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XII PHYSICS RBSE – FLASHBACK 2013 Time: 3.15 Hrs. Max. Marks: 56

General Instructions to the Examinees:

1. Candidates must write first his/ her Roll No. on the question paper compulsorily.

2. All the questions are compulsory.

3. Write the answer to each question in the given answer-book only.

4. For questions having more than one part the answers to those parts are to be written together in

continuity.

5. If there is any error/ difference / contradiction in Hindi & English Versions of the question paper, the

question of Hindi version should be treated.

6. Q. Nos. Marks per question

1 – 13 1

14 – 24 2

25 – 27 3

28 – 30 4

7. There are internal choices in Q. Nos. 21 and 27 to 30.

8. Use of calculator is not allowed in the examination.

Q.1 Write any two properties of electric field lines.

A.1 (i) The electric lines of force appear to start from positive charge and to terminate at negative

charge.

(ii) The tangent drawn at any point on the line of force gives the direction of electric filed

strength at that point and the direction of force acting on a positive charge at that point.

Q.2 Define equipotential surface.

A.2 An equipotential Surface is the surface which have the same potential at each point.

Q.3 Write the corresponding values of X and Y for which the lengths of conductors X = 4 Ω and

Y = 48 × 10-8

Ω-m are reduced to half.

A.3 Since resistance is proportional to length therefore, X becomes half i.e. X = 2 Ω. However, resistivity

is independent of length, therefore Y remains same i.e. 8Y 48 10 m

Q.4 In given diagrams write the direction of magnetic field produced at point P in form of and .

A.4

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Q.5 Write Faraday‟s law of electromagnetic induction.

A.5 According to Faraday‟s law of electromagnetic induction, the magnitude of induced emf is equal to

the rate of change of magnetic flux with respect to time.

Q.6 In an alternating circuit applied voltage is 220V. If R = 8Ω, XL = XC = 6Ω then write the values of

the following:

(a) Root mean square value of voltage

(b) Impedance of circuit.

A.6 (a) Vrms = 220 V

(b) Impedance Z = 22

L CR X X

228 6 6

Z 8

Q.7 (a) Write Gauss‟ law for magnetism in the form of Maxwell equation.

(b) Write the value of0 0

1

.

A.7 (a) Gauss Law for magnetism

Total magnetic flux through a closed surface is equal to zero.

B.ds 0

(b) 8c 3 10 m / s

Q.8 Which type of lens is used to correct astigmatism?

A.8 Cylindrical lens is used to correct astigmatism.

Q.9 Define the phenomenon of total internal reflection.

A.9 when a ray of light is incident on the interface from denser medium to rarer medium, it is deviated

away from the normal. When angle of incidence is increased, angle of refraction increases and at a

stage it becomes 90o.

The angle of incidence in denser medium for which the angle of refraction in rarer medium is 90ois

called the critical angle (c).

When angle of incidence of the ray, incident on rarer medium from denser medium is greater than

the critical angle, the incident ray does not refract into a rarer medium but is reflected back into the

denser medium. This phenomenon is called total internal reflection. The condition for total

internal reflection are

(i) The ray must travel from a denser into a rarer medium.

(ii) The angle of incidence i > critical angle C.

Q.10 An element A decays into element C by the following two step processes:

4

2

0

1

A He B

B 2( e ) C

Select pair for isobars from elements A, B and C.

A.10 B and C are isobars.

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Q.11

Write the name of logic gates related to figure P and table Q.

A.11 PAND gate, QNOR gate

Q.12 Heights of transmitting and receiving antennas from earth surface are hT and hR respectively. Write

the relation for maximum line-of-sight (LOS) distance between two antennas.

A.12 Maximum line –of-sight (LOS) distance between two antennas, M T Rd 2Rh 2Rh

Q.13 What is modulation? In the given block diagram of a receiver for detection of amplitude modulated

wave, write the name of part X.

A.13 The original low frequency information/ message signal cannot be transmitted over long distances.

Therefore, it is essential to superimpose the information signal in high frequency radiowave called

the carrier wave. The superposition signal on high frequency radiowave (carrier wave) is called the

modulation.

Q.14 (a) In the given circuit diagram, find the value of unknown resistance S, in the balancing condition of

meter bridge.

(b) In the given circuit write the value of resultant resistance in between X and Y when key K is

(i) Opened

(ii) Closed

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A.14 (a) For meter bridge

R

S 100

Here, = 40 cm

R = 4 Ω

S = R 100

4 100 40

40

4 60

40

S 6

(b) (i) When key is opened

Req =

6 4 3 126

6 4 3 12

(ii) When key is closed

Req =6 3 4 12

56 3 4 12

Q.15 Write Kirchhoff‟s first rule (law of junction). Drawing a circuit diagram of Wheatstone bridge,

derive condition for zero deflection in the bridge.

