PHYSICS Complete Syllabus of Class XII

Model Test Paper-1(for School / Board Exams.)

PHYSICS

(Complete Syllabus of Class XII)

MM : 70 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

GENERAL INSTRUCTIONS :

(i) All questions are compulsory.

(ii) Questions number 1 to 5 are very short answer type questions and carry 1 mark each.

(iii) Questions number 6 to 10 are short answer type questions and carry 2 marks each.

(iv) Questions number 11 to 22 are also short answer type questions and carry 3 marks each.

(v) Questions number 23 is a value based questions and carry 4 mark.

(vi) Questions number 24 to 26 are long answer type questions and carry 5 marks each.

(vii) There is no overall choice. However, an internal choice has been provided in one question of 2 marks,

one question of 3 marks and all three questions of 5 marks each. You have to attempt only one of the

given choices in such questions.

(viii) Use log tables if necessary, use of calculator is not allowed.

SECTION-A

1. State Gauss Law. [1]

2. Can two similar nature charges attract? Explain with reason. [1]

3. What is Total Internal Reflection. [1]

4. Explain Lorentz force. [1]

5. State Law of Radioactivity. [1]

SECTION-B

6. A ray of light strikes a flat glass plate of thickness ‘t’ and refractive index ‘’ at a small angle ‘’. Obtain an

expression for the lateral displacement ‘d’. [2]

7. Define the following terms in alternating current.

(i) Quality factor

(ii) Active and reactive power. [2]

8. Differentiate between primary and secondary rainbow. [2]

(2)

Model Test Paper-1 Subjective Test for Class-XII (Physics)

9. Find equivalent resistance between X and Y.

2R

2R

2R

R

YX [2]

10. Differentiate between half life and mean life. [2]

OR

The activity of a sample of radioactive material is A1 at time T

1 of A

2 at time t

2 [t

2 > t

1]. Obtain an expression

for its mean life.

SECTION-C

11. The specifications of an ac device are 240 V and 960 W. [3]

(i) How can it be operated at 120 V?

(ii) What is the current that the device takes from the 120 V line?

(iii) What is the resistance of device?

12. A slab of refractive index 2 = 3/2

is placed at bottom of water of refractive index

1 = 4/3. A coin is observed

vertically above. Find apparent shift in the position of coin. Also find the refractive index of the combination

of glass and water.

2 = 3/2

8 cm

4.5 cm

Coin

Eye

13. Explain working of moving coil galvanometer. [3]

14. Prove condition of balanced Wheatstone bridge. [3]

15. A metal rod of length ‘l’ rotates with constant angular speed ‘’ about one end in a uniform magnetic field

‘B’. Find the induced emf between the ends along with polarity. [3]

16. Explain why classical theory fails to explain photoelectric effect. [3]

17. Explain working of semi conductor diode as full wave rectifier. [3]

18. NOR gate is universal gate. Explain with example. [3]

19. Explain Young’s double slit experiment. [3]

20. Calculate energy released in the following reaction [3]

3Li6 +

0n1

2He4 +

1H3

21. Derive Len’s Maker’s formula. [3]

22. What are the essential elements of the communication system? [3]

OR

Why it is necessary to convert low frequency base- band signal to high before transmission? [3]

(3)

Subjective Test for Class-XII (Physics) Model Test Paper-1

23. A thin equi-convex lens of glass ( =1.5) with radius of curvature ‘R’ is placed on a horizontal plane mirror.

When the space between the lens and mirror is filled with liquid, an object held vertically above the lens is

found to coincide with its own image. Obtain an expression for the refractive Index of the liquid. [4]

SECTION-D

24. Explain construction and working of Van de Graaff generator. [5]

OR

Explain construction and working of cyclotron.

25. Calculate charge on each capacitor. [5]

3 F

3 F

3 F6 F

6 F

10 V

OR

A capacitor of 2F charged to a potential difference of 100 V is connected to an uncharged capacitor of 3F

through switch S. When ‘S’ is closed then

(a) Determine the final charge on each capacitor.

(b) Find the fraction of energy lost.

S

100 V3 F

2 F

26. Explain the properties of [5]

(i) Di-magnetic material

(ii) Para-magnetic material

(iii) Ferro-magnetic material

OR

Define the following terms :

(i) Intensity of magnetising field

(ii) Intensity of magnetisation

(iii) Magnetic susