IJHSS, Tangasseri, Model Examination,Jan.2013 Solved PHYSICS Paper-I(Theory)

STD:- XII (Three hours) (Candidates are allowed additional 15 minutes for only reading the paper,

They must NOT start writing during this time) Answer all questions in Part I and six questions from Part II, Choosing two questions

from each of the Sections A, B and C. All working including rough work should be done on the same sheet as, and adjacent to,

the rest of the answer. The intended marks for questions or parts of questions are given in brackets [ ].

(Material to be supplied: Log tables including Trigonometric functions) A list of useful physical constants is given at the end of this paper.

PART I(Answer all questions) Question 1. A Choose the correct alternative A, B , C or D for each of the questions given below: [5] (i) A charge ‘q’ is placed at the middle point of the line joining two charges 12µC each.

The system of three charges will lie in equilibrium if ‘q’ is equal to: (A) 8µC (B) 3µC (C) –3µC (D) +4µC

Ans:- (C) In this case we have to ensure that both 12µC charges as well as ‘q’ must be static. The force between the two charges 12µC = k*12*12/r2. This force moves each of the charges outward. In order to hold them at their place there must be a force generated by the charge ‘q’ = – k*12*q/(r/2)2. For equilibrium, k*12*12/r2= – k*12*q/(r/2)2. Hence q= –3µC.

(ii) A uniform wire of length ‘l’ and radius ‘r’ has resistance 100Ω. It is recast into thin wire of length 2l. The resistance of the new wire is: (A) 200 Ω (B) 300 Ω (C) 400 Ω (D) 500 Ω

Ans:- (C). Use the concepts that the resistivity of the material remains the same and the volume of the wire remains the same. For volume equality: A1* l =A2* 2l. A1/A2 =2 For resistivity equality: ρ = R1A1/L1=R2A2/L2. R1/R2=( A2/A1)( l/2l)=1/4 R2 = 4R1 =400Ω.

(iii) The rate of change of current 10 As-1 in a coil produces an e.m.f of 5 V. Then the self inductance of the coil in henry is: (A) 0.5 (B) 0.25 (C) 1 (D) 1.25

Ans:- (A). Use the formula for self inductance: L = e/(dI/dt). L = 5/10 =0.5 henry. (Writing the unit as Vs A-1 is clumsy.

(iv) The focal length of thin lens of refractive index (µ = 1.5) in air is 10 cm. If air is replaced by water of µ =4/3, its focal length is : (A) 20 cm (B) 30 cm (C) 40 cm (D) 25 cm

Ans:- Use the Lens maker’s equation: ⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

21

11)1(

1RR

nf

. Remember that n is the absolute

refractive index of the material (here glass). In general it is the relative refractive index we have to consider. The (n-1) part of the equation is applied in this case. This part becomes: ng/nw.

Thus: )1(1−= nk

f…..(1) )1(1

1 −=w

g

nn

kf

……(2)

1

From equations (1) and (2) f1=40cm.

(v) If e/m of electron is 1.76 x 1011 C kg-1 and the stopping potential is 0.71 V, then the maximum velocity of the photoelectron is : (A) 150 kms-1. (B) 200 kms-1. (C) 500 kms-1. (D) 250 kms-1.

Ans:- Remember the relation between the kinetic energy and the potential energy of the photoelectron: ½ mv2= e Vo. v = √(2e Vo)/m. Use the given e/m and find v = 500kms-1.

B Answer all questions briefly and to the point: [15] (i) An oil drop weighing 1.0 x 10-15 N and carrying a charge of 8 x 10-19 C is found to

remain at rest in a uniform electric field of intensity ‘E’. Find ‘E’. Ans:- At the equilibrium state, the electric force is balanced by the gravity.

Use the equation and find E : qE =mg E = 1.25 *103 NC-1. The units of E are : NC-1 or Vm-1.

(ii) Which conservation principle is involved in Kirchoff’s second law? Ans:- Law of conservation of energy.

(Remember that in this law we are discussing about the product of current and resistance (or Voltages) which is the electric potential energy per unit charge from definition)

(iii) State any one difference between Joule effect and Peltier’s effect (iv) What is the resistance and tolerance of yellow, violet, red colour coded resistor

respectively? Ans:- Yellow represent the digit 4, violet; 7 and red represent the power of ten

multiplication factor. If the forth band is missing, it means that the tolerance is 20%. Thus the resistance value =(47 * 102+ 20% ) Ω

(v) An electron of mass ‘m’ and charge ‘e’ is moving around the nucleus in a hydrogen atom with an angular momentum ‘l’. What is its magnetic moment?

Ans:- Use the concept of magnetic moment: M =IA But I =Charge /time = e/T =

mrel

mremvr

rmemv

rev

vre

22 22222 πππππ==== : (mvr=angular

momentum.)

mel

mrrelA

mrelM

222 2

2

2 ===ππ

π

(vi) A metal conductor of length 1m rotates vertically about one of its ends at angular velocity 5 rad/s. If the horizontal component of earth’s magnetic field is 0.2 x 10-4 T, then what is the emf developed between the ends of the conductor?

Ans:- Use e =blv =bl (lω)= 50µV. ( since the radius of the circle is formed by the conductor; v =radius * angular velocity)

(vii) The distance of coverage of a transmitting antenna is 12.8 km. Then what is the height of the antenna?

Ans The standard formula is : d =√(2hR). h =12.8 m.

(viii) The least distance of distinct vision of a person is 75cm. What is the focal length of the reading spectacles for such a person?

Ans:- Use the equation: vuf111

+−= . Here v=-D=-25 cm and u=-75cm.The New Cartesian

sign convention is used in this case. F = +37.5 cm ; the plus sign shows that the person must sue a convex lens.

(ix) What is the de’Broglie wavelength associated with an electron accelerated by a potential of 64V.

2

Ans:- Use the standard equation for wave length of electron: mV

101027.12 −×=λ

λ =1.53 * 10-10 m OR 1.53 angstrom unit. (x) What is the fraction of the radioactive sample that will remain undecayed after 4-

half life periods? Ans:-

Use the formula: n

oNN ⎟⎠⎞

⎜⎝⎛=

21

where n is the number of half periods:

It gives N = No/16. (You may use simple logic of ‘half of half each time’ instead of the formula)

(xi) Find the momentum of a photon of energy 3.0eV. Ans:- The relation connecting energy and momentum is : p= E/c.

P = (1.9*10-19 *3)/3*108 =1.6*10-27 Ns. (xii) If a hydrogen atom goes from a third excited state to second excited state, what kind

of radiation (visible light, ultraviolet, infrared etc) is emitted? Ans:- Instead of using the standard formula for calculating the wave length remember the

various series. Here it is Balmer series. Therefore the radiation is either visible of infrared.

(xiii) Where in our Universe is the thermo nuclear energy being released naturally?

Ans:- In stars ( Sun) (xiv) In which of the solids (semi conductors, conductors or insulators) do conduction and

valance band overlap? Ans:- In conductors. (xv) Write down the truth table of a NAND gate. A B Y

0 0 1 0 1 1 1 0 1 1 1 0

PART II . SECTION A (Answer any two questions) Question 2 (a) With the help of a labeled diagram, obtain an expression for the electric field

intensity ‘E’ at a point ‘P’ in broad side position (ie equatorial plane) of an electric dipole.

[4]

Ans:-

The point on the equatorial line is marked P. A unit positive charge is

placed at P.

r

EA

EB

EA Cosθ + EB Cosθ

EB Sinθ

EA Sinθ

θ P

-q θ O θ

A B2a

+q

3

EA represent the force between the unit positive charge and –q, and EB

represent that between the unit charge and +q.

24 APqEo

A πε= …………(1)

)2...(..........4 2BP

qEo

B πε=

And BA EE =

Resolving EA and EB into components:

EA sinθ and EB Sinθ are equal in magnitude and opposite in direction,

and hence these components will cancel each other.

The resultant intensity at P will be E=EA Cosθ + EB Cosθ …….(3)

E=2EA Cos θ …………………………………………(4)

From the triangle APO:

Cos θ =OA/AP =22 ar

a+

……….(5)

Equation (4) becomes:

23222222

22222

)(4

2)(

14

2

422

ar

qaar

aar

qE

ara

ACq

araEE

oo

oA

+=

++=

+=

+=

πεπε

πε

But the dipole moment, p=2qa.

The above equation becomes:

)6(..........)(4 2

322 ar

pEo +

=πε

For r>>a, the above equation becomes: 34 rpE

oπε=

Thus the electric field decreases as the cube of the distance from the

centre of the dipole along the equatorial line.

(b) Find the electric charge Q1 on plates of capacitor C1 in the figure shown below:

[2]

Ans:- Since the capacitors are connected in series, the charge on each capacitor is the same. Effective capacitance C needs to be found out.

FC

CCCC

μ732

327

321

161

811111

321

=∴

=++=++=

110V 8µF

C3

32µF16µF

C1 C2

4

Charge Q = Q1=CV=(32/7)*110 =502.9µC

(c) What is meant by temperature coefficient of resistance? Write down Ohm’s law in vector form , stating the meaning of every symbol used

[3]

Ans:- The temperature constant α is given by:

tRRR

o

ot −=α , where Rt –resistance at toC, Ro- resistance at 0oC and t the

temperature. Ohms law is given by: I = V/R, Where I is the current, V the voltage and R the resistance respectively.

,RAV

AI=∴ where A is the area of cross section of the wire.

,*RA

lEAI=∴ where l is the length of the wire

EE

lRA

EJr

rr

σρ===∴ , where ρ is the resistivity and σ , the conductivity of the wire.

Question 3 (a) Draw a labeled diagram of a balanced Wheatstone’s bridge. Using either Ohm’s law

or Kirchofff’s laws, obtain the relation between four resistors forming the bridge. [3]

Ans:-

When the bridge is balanced, the current I5 =0. Potential difference across P = Potential difference across R I1P = I2 R ……..(1) Potential difference across Q = Potential difference across S I3 Q = I4 S …….(2) Dividing equations(1) by (2) gives: I1P/I3Q = I2R/I4S. P/Q = R/S ; since I5= 0; I1=I3 and I2 = I4.

(b) Obtain an expression for magnetic flux density ‘B’ at the centre of a circular coil of radius ‘R’ and having ‘N’ turns, when a current ‘I’ is flowing through it.

[3]

Ans:- Preferred method:- Use Biot-Savart’s law and Ampere’s circuital law.

Magnetic field at the centre of the circle due to the current element I dl :

)1(..........4

; 90 ,4 2

o2 R

IdldBSinceR

IdlSindB oo

πμ

θθπμ

===

The circular coil is divided into current elements. Total field at the centre of the coil due to all these elements:

I4

I

E

R S

P Q

G

I5

I3

I2

I1

I

dl O R

5

RIN

coiltheofcentretheateldmagneticfi

turnsofnumberNForRI

B

RRI

B

dlRI

RIdlB

o

o

o

Ro

Ro

πμ

πμππμ

πμ

πμ ππ

4

: ''4

42

44

2

2

02

2

02

=

∴

=

=

== ∫∫

Second Method: This method is not suitable because of lengthy and complicated steps.

Let R be the radius of the current carrying loop with current I. Let P be a point on the axis of the coil at a distance of x from its centre.

Magnetic field at P due to the current element dl at A :

o2

2

90 since),2.(....................4

)1....(..........4

sin

==

=

θπ

μπ

θμ

rIdl

dB

rIdl

dB

o

o

The above magnetic field acts along PQ perpendicular to the line AP. Resolving dB into components along PN and PM: PS=dB Cos α & PM= dB sin α………..(3) Consider a diametrically opposite element at A’. Due to the symmetry, the

magnetic field at P due to this current element is having the same magnitude but along the direction PQ’. Resolving this will give:

PS’=dB Cos α & PM= dB sin α……….(4) Thus components PS & PS’ cancels and the field at P is due to the component

dB sin α. Therefore, the field due to all the elements of the coil is:

( ) :(5)Eqn in (6) gSustitutin )6........( rRsin

direction.- xpositive thealong actsit ,)5.....(sin4

21

22

2

Rxrand

rIdl

dBB o

+==

== ∫

α

απ

μ

( ) ( ) ( )

( ))7.(..........

2

2444

23

22

2

23

2223

2223

22

Rx

IRB

RRx

IRdl

Rx

IRdl

Rx

IRB

o

ooo

+=

×+

=+

=+

= ∫∫

μ

ππ

μ

π

μ

π

μ

At the centre of the coil x=0

Q’

Q

M

S’ A’

dlA

x

R r S

P O I α α

6

Equation (7) becomes:

RI

B

RIR

B

o

o

2

2 3

2

μ

μ

=

=

For ‘N’ turns, field RIN

B o

2μ

=

(c) (i) State any two differences between a moving coil galvanometer and a tangent galvanometer. (ii) What is the use of a Cyclotron?

[3]

Ans:- (i) (1)The operation of moving coil galvanometer depends on the force acting on a current carrying conductor in a magnetic field where as tangent galvanometer works on the principle of the tangent law. (2) A moving coil galvanometer with its dead beat nature can be used to measure the current directly and faster than the tangent galvanometer. (ii) A cyclotron is used to accelerate protons and deuterons for artificial transmutation of elements.

Question 4 (a) What is meant by ‘the time constant’ of an L-R circuit?

When the current flowing through a coil ‘P’ decreases from 5A to 0 in 0.2 s, an emf of 60V is induced across the terminals of an adjacent coil ‘Q’. Calculate the coefficient of mutual inductance of the two coils P and Q

[3]

Ans:- The time constant is defined as the time required by the current in an L-R circuit to fall to 0.368th of its steady value.

OR The time constant is defined as the time required by the current in an L-R circuit to grow to 0.632th of its steady value.

Hdt

dIeM 4.12

52.060

2.0560

=×

===

(b) When an alternating emf e = 300 Sin(100πt + π/6) volt is applied to a circuit, the current ‘I’ through it is I=5.0Sin(100πt – π/6)ampere. Find the (i) phase difference between the emf and the current (ii) average power consumed by the circuit.

[3]

Ans:- (i) Phase difference φ = π/6 – (–π/6) = π/3 rad. (ii) Pa = IrmsVrms Cos φ

WCosCos

VI

CosVI

5.37325.0300

2

2200

00

=×

==

×=

πφ

φ

(c) Obtain an expression for the resonant frequency fo of a series LCR circuit.

[3]

Ans:- The impedance (Z) of an L-C-R circuit is given by:

( )2

222 1⎟⎠⎞

⎜⎝⎛ −+=−+=

CLRXXRZ CL ω

ω , where L is the self inductance, C the

capacitance and ω, the angular frequency. The phase difference between the current and voltage in the circuit φ is given by:

RCL

RXX CL ωω

φ1

tan−

=−

= . When XL = XC, the current and voltage will be in the

same phase and Z will have the minimum value and the current becomes maximum. This condition is called the resonance. At resonance: XL = XC.

7

LCf

LC

CL

o ππω

ω

ωω

21

2

1

1

==∴

=∴

=

‘fo’ is called the resonant frequency. SECTION B (Answer any two questions)

Question 5 (a) Draw a neat labeled diagram and explain the working of Michelson’s method for

the determination of speed of light in air. The speed of red light in vacuum is 299792458 ms-1. What is the speed of red light in glass of refractive index 1.5?

[4]

Ans:-

The most important part of the experimental setup is an octagonal mirror M which can be uniformly rotated about its centre. A well collimated beam of light from a powerful source is incident on the first face of the rotating mirror M and the reflected beam is received at the eye piece E after traveling through a distance of 70 km. The multiple reflections are carried out by plane mirrors M1 to M5 and two concave mirrors C1 and C2 as kept in the diagram. If the mirror M is at rest and the beam of light is reflected from the first face is will be received after reflection from the fifth face. As the mirror is rotated the beam of light received at the eye piece flickers and becomes once again steady when the sixth face replaces the fifth face. It means that the time required for the beam to travel twice the distance between the stations at C1 and C2 is the same as the time required for the mirror to turn one of its faces ( or one by eight of a rotation) Let n be the frequency of rotation of the mirror M and d, the distance between the stations ,then: Speed of light in air = nd

n

dv 168

12

== .

Refractive index = Speed of light in air/Speed of light in the medium. Speed of light in the medium= Speed of light in air/ Refractive index = 299792458 /1.5 =199861639ms-1.

(b) How are electromagnetic waves produced? Which vector of the electromagnetic wave is responsible for the optical effects?

[2]

Ans:- An oscillating (or accelerated) electron beam produces electromagnetic waves. The electric vector is responsible for the optical effects.

C2

TPE

M3

Slit

Source of light

M5

8

7

6

1 2

3

4 5

M2

M4

M1

C1 M

8

(c) Draw labeled graphs showing the variation in intensity with angular position of (i) diffracted light in a single slit diffraction experiment and (ii) interfered light in double slit interference experiment.

[2]

Ans:-

(i)

Intensity of fringes

Intensity of fringes

Angular distance from the centre (θ)

(ii)

Distance from the centre (x)

Question 6 (a) Explain the phenomenon of total internal reflection.

What is the refractive index of the medium for which its critical angle with air is 42o? Explain one practical application in communication.

[3]

Ans:- When a ray of light traveling from optically denser to rarer medium, if the angle of incidence in the denser medium is greater than the critical angle for the pare of media, the light ray is reflected back into the denser medium. This phenomenon is called total internal reflection.

.49.14211

=∴

==

m

oa

m

SinSinCμ

μμ

Total internal reflection is applied in optical fibers which are used for data transfer in communication.

(b) With the help of a neat labeled ray diagram derive lens maker’s formula for refraction through a thin bi-convex lens.

[3]

Ans:-

The ray OA from the medium of refractive index n1 is incident at the convex surface and bend (AB) towards the normal C1N1 and forms an image I1 in the denser medium n2. Thus:

v1u

O C2

A

C C1 I1

N2N1 B

n1

n2

I

R2 R1

n1

9

)......(1

121

1

2 iR

nnun

vn −

=−

Since the second surface is concave for the ray AB in the denser medium n2, the refracted ray BI bends away from the normal C2N2 and the final image I is formed in the rarer medium n1.Thus:

)......(2

21

1

21 iiR

nnvn

vn −

=−

Adding equations (i) and (ii):

)(..........11)1(1

lens; theoflength focal thef, vand u Ifplaced. isit in which medium therespect towith

medium lens theofindex refractive theisn where11)1(11

11)1(11

11)(

21

21

211

2

2112

11

iiiRR

nf

RRn

uv

RRnn

uv

RRnn

un

vn

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=∞=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−

The above equation is called the lens maker’s equation. (c) When two thin lenses of focal lengths f1 and f2 are kept in contact, prove that their

combined focal length ‘f’ is given by 21

111fff

+= [2]

Ans:-

Let f1 and f2 be the focal lengths of the two thin lenses respectively. The first lens L1 forms a real image at I1 of an object kept at O.

)(..........111

11

iuvf

−=∴

This real image acts as a virtual object for the lens L1 and a real image is formed at I.

)(..........111

12

ivvf

+−=∴

When lenses are combined:

22

22

111

1.1111

fff

fvuff

+=∴

=+−=+∴

Where f is the combined focal length of the combination.

O I I1

vu v1

L1 L2

Question 7 (a) (i) Define the term dispersive power of a prism.

(ii) Derive an expression for the condition for achromatic combination of two thin lenses in contact.

[3]

Ans:- (i) Dispersive power is defined as the ratio of the angular dispersion between two colours to the mean deviation caused to the emergent rays after dispersion of white light from a prism.

(ii) Let fR,fV and fY be the focal lengths of first lens for red, violet and yellow colours respectively. Let ω be its dispersive power. Then:

10

)(..........11 ifff YRV

ω=−∴

Let f1R,f1

V and f1Y be the focal lengths of second lens for red, violet and

yellow colours respectively. Let ω1 be its dispersive power. Then:

)(..........111

1

11 iifff YRV

ω=−∴

When both lenses are combined. Adding (i) and (ii) gives:

1

1

YY ffωω

+

Where FV and FR are the focal length of the combination of lenses for colours violet and red respectively. For achromatic combination, the focal length FV = FR.

1

1

1

1

0

YY

YY

ff

ff

ωω

ωω

−=∴

=+

This is the condition for achromatic combination. (b) What is shortsightedness? How is it corrected?

A person has his near point at distance 0.5m. What type of lens he must use for correction? Calculate the power of that lens.

[3]

Ans:- It is a defect of the eye in which the person can see the nearby objects but not able to see far away objects. It is corrected by using a concave lens of suitable power. Here : u = D= - 0.25m; v= -0.50m

.2245.0

125.01

5.01

25.01

111

dioptreP

vuf

+=−=−=−

+−

−=

+−=

The positive sign shows that the person must use a convex lens.

(c) (i) Define resolving power of telescope. (ii) State one advantage of a reflecting telescope over refracting telescope

[2]

Ans:- (i)It is the ability of the telescope to produce two distant closely placed objects as two distinct objects. It is given by:

1

22.1. −= raddPR

λ Where d is the diameter of the objective and λ , the wavelength

of the light used to observe the distant objects. (ii) Reflecting telescopes are free from chromatic aberration.

SECTION C(Answer any two questions) Question 8 (a) (i) How are cathode rays produced? Write down any two properties of cathode

rays. (ii) What is the principle of e/m of electrons by Thomson’s method?

[2]

Ans:- (i)Cathode rays are produced in a highly evacuated discharge tube under very high potential when highly accelerated positive ions hits the cathode of the discharge tube. Cathode rays can ionize gases. Cathode rays are deflected in electric and magnetic fields. (ii) Electrons can be deflected in a magnetic and an electric field. When both fields are crossed the velocity of electrons = Electric field / magnetic field.

(b) Write down Einstein’s photoelectric equation and explain the terms: threshold frequency and work function of an emitter.

[3]

11

Ans:- Einstein’s photoelectric equation is : ( )22max2

1ohmv υυ −=

Where; m- mass of electron vmax-maximum velocity acquired by the photoelectron ν- the frequency of incident radiation νo- the threshold frequency, and h- the Planck’s constant. Threshold frequency:- It is the lowest frequency for the incident radiation on the emitter below which no photoelectric emission is possible. Work function:- It is the minimum energy required for the incident radiation to release electrons from the emitter.

(c) Explain Geiger-Marsden experiment of alpha ray scattering with proper illustration and derive an expression for the nuclear radius in terms of closest approach of alpha particles to the nucleus.

[3]

Ans:- In the alpha ray scattering experiment, Geiger and Marsden found that the scattering of alpha particles by the nucleus is in accordance with the Coulomb’s law of interaction between charges.

A few alpha particles will collide with the nucleus of the atom head-on and scattered at an angle of nearly 180o. At the distance of closest approach the potential energy of the α-particle is:

( )( ) ).....(24

1 ir

eZeUooπε

= . This energy is derived from the kinetic energy of the α-

particle, ‘K’.

( )( )

).........(24

1

24

1

2

iiKZer

reZeKU

oo

oo

πε

πε

=∴

==∴

Equation (ii) gives the value of distance of closest approach to the nucleus and its gives an estimate of the size of the nucleus.