A.15 Kirchoff‟s Current Law,

“The sum of incoming currents is equal to sum of outgoing currents at a junction.”

Let the current given by battery in the balanced position be I. This current on reaching point A is

divided into two parts I1 and I2. As there is no current in galvanometer in balanced state, current in

resistances P and Q is I1 and in resistances R and S it is I2.

Applying Kirchoff‟s I low point A

I-I1-I2=0 or I=I1+I2 …. (1)

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Applying Kirchoff‟s II low to closed mesh ABDA

-I1P+I2R=0 or I1P=I2R …. (2)

Applying Kirchoff‟s II low to closed mesh BCDB

-I1Q+I2S=0 or I1Q=I2S …. (3)

Divided equation (2) by (3), we get

1 2

1 2

or I P I R P R

I Q I S R S …. (4)

Q.16 (a) In free space at a point magnitude of electric field vector E

is 9.3 V/m. Find the magnitude of

magnetic field vector B

at the point.

(b) Out of ultraviolet, infrared and X-rays whose wavelength is maximum?

A.16 (a) 00

EB

C

8

0 8

9.3B 3.1 10 T

3 10

(b) Infrared

Q.17 Match the following:

Column I Column II

μ = tan ip

c

1

sin i

sin i

sin r

mA Dsin

2

Asin

2

Snell‟s law

Brewster‟s law

Prism

Total internal reflection

A.17 μ = tan ip Brewster‟s Law

c

1

sin i Total Internal Reflection

sin i

sin r Snell‟s Law

mA Dsin

2Prism

Asin

2

Q.18 Define the following in photoelectric effect phenomenon:

a) i) Work function

ii) Stopping potential (cut-off potential)

b) Calculate energy of photon of wavelength 3.31o

A .

A.18 a) i) Work function – The valance electrons in the atoms of a metal revolving freely

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with different kinetic energies near the surface experience a force directed inwards. These

electrons require additional energy to become free and come out of the surface.

At 0 K, the minimum energy required to make an electron of maximum kinetic energy free

from the metal is known as work function of the metal.

ii) Stopping potential (cut-off potential) – When the collecting plate is given negative

potential, the photoelectric current is found to decrease rapidly till for a sharply defined

negative potential V0 of the plate, the current becomes zero. The minimum retarding (or

negative) potential V0given to the collecting plate at which the photoelectric current just

becomes zero is called the stopping potential or the cut – off potential for that particular

frequency of incident light.

b) E = hν

E = 8

10

3 106.62

3.31 10

E = 6 × 10-16

J

Q.19 Calculate de Broglie wavelength associated with an electron, accelerated through a potential

difference of 400 V.

A.19 V = 400 V

λ = 1.227

nmV

λ = 1.227

nm400

λ = 0.061 nm

Q.20 Explain the working to obtain V-I characteristics curve in forward biasing of P-N junction diode.

Draw circuit diagram of experimental arrangement.

A.20 A graph showing the variation of current flowing through a p-n junction with the voltage applied

across it is called V-I characteristic of a p-n junction.

Circuit shows the experimental arrangement to study characteristic curve of a p-n junction when it is

forward biased. A battery is connected across p-n junction through a rheostat so that voltage across

the diode can be changed. The milliammeter measures the current through the diode and voltmeter

measures the voltage across it. For different values of voltages the values of current is noted. A graph

is plotted between V and I as shown called as forward characteristic curve.

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Important features of Graph –

(1) The V –I graph is not a straight line so it does not obey ohm‟s law.

(2) Initially current increases very slowly till the voltage Crosses a voltage certain value called as

threshold voltage or cut-in-voltage. Below this voltage depletion layer plays dominant role.

(3) After this cut – in – voltage, the diode current increases rapidly (exponentially) even for small

increase in diode bias voltage. Here the majority charge carries feel negligible resistance at the

junction.

Q.21 Write Biot-Savart law.

Write the path of motion of an electron when it enters in magnetic field at

(a) perpendicular

(b) an angle θ.

OR

Define angle of dip. Find the value of dip angle at any point for which the horizontal component and

vertical component of earth‟s magnetic field are equal.

A.21 Biot-savart‟s law is an experimental law predicted by Biot and savart. This laws deals with the

magnetic field induction at a point due to a small current element.

According to Biot-savart law, the magnitude of the field B

d is:

(i) directly proportional to the current I through the conductor

B Id

(ii) directly proportional to the length dl of the current element,

Bd dl

(iii) directly proportional to sinwhere „ ‟ is the angle between

dl and r

B sin d

(iv) Inversely proportional to the square of distance „r‟ of the point P form the current element.

2

1Bd

r

Combining all these four factors we get

2 2

0

2

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I sin KIdlsinB or B =

I sinor, B =

4

K = 10 w / A m4

where

dld d

r r

dld

r

b,

The path of motion of an electron when it enters in magnetic field at –

(a) perpendicular – circular

(b) an angle θ – helical.

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OR

Angle of Dip – The angle made by the direction of the total intensity of the earth‟s magnetic field,

i.e. the axis of a freely suspended magnet, with the horizontal is called angle of Dip at that place. It is

denoted by θ or .

Let angle of dip is „I‟ at any point where horizontal component and vertical component of earth‟s

magnetic field B

are equal.

BH = B cos I and Bv = B sin I

But BH = Bv So, B cos I = B sin I

or, tan I = 1 = tan 45o or I = 45

o

So, at a place where dip angle is 45o the two components BH and Bv are equal.

Q.22 Write Bohr‟s any two postulates for hydrogen atom.

A.22 Bohr combined classical electromagnetic wave theory and early quantum concepts and gave his

theory in form of three postulates.

Bohr’s First Postulate – It was that an electron in an atom could revolve in certain stable orbits

without emission of radiant energy, contrary to the prediction of electro-magnetic theory. Necessary

centripetal force is obtained from the electric attractive force, i.e.

Fe=Fc

or 2 2

2

kze mv

r r

Where0

1k

4

, m-mass of electron

Bohr’s Second Postulate-It defines these stable orbits. The electron revolves around the nucleus

only in those orbits for which the angular momentum is some integral multiple of h

2where h is the

Planck's constant = 6.62 x 10-34

J -s.

Thus, the angular momentum of the orbiting electron is quantised i.e.

nhL mvr

2

nhor mvr

2

Q.23 Write the law of radioactive decay. Decay constant of a radioactive substance is 10-3

per year.

Calculate its half-life time in year.

A.23 Law of Radioactivity states that, no. of nuclei undergoing decay per unit time is proportional to total

no. of nuclei in the sample.

λ = 10-3

/ year

T1/2 = 3

ln 2 0.693693years

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Q.24 Define the following:

(a) Transducer

(b) Bandwidth.

A.24 (a) Transducer−Any advice that converts one form of energy into another is called transducer. An

electrical transducer converts some physical variable like pressure, temperature, displacement

etc. into corresponding variation‟s in the electric signal at its output.

(c) Bandwidth−The frequency range over which an equipment operates or the portion of spectrum

occupied by the signal.

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Q.25 Write Ampere‟s circuital law. A long straight wire of a circular cross-section (radius a) carrying

steady current. Current is uniformly distributed in the wire. Calculate magnetic field inside the region

(r < a) in the wire.

A.25 According to Ampere Circuital law, the circulation of magnetic field around a closed loop is equal

to0 times the net current crossing the surface bounded by the loop.

Q.26 Draw a ray diagram for image formation by a concave mirror and establish a relation between object

distance (u), image distance (v) and focal length (f).

A.26 Consider an object AB Placed on principal axis

beyond centre of curvature C of a concave mirror of

small aperture. A ray AM from the object travels

parallel to the principal axis and after reflection

from the mirror it passes through focus F.

Another ray AP is incident at pole P of the mirror

and is reflected along PA' in accordance with laws

of reflection so that APB BPA' .

The two reflected rays meet at point A' forming a

real image A'B'

Using cartesian sign convention,

Object distance,

BP = -U

Image distance, B'P = -v

Focal length FP = -f

Radius of Curvature CP = - R = - 2f

Now, A'BP ABP

A'B' PB' v –v

   = = AB  BP U U

…. (i)

Draw perpendicular MN from M on principle axis.

MNF A'B'F

A 'B' B'F

MN NF

A 'B' PB' PF

AB PF

(as mirror have small aperture NF PF)

v fA 'B'  v f

  AB  f f

…. (ii)

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Form equations (i) and (ii), we get

v v f=

U f

or, vf' = Uv – Uf

Divided both sides by Uvf we get

1 1 1

U f v

or, 1 1 1

v U f

This proves mirror formula for a concave mirror.

Q.27 What is rectification? Explain the working of a fullwave rectifier. Draw necessary circuit diagram.

OR

What is meant by amplification? Draw a simple circuit of n-p-n common emitter transistor amplifier.

On the basis of band theory, distinguish between conductor, insulator and semi-conductor.

A.27 Rectification- Conversion of ac into dc using a rectifier is called rectification. A full wave rectifier

is consists of a transformer two junction diodes D1 and D2 and load resistance RL. The input ac signal

is fed to the primary coil P of a transformer.

The two ends A and B of the secondary are connected to P-ends of D1 and D2. The secondary is

tapped at central point T which is connected to the n-ends of two diodes through load resistance RL.

Working- At any instant, the voltage at end A(input of D1) and end B (input of D2) of the secondary

with respect to centre Tap T will be out of phase with each other. Suppose during the positive half

cycle of ac, the end A is positive and end B is negative with respect to the centre tap T. The diode D1

gets forward biased and conducts current along the path 1AD YTA . The diode D2 is reverse biased

and does not conduct.

During negative half cycle, the end A becomes negative and end B becomes positive with respect to

the centre tap T. The diode D1 gets reverse biased and does not conduct. The diode D2 conducts

current along the path 2HD YTB . As during both half cycles of input ac the current through load

RL flows in the same direction X to Y, so we get a pulsating dc voltage across RL. Since output

voltage across RL is obtain for both half cycles of input ac, the process is full wave rectification.

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OR

The process of increasing or magnifying current, voltage or power of input signal is called

amplification.

Q.28 Write the statement of Gauss‟ law for electrostatics.

Draw a diagram and derive an expression for electric field due to a uniformly charged infinite plane

sheet at a point near the sheet.

In the given diagram write the value of electric flux passing from the surface.

OR

Define capacitor.

Draw a circuit diagram and obtain a relation for equivalent capacitance in the series combination of

capacitors.

In given circuit diagram write the value of V1.

A.28 Total electric flux through a closed surface is equal to 0

1

times the net charge enclosed by the closed

surface. Let electric charge be uniformly distributed over the surface of a thin, non-conducting

infinite sheet. Let the surface charge density (i.e., charge per unit surface area) be . We have

calculate the electric field strength at any point distance r from the sheet of charge. To calculate the

electric field strength near the sheet, we now consider a cylindrical Gaussian surface bounded by two

plane faces A and B lying on the opposite sides and parallel to the charged sheet and the cylindrical

surface perpendicular to the sheet (fig). By symmetry the electric field strength at every point on the

that surface is the same and its direction is normal outwards at the point on the two plane surfaces

and parallel to the curved surface.

Total electric flux

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1 2 3

1 2 3

1 2 3S S S S

o o o

1 2 3S S S S

E .dS E.dS E.dS E.dS

or E .dS E.dS cos0 E.dS cos0 E.dS cos0

1 2dS dS 2

Totalelectricflux 2 .

E E E a E a E a

E a

As is charge per unit area of sheet and a is the Intersecting area, the charge enclosed by Gaussian

surface =a

According to Gauss's theorem.

Total electric flux 0

1

(total charge enclosed by the surface)

0

0

1. ., 2E ( )

E2

i e a a

Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of

the point and is directed normally away from the charge. If the surface charge density is negative

the electric field is directed towards the surface charge.

Total Flux passing from the surface

netE

0

6 26

12

q

1 10 Nm1.129 10

8.85 10 C

OR

Capacitor− An arrangement of two conductors separated by an insulating medium is called

capacitor. It is used to store charge and electrical energy.

Equivalent capacitance in series combination−Figure shows three capacitance C1, C2 and

C3connected in series. A potential

difference V is applied across the

combination. This sets up charges Q on

the two plates of each capacitor.

The p.d across the various capacitors are−

1 2 3

1 2 3

Q Q QV ,V ,V

C C C

The sum of these p.d must be equal to the

applied p.d

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1 2 3

1 2 3

1 2 3

Q Q QV V V V

C C C

V 1 1 1or,

Q C C C

If Cs is the equivalent capacitance of the series combination then SC

Q

V

S 1 2 3

1 1 1 1or,

C C C C

In parallel voltage on C2 and C3 are equal.

So V =V1 + V2

10V=V1+4V V 1 = l0V - 4V = 6V Ans.

Q.29 Draw phasor diagram for a series LCR circuit with alternating voltage source, determine the

expression for the impedance of the circuit.

OR

Draw a diagram of AC generator and describe it. Derive an expression for instantaneous value of

induced emf.

A.29 Suppose resistance R, inductance L and capacitance C are connected in series and an alternating

source of voltage V = V0 sin t is applied across it. (fig. a) On account of being in series, the current

(i) flowing through all of them is the same.

Suppose the voltage across resistance R is VR voltage across inductance L is VL and voltage across

capacitance C is Vc. The voltage VR and current i are in the same phase, the voltage VL will lead the

current by angle 90o while the voltage VC will lag behind the current by angle 90° (fig. b). Clearly VC

and VL, are in opposite directions, therefore their resultant potential difference

= VC - VL (if VC > VC)

Thus VR and (VC - VL) are mutually perpendicular and the phase difference between them is 90°. As

applied voltage across the circuit is V, the resultant of VR and (VC - VL) will also be V. From fig.

V 2

= VR 2 + (VC - VL)

2 => V= VR

2 +(VC - VL)

2 .... (1)

But VR =R i, VC =XC i and VL =XL i .... (2)

where C

1X

C

capacitance reactance and XL =L = inductive reactance

C

22

LR i X Xi iV

2

C L

2VImpedanceof circu R Xt, Z X

i

2 2

L

22

Ci.e. R X R1

Z X LC

OR

AC generator - A dynamo or generator is a device which converts mechanical energy into electrical

energy. It is based on the principle of electromagnetic induction.

Construction: It consists of the four main parts:

(i) Field Magnet: It produces the magnetic field. In the case of a low power dynamo, the

magnetic field is generated by a permanent magnet, while in the case of large power dynamo,

the magnetic field is produced by an electromagnet.

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(ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or

ring. It can revolve round an axle between the two poles of the field magnet. The drum or

ring serves the two purposes: (i) 1t serves as a: support to coils and (ii) It increases the

magnetic field due to air core being replaced by an iron core.

(iii) Slip Rings: The s lip rings R1 and R2 are the two metal rings to which the ends of armature

coil are connected. These rings are fixed to the shaft which rotates the armature coil so that

the rings also rotate along with the armature.

(iv) Brushes: These are two flexible metal plates or carbon rods (B1 and B2) which are fixed are

constantly touch the revolving rings. The output current in external load RL is taken through

these brushes.

Working: When the armature coil is rotated in strong magnetic field, the magnetic glux linked with

the coil changes and the current is induced in the coil, its direction being given by Fleming's right

hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise

direction, the wire ab moves upward and cd downward, so that the direction of induced current is

shown in fig. In the external circuit, the current flows along B1RLB2·The direction of current remains

unchanged during the first half turn of armature. During the second half revolution, the wire ab

moves downward and cd upward, so the direction of current is reversed and in external circuit it

flows along B2RLB1. Thus the direction of induced emf and current changes the external circuit after

each half revolution.

If N is the number of turns in coil, the frequency of rotation, A area of coil and B the magnetic

induction, then induced emf

{ cos 2 }

2 sin 2

d de NBA ft

dt dt

NBA f ft

Obviously, the emf produced is alternating and hence the current is also alternating.

Q.30 To produce interference fringe pattern, draw a ray diagram of Young‟s double slit experiment.

Derive an expression of fringe width for bright interference fringes. If fringes width for bright fringes

is 2 mm, write the fringe width for dark fringes. Whether the centre point of interference fringe

pattern is bright or dark? Explain it.

OR

Draw a ray diagram for a concave mirror and prove that focal length of the mirror is half of its radius

of curvature. Write mirror equation for a concave mirror. When light enters from rare to denser

media, what effect occurs on the values of wavelength and frequency of it?

A.30 Observing Interference pattern in Young's Double Slit Experiment –

Fringe of Bright Fringe - Suppose S1 and S2 are two fine slits, a small distance d apart and screen at

a distance D from the slits. Perpendicular bisector lines are joined at O point of screen. Hence central

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bright fringe be at O. P held apart xn from O, where nth

bright fringe is formed. The path difference

between two waves arriving at P is equal to S2P- S1P. By Right angleS2 PN,

2

2 2 2 2

2 2 n

DS P S N PN D X

2

Similarly by right angle S1PM,

OR

Consider a ray AB parallel to principal axis incident at point B of concave mirror of small aperture.

After reflection from the mirror, this ray converges to point F obeying the laws of reflection. F is the

focus of the mirror, C is the centre of curvature, CP = Radius of curvature and BC is normal to the

mirror at point B. According to law of reflection i r

As AB is parallel to P PC i

In BFC r Hence, CF = FB

For mirror of small aperture

FB = FP

CF = CF + FP

Hence = FP + FP = FP

or R = 2F

or R

2f

Thus focal length = 1

2 radius of curvature.

Mirrors equation

1 1 1

v u f

In moving from rarer to denser medium wavelength decreases but frequency remains constant.