nh

2 (from eqn 2) v =

nh

2 mr......(4)

1155/1/1 P.T.O.

Also, electrostatic force between electron (e–) and nucleus of charge (+Ze) is

F = 2

0

1 |Ze||e|

4 r......(5)

From (1) and (5), 2 2

20

mv 1 Ze

r 4 r

......(6)

or r = 2

20

1 Ze

4 mvSubstitute from (4),

r = 2 2 2 2

2 20

1 Ze 4 m r

4 m n h

r = 2 2

02 2

(4 )n h

4 mZe

r = 2 2

02

n h

me .Z

For H, Z = 1

r = 2 2

02 2

(4 ) n h

4 me

......(7)

r n2

Since energy of moving electron is its kinetic energy, hence from (6),

2 2

20

mv 1 Ze

r 4 r

mv2 = 2

20

1 Ze

4 r

K. E. = 2

2

0

1 1 1 Zemv

2 2 4 r

......(8)

Potential energy = Potential × Charge

Since potential of electron is due to nucleus 0

1 Zei.e.,

4 r

P.E. = 0

1 Ze( e)

4 r

=

2

0

1 Ze

4 r

.....(9)

Total energy of electron = KE + PE

E = 2 2

0 0

1 Ze 1 Ze

2 4 r 4 r

E = 2

0

1 Ze

4 2r

.....(10)n = 4

n = 3

n = 2

n = 1

n = 5

Put (7) in (10) we get, E = 2 2 4

2 2 20

2 mZ e

(4 ) n h

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For H-atom, substitute standard values,

We get, E = 2

13.6

n

eV

When electron jumps from outer to second orbit then 2 2 2

1 1R

2 k

where k = 3, 4,

5....... and energy of electron is E2 = 2

13.6

2

= – 3.4 eV. This is Balmer Series.

24. (a) In what ways is diffraction from each slit related to the interference pattern in a doubleslit experiment?

(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study thediffraction taking place at a single slit of aperture 2 × 10–4 m. The distance between theslit and the screen is 1.5 m. Calculate the separation between the positions of the firstmaxima of the diffraction pattern obtained in the two cases.

Ans. (a) When there are only a few sources, say two interfering sources, then the result isusually called interference, but if there is a large number of them, it seems that theword diffraction is more often used.

In the double-slit experiment, we must note that the pattern on the screen is actuallya superposition of single-slit diffraction from each slit. This is shown in figure.

The actual double-slit interference pattern. The envelope shows the single slit diffraction.

(b) d = 2 × 10–4 m

D = 1.5 m

1 = 590 × 10–9 m

2 = 596 × 10–9 m

For first maxima,

d sin = 13

2

For small angle (sin tan )

So, d tan = 13

2

1 1y 3d.

D 2

or, y1 = 13 D

2d

For second wavelength

y2 =

23 D

2d

So, separation y = 2 1

3D

2d

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y = 94

3 1.5596 590 10

2 2 10

y = 6.75 × 10–5 m

25. In a series LCR circuit connected to an ac source of variable frequency and voltage = vm sin

t, draw a plot showing the variation of current (I) with angular frequency () for two differentvalues of resistance R

1 and R

2 (R

1 > R

2). Write the condition under which the phenomenon of

resonance occurs. For which value of the resistance out of the two curves, a sharper resonanceis produced? Define Q-factor of the circuit and give its significance.

Ans. (a) Condition for resonance to occur is

XL = X

C

(b) Sharper resonance is produced for R2

I

R2

R1

0

(c) Quality factor is the ratio of voltage drop across theinductor or capacitor to the voltage drop acrossresistor.

(d) Significance it shows the sharpness of the circuit.

Sharper the curve is more sensitive the circuit will be.

26. While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm.It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenlyhe noticed a child walking alone on the road. He asked the boy to come inside the car till thethunderstorm stopped. Dr. Pathak dropped the boy at his residence. The boy insisted that Dr.Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for hisconcern for safety of the child.

Answer the following questions based on the above information:

(a) Why is it safer to sit inside a car during a thunderstorm?

(b) Which two values are displayed by Dr. Pathak in his actions?

(c) Which values are reflected in parents’ response to Dr. Pathak?

(d) Give an example of a similar action on your part in the past from everyday life.

Ans. (a) The body of the car provides a electrostatic shield. This prevents the lightning strikeon the people in the car.

(b) Concern for others, values human life,.

(c) Obliged with the gesture of Dr. Pathak, highly thankful.

(d) I saw an old couple stranded in the street due to puncture of their vehicle tyre. I guided

them to a mechanic, got the tyre rectified and guided them home.

27. (a) Draw a ray diagram showing the image formation by a compound microscope. Henceobtain expression for total magnification when the image is formed at infinity.

(b) Distinguish between myopia and hypermetropia. Show diagrammatically how thesedefects can be corrected.

OR

(a) State Huygen’s principle. Using this principle draw a diagram to show how a planewave front incident at the interface of the two media gets refracted when it propagatesfrom a rarer to a denser medium. Hence verify Snell’s law of refraction.

1455/1/1 P.T.O.

(b) When monochromatic light travels from a rarer to a denser medium, explain thefollowing giving reasons:

(i) Is the frequency of reflected and refracted light same as the frequency of incidentlight?

(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?

Ans. (a) Compound Microscope: Compound microscope is used for observing highly magnifiedimages of tiny objects. It makes use of two converging lenses. The objective lens is ofvery small focal length and short aperture and the eye-piece is of small focal length andlarge aperture. The image formed by the objective lens becomes the object for theeyepiece. The eyepiece acting as a magnifying glass produces enlarged virtual imageof it

d

u f0

fe

EyepieceB’

h’hA

BB”

ObjectiveA’

E

A”

Magnifying power of a Compund microscope is given by (i), 0

0 e

v dm 1

|u | f

u0 = object distance for objective lens

v0 = image distance for objective lens

f0 = focal length of objective lens

fe = focal length of eyepiece.

Since the object lies very close to f0, the image formed by the objective lens is veryclose to the focal length (fe) of the eyepiece.

0 0 0u f , v L where L is the length of the microscope tube.

Magnifying Power, e0

L dm 1

ff |

.

When image is formed at infinity

00

0 0

v Lm

u f ; e

e

dm

f

Magnifying powerm m = m0 × me m = 0 e

L d

f f

The magnifying power of compound microscope is negative hence the image formed isinverted.

(b) The light from a distant object arriving at the eye-lens may get converged at a point infront of the retina. This type of defect is called nearsightedness or myopia. This meansthat the eye is producing too much convergence in the incident beam. To compensatethis, we interpose a concave lens between the eye and the object, with the divergingeffect desired to get the image focussed on the retina.

If the eye-lens focusses the incoming light at a point behind the retina, a convergent

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lens is needed to compensate for the defect in vision. This defect is called farsightedness

or hypermetropia.

Corrected Myopic eye. Corrected Hypermetropic eye.

OR

Ans. (a) HUYGENS PRINCIPLE

(i) Each point of the wavefront is a source of a secondary disturbance and secondarywavelets emanating from these points spread out in all directions with the speedof the wave.

(ii) The new position of the wavefront at a later time can be found using the product vtwhere v is the speed and t is the time.

(iii) Energy propagates in the direction perpendicular to the wavefront.

A

B

C

D

A '

B'

Secondarywavelents

New Position ofwavefront

Primarywavefront

A

B

C

D

C'

D'

REFRACTION OF PLANE WAVES USING HUYGENS PRINCIPLE

Refraction of a plane wave:– Let v1 and v2 be the speed of light in medium 1 and medium2. Suppose a plane wavefront AB is incident on PP' at an angle i.

Incident wavefront

A'

B

A

E

C

ii

rr

v2

v1

v1medium 1

medium 2Refractedwavefront

P'

v v2 1<

P

Let t be the time taken by the wavefront to travel distance BC in medium 1. Thus, BC =v1t.

1655/1/1 P.T.O.

In the same time the wavefront travels distance AE in medium 2. Thus, AE = v2t and CEwould represent the refracted wavefront.

In ABC , 1sinv tBC

iAC AC

; In AEC , 2sin

v tAEr

AC AC

where i and r are the angles of incidence of refraction.

1

2

sin

sin

vi

r v ...(i)

Now, 1 21 2

andc c

n nv v

2 1

1 2

n v

n v or n21 =

1

2

v

v ...(ii)

From (i) and (ii), 21sin

sin

in

r

This is the Snell's law of refraction.

(b) (i) Frequency is the property of source. Hence, frequency does not change when lightis reflected or refracted

(ii) No, decrease in speed does not imply reduction in energy carried by light wave.This is because the frequency does not change and according to the formula E =h, energy will be independent of speed.

28. (a) State the working principle of a potentiometer. With the help of the circuit diagramexplain how a potentiometer is used to compare the emf’s of two primary cells. Obtainthe required expression used for comparing the emfs.

(b) Write two possible causes for one sided deflection in a potentiometer experiment.

OR

(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules. obtain the balancecondition in terms of the resistances of four arms of Wheatstone bridge.

(b) In the meterbridge experimental set up, shown in the figure, the null point ‘D’ is obtainedat a distance of 40 cm from end A of the meterbridge wire. If a resistance of 10 isconnected in series with R

1, null point is obtained at AD = 60 cm. Calculate the values

of R1 and R

2.

R1 R2

AD

C

G

B

KAns. (a) Principle: The fall of potential across any portion of the potentiometer wire is directly

proportional to the length of that portion provided the wire is of uniform cross–sectionalarea and a constant current is flowing through it.

1755/1/1 P.T.O.

+–

Rh K

A J B

3

2

E1

E2

J

A

G

Working:

A – The area of cross-section of the wire.

r – Specific resistance of the material of the wire

V – Potential drop across the portion of the wire of length l.

R – Resistance

I – Current flowing through the wire.

IV IR k

A

V l where I and A are constants.

Vk

potential gradient or falls of potential per unit length of the wire.

To find the e.m.f of a cell the following circuit is used.

Let the null point is found at J and AJ = l.

The potential drop across the potentiometer wire AJ is equal to the unknowne.m.f E.

E = kl where k is the potential gradient of the potentiometer wire. Knowing k andmeasuring l from the attached scale E can be calculated.

+–

Rh K

A J B

E

A

G

l

+–

Rh K

A J B

32

E1

E2

J

A

G

1

Comparison of e.m.f of two cells

When the key is put between 1 and 3 terminals, cell of e.m.f E1 is connected in the

circuit and the corresponding balancing length is found to be l1.

E1 = kl

1 [where k = Potential gradient of the potentiometer wire]

When the key is put between 2 and 3 terminals, cell of e.m.f E2 is connected in the

circuit and the corresponding balancing length is found to be l2.

1855/1/1 P.T.O.

E2 = kl

2

1 1

2 2

E

E

Measuring l1 and l

2 from the attached scale the ratio of E

1 and E

2 can be calculated.

(b) Causes for one sided deflection

(i) Potential difference between the ends of the potentiometer wire is less than theemf of the cell in the secondary circuit.

(ii) The positive side of the driving cell is connected to the negative terminal of thecells in the secondary circuit.

OR

Ans.(a) Kirchhoff’s Rules:

Junction Rule ( I = 0). The sum of the charge/currents entering any junction mustequal the sum of the charge/currents leaving that junction.

This is a statement of Conservation of Charge.

Loop Rule ( V = 0). The sum of the potential differences across all the elements aroundany closed circuit loop must be zero.

This is a statement of Conservation of Energy.

Wheatstone Bridge Principle: Wheatstone bridge is a very accurate and precise methodof determining an unknown resistance.

G

B

P

D

A C

Q

R S

(I – I )1 g

(I – I1)

Ig

I – I + I1 g

I

I1

I

Applying loop rule to mesh ADBA , we get

–(I – I1)R + I

gG + I

1P = 0 …(1)

On applying loop rule to mesh DCBD, we get

–(I – I1 + I

g)S + (I

1 – I

g)Q – I

gG = 0 …(2)

The value of R is so adjusted that the points B and D are at the same potential so that thecurrent I

g through the galvanometer is zero. In this position the bridge is said to be balanced.

When Ig = 0, eqns, (1) and (2) reduce to the form

I1P = (I – I

1)R

and I1Q = (I – I

1)S

Dividing one by the other

(Condition for the balance of bridge)P R

Q S

1955/1/1 P.T.O.

(b)1

2

R 40

R 60 ...(i)

1

2

R 10 40

R 60

Dividing

1

1

R 10 40 40 4

R 10 60 60 9

9R1 = 4R

1 + 40

5R1 = 40

R1 = 8

Using (i) we get,

R2 =

60

40

R1 =

608

40 = 12

29. (a) Derive the expression for the torque on a rectangular current carrying loop suspendedin a uniform magnetic field.

(b) A proton and a deuteron having equal momenta enter in a region of uniform magneticfield at right angle to the direction of the field. Depict their trajectories in the field.

OR

(a) A small compass needle of magnetic moment ‘m’ is free to turn about an axisperpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia ofthe needle about the axis is ‘I’. The needle is slightly disturbed from its stable positionand then released. Prove that it executes simple harmonic motion. Hence deduce theexpression for its time period.

(b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical ata certain place on the earth. Find out the values of (i) horizontal component of earth’smagnetic field and (ii) angle of dip at the place.

Ans. (a) Torque experienced by a Current Loop (Rectangular) in a uniform Magnetic Field:

Let be the angle between the plane of the loop and thedirection of the magnetic field. The axis of rotation of thecoil is perpendicular to the magnetic field.

SPF I(b B)

| FSP

| = I b B sin

QRF I(b B)

| FQR

| = I b B sin

P

l

I

Q

R

B×

S

FSP

FRS

FPQ

FQR

b

I

Forces

SPF and

QRF are equal in magnitude but opposite

in direction and they cancel out each other. Moreoverthey act along the same line of action (axis) and hence donot produce torque.

QRF I( B)l

| FPQ

| = I l B sin 90° = I l B

2055/1/1 P.T.O.

RSF I( B)l

| FRs

| = I l B sin 90° = I l B

Forces

PQF and

RSF being equal in magnitude but opposite in direction cancel out

each other and do not produce any translational motion. But they act along differentlines of action and hence produce torque about the axis of the coil.

Torque experienced by the coil is

= FPQ

× PN (in magnitude)

= I l B (b cos )

= I lb B cos

= I A B cos (A = lb)

= N I A B cos (where N is the no. of turns)

PI

Q

R

B

× S

FRS

FPQ

b

I

B

Nn

n

If is the angle between the normal to the coil and

the direction of the magnetic field, then

+ = 90° i.e. = 90° –

So, = I A B cos (90° - )

= N I A B sin

(b) Since mv p

rqB qB

and the charge on both the proton and deutron are the same, the

particles will follow a circular path with their radius in the ratio 1: 1

OR

Ans. (a) This is done by placing a small compass needle of known magnetic moment m andmoment of inertia I and allowing it to oscillate in the magnetic field. This arrangementis shown in figure.

The torque on the needle is,

= m × B

In magnitude = mB sin

Here is restoring torque and is the angle between m and B.

Therefore, in equilibrium2

2

dI mBsindt

Negative sign with mB sin implies that restoring torque is in opposition to deflectingtorque.

This represents a simple harmonic motion. The square of the angular frequency is 2

= mB/I and the time period is,

IT 2

mB

(b) Since, the compass needle is oriented vertically,

(i) Horizontal component of earth’s field will be zero.

(ii) Angle of dip will be 90° as tan = V

H

B

B .

× · × · × · × · ×

2155/1/1 P.T.O.

1. A cell of emf ‘E’ and internal resistance ‘r’ draws a current ‘I’. Write the relation betweenterminal voltage ‘V’ in terms of E, I and r.

Ans. E = V+Ir where r is the internal reistance.

2. Which of the following substances are diamagnetic?

Ans. Bi and Cu

3. A heating element is marked 210 V, 630 W. What is the value of the current drawn by theelement when connected to a 210 V dc source?

Ans. Current drawn = I = P/V = 630/210 = 3 A.

10. An ammeter of resistance 1 W can measure current upto 1.0 A (i) What must be the value ofthe shunt resistance to enable the ammeter to measure upto 5.0A? (ii) What is the combinedresistance of the ammeter and the shunt?

Ans.g

g

I GS

I I

[Here, Ig represents the current for full scale deflection in the ammeter of resistance 1.]

[G resistance of ammeter that measures 1 Amp]

Ig = 1A

G = 1

I = 5A

(1) (1)S 0.25

(5 1)

Shunt resistance S = 0.25

Req

GS (1)0.25

G S 1 0.25

Req

= 0.25 1

0.21.25 5

13. In the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would (a) thebrightness of the lamp and (b) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’is increased? Justify your answer.

R

V L+

9V–

Code No. 55/1/2

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTION ONLY

Series : SKS/1

2255/1/2 P.T.O.

Ans. As R is increased, Ib decreases. So, I

c decreases as =

c

b

I

I. With decrease in I

c, power dissipated

by bulb is less. So, voltmeter reading and brightness of the bulb, both decrease.

14. A convex lens of focal length 20 cm is placed coaxially in contact with a concave lens of focallength 25 cm. Determine the power of the combination. Will the system be converging ofdiverging in nature?

Ans. f1 = + 20; f

2 = – 25 cm

P = P1 + P

2

1 1 25 20 5

20 25 500 500

1 11 1cm 100m 1D

100 100

The system will behave as a converging lens because its power is positive.

19. Using Bohr’s postulates, obtain the expression for (i) kinetic energy and (ii) potential energyof the electron in stationary state of hydrogen atom.

Draw the energy level diagram showing how the transitions between energy levels result inthe appearance of Lyman series.

Ans. The basic postulates of Bohr’s theory are :

(a) The central core of every atom is called nucleus and it carries only positive charge andalmost the entire mass of the atom. Negatively charged electrons revolve around it witha centripetal force which is provided by the electrostatic force of attraction betweenelectron and nucleus.

F = 2mv

r

where m and v are mass and velocity of electron moving in orbit of radius r. .....(1)

(b) Electron revolve in discrete non-radiating orbits (stationary orbits) with angular

momentum equal to integral multiple of h

2 (h is Planck’s constant)

i.e., mvr = nh

2.....(2)

where mvr = angular momentum of e–

(c) Radiation of energy takes place when electron jumps from one orbit to another and theamount of energy radiated is equal to the difference in energies of the two permittedorbits i.e.,

hv = E2 – E1 .....(3)

where v = frequency of radiation emitted

We know, mvr = nh

2 (from eqn 2) v =

nh

2 mr......(4)

Also, electrostatic force between electron (e–) and nucleus of charge (+Ze) is

F = 2

0

1 |Ze||e|

4 r......(5)

From (1) and (5), 2 2

20

mv 1 Ze

r 4 r

......(6)

2355/1/2 P.T.O.

or r = 2

20

1 Ze

4 mvSubstitute from (4),

r = 2 2 2 2

2 20

1 Ze 4 m r

4 m n h

r = 2 2

02 2

(4 )n h

4 mZe

r = 2 2

02

n h

me .Z

For H, Z = 1

r = 2 2

02 2

(4 ) n h

4 me

......(7)

r n2

Since energy of moving electron is its kinetic energy, hence from (6),

2 2

20

mv 1 Ze

r 4 r

mv2 = 2

20

1 Ze

4 r

K. E. = 2

2

0

1 1 1 Zemv

2 2 4 r

......(8)

Potential energy = Potential × Charge

Since potential of electron is due to nucleus 0

1 Zei.e.,

4 r

P.E. = 0

1 Ze( e)

4 r

=

2

0

1 Ze

4 r

.....(9)

Total energy of electron = KE + PEn = 4

n = 3

n = 2

n = 1

n = 5

For H-atom, substitute standard values,

We get, E = 2

13.6

n

eV

When electron jumps from outer to first orbit then 1 2 2

1 1R

1 k

where k = 2, 3, 4,

5....... and energy of electron is E1 = 2

13.6

1

= –13.6 eV. This is Lymann Series.

22. Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. Auniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ ismoved with a velocity v towards the arm RS. Assuming that the arms QR, RS and SP havenegligible resistances and the moving arm PQ has the resistance r, obtain the expression for(i) the current in the loop (ii) the force and (iii) the power required to move the arm PQ.

2455/1/2 P.T.O.

× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×× × × × × × × × ×

S

R

P

Q

v

Ans. As the arm PQ = l is moved to the left as shown with a velocity v, the flux linked will decreased

as B = Blx = Blvt. Therefore, the emf induced will be e =

Bd

Blvdt

(a) The current flowing in the loop will be I = e

r

(b) The force experienced will be F = I(l × B) = e

Blr

(c) The power required to move the arm will be I2r = 2 2 2 2 2

2

e e B l vr

r rr

23. Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in communicationsystem.

(a) Why is sky wave mode propagation restricted to frequencies upto 40 MHz?

(b) Give two examples where space wave mode of propagation is used.

Ans. Sky Wave Propagation

• Waves after being emitted are reflected by the atmosphere to reach the receiver.

• Ionosphere reflects waves within the region (1710 kHz to 40 Mhz).

• The phenomenon of bending of em waves so that they are diverted towards the earth issimilar to total internal reflection in optics.

Space wave propagation (LOS communication)

• Space waves (54Mhz – 4.2 Ghz) travel in a straight line.

• Hence, the transmitter and receiver must be on the line of sight (LOS) together.

(a) 40 Mhz is critical frequency. Beyond this frequency, rays penetrate and escape theionosphere.

(b) Space waves are used for television broadcast and satellite communication.

× · × · × · × · ×

2555/1/1 P.T.O.

6. A 5V battery of negligible internal resistance is connected across a 200 V battery and aresistance of 39 as shown in figure. Find the value of the current.

5V

39 200V

Ans. Let, current in circuit = I

Effective E.M.F of circuit E = 200 – 5

= 195V

Total resistance of circuit,

R = 0 + 39 = 39

I = = 195

539

EA

R

7. Which of the following substances are para-magnetic?Bi, Al, Cu, Ca, Pb, Ni

Ans. Al, Ca

8. A heating element is marked 210 V, 630 W. Find the resistance of the element when connectedto a 210 V dc source.

Ans. V = 210 V, P = 630 W

R = ?

P = 2V

R

630 = 2(210)

R

R = 210 210

70630

R = 70

9. An ammeter of resistance 06W can measure current upto 1.0 A. Calculate (i) The shuntresistance required to enable the ammeter to measure current upto 5.0 A (ii) The combinedresistance of the ammeter and the shunt.

Ans. IA = 1.0A, I = 5.0 A,

RA = 0.6

(i) rs =

AA

A

IR

I I

= 1

0.65 1

Code No. 55/1/3

Studymate Solutions to CBSE Board Examination 2012-2013

UNCOMMON QUESTION ONLY

Series : SKS/1

2655/1/3 P.T.O.

= 1

0.64

rs = 0.15

(ii) Reff

= 0.6 0.15

0.6 0.15A s

A s

R r

R r

Reff

= 0.6 0.15

0.75

= 0.12

15. A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens of focallength 40 cm. Determine the power of combination. Will the system be converging or divergingin nature?

Ans. f1 = + 30 cm, f

2 = –40 cm

P = P1 + P

2

= 1 2

1 1

f f

= 1 1 10 10

0.30 ( 0.40) 3 4

P = 4 3 10

10 0.83D12 12

System will be converging in nature.

18. A capacitor of unknown capacitance is connected across a battery of V volts. The chargestored in it is 300 µC. When potential across the capacitor is reduced by 100 V, the chargestored in it becomes 100 µC. Calculate the potential V and the unknown capacitance. Whatwill be the charge stored in the capacitor if the voltage applied had increased by 100V?

OR

A hollow cylindrical box of length 0.5 m and area of cross-section 20 cm2 is placed in a threedimensional coordinate system as shown in the figure. The electric field in the region is

given by ˆ20 ,E x i

where E is NC–1 and x is in metres. Find.

(i) Net flux through the cylinder.

(ii) Charge enclosed in the cylinder.

Y

Z

XO0.5 m

Ans. C = ?, battery voltage, V1 = V volts

Q1 = 300 µC, Q

2 = 100 µC, V

2 = (V – 100)

Q1 = CV

1

300 × 10–6 = C × V ...(i)

Q2 = CV

2

100 × 10–6 = C × (V – 100) ...(ii)

2755/1/1 P.T.O.

Eqn. (ii) ÷ Eqn. (i)

1 100

3

V

V

V = 3V – 300

2V = 300

V = 150 volts

From (i)

300 × 10–6 = C × 150

C = 2 µF

V3 = V + 100 = 250 volts

Q3 = CV

3 = 2 × 250 = 500 µC

V = 150 volts, C = 2µF and Q3 = 500 µC

OR

L = 0.5 m (box length)

A = 20 cm2 = 20 × 10–4 m2

ˆ20E x i

(i) Net flux through the cylinder,

Y

Z

XO0.5 m

I

II

III

= 1 +

2 +

3

= E1. A

1 cos180° + E

2A

2 cos90° + E

3A

3 cos0°

= (20 × 0.5) × (20 × 10–4) (–1) + 0 + (20 × 1.0) × (20 × 10–4) × (1)

= –2 × 10–2 + 4 × 10–2

= 2 × 10–2

= 0.02 V-m

(ii) By Gauss theorem, 0

1enQ

0.02 = 12

1

8.85 10 enQ

Qen

= 1.772 × 10–13 C

21. (a) In a nuclear reaction

3 3 4 1 12 2 2 1 1He He He H H 12.86MeV, though the number of nucleons is conserved on

both sides of the reaction, yet the energy is released. How? Explain.

(b) Draw a plot of potential energy between a pair of nucleons as a function of theirseparation. Mark the regions where potential energy is (i) positive and (ii) negative.

Ans. (a) 3 3 4 1 12 2 2 1 1He He He H H 12.86 MeV

though the number of nucleons is conserved on both sides of the reaction, but totalmass of products is less then total mass of reactants. Energy is released due to thismass defect.

E = m × C2

2855/1/3 P.T.O.

× · × · × · × · ×

(b)

Y

r0

+P.E

–P.E

SeparationX

25. (a) Using Bohr’s postulate, obtain the expression for total energy of the electron in the nth

orbit of hydrogen atom.

(b) What is the significance of negative sign in the expression for the energy?

(c) Draw the energy level diagram showing how the line spectra corresponding to Paschenseries occur due to transition between energy levels.

Ans. (a) See solution question no. 23 set-1

(b) Negative sign of total energy signifies that if equal amount of energy is provided torevolving electron it will move out of atom. Hence it denotes binding energy of electronin the orbit.

(c)n = 6n = 5n = 4n = 3n = 2n = 1