Nucleus α- particles (+2e)

+Ze

rO

Question 9 (a) What is meant by ‘mass defect’?

Draw sketch graph showing binding energy per nucleon against mass number of elements and based on the graph explain nuclear fission.

[3]

Ans:- The difference between the masses of the nucleons of a nucleus and the rest-mass of the nucleus is called the mass defect.

From the above graph it is clear those elements having greater mass numbers have lower binding energy per nucleon. Hence they split into two lighter nuclei to increase their binding energy for better stability. Thus nuclear fission takes place.

(b) How are X-rays produced? With the help of a sketch graph explain the continuous and characteristic nature of X-ray spectrum.

[3]

12

B.E per nucleon (MeV)

Mass number (A)

Fe56

U238

8.8 O2

Li7

H2 1.0 0

Ans:- X-rays are produced when fast moving cathode rays bombard heavy metals having high melting points in a Coolidge tube.

The X-ray emitted by a target contains two types of spectra:

(i)A continuous spectra and (ii) line spectra superimposed on the continuous spectra. The positions of intensity peaks are characteristics of the target used. These are called characteristic X-rays. The continuous spectrum has a minimum wavelength depends on the voltage applied to the X-ray tube. As the voltage is increased the continuous spectrum moves towards the right.

(c) What are matter waves? Calculate the wavelength of electrons of accelerated by a potential of 150V.

[2]

Ans:- The wave associated with moving particle is called matter waves. It is independent of the charge carried by the particle. Its wave length is given by:

mvh

=λ

We have for wave length of electron:

AO

V0.1

150150150

===λ

K-Series

Intensity L-Series

O λminWavelength λ

Question 10 (a) Draw a labeled energy band diagrams for a conductor, semiconductor and an

insulator. [3]

Ans:-

(b) With the help of a labeled common emitter n-p-n transistor circuit explain how electromagnetic oscillations are produced?

[3]

Ans:- Principle:- An oscillator is an electronic device having an amplifier and a positive feedback circuit. It produces sustained electrical oscillations of constant frequency and amplitude without an input.

Conduction Band Conduction Band Conduction Band Energy Gap

Valance Band Energy Gap (1.1eV) Overlaps Energy Gap ( >6eV)

Valance BandConductor

Semiconductor Valance Band Insulator

L1

n-p-n

13

L C1

B2+

-

B1

-K

C

B

E

+

The emitter- base is forward biased and the collector-emitter is reverse biased. When the key K is closed, a small collector current flows through the circuit. Since L and L1 are inductively coupled this current aids the forward biasing and the collector current increases. The electrical energy will be stored in the magnetic field of the coil L. When it reaches the maximum value no more current flows through it and the magnetic field decreases. This change in magnetic flux produces an induced current which charges the capacitor. The alternate charging and discharging of the capacitor through the coil produces electrical oscillations of constant magnitude and frequency. The frequency of oscillation depends on the values of inductance of the coil and the capacitance of the capacitor by the following relation:

LCf

π21

= .

(c) Prepare a table for the combination of gates shown in Figure 8 below:

[2]

Ans:- A B Y 0 0 0 0 1 1 1 0 1 1 1 1

14

Useful Constants and Relations: 1 Sped of light in vacuum (c) = 3.0 x 108 ms-1 2 Plank’s constant (h) = 6.6 x 10-34 Js 3 Constant for Columb’s law

⎟⎟⎠

⎞⎜⎜⎝

⎛

oπε41 = 9 x 109 mF-1

4 Charge of an electron (–e) = -1.6 x 10-19 C 5 Mass of an electron (me) = 9x10-31kg. 6 1 electron volt (1 eV) = 1.6 x10-19 J 7 1 nm = 10-9 m 8 The radius of earth = 6400km

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A Y

B

In the combination of gates, the first one is a NOR gate and the second one a NAND gate. The NAND gate is made into a NOT gate. Thus a NOR is NOTed to get an OR gate. The combination is an OR gate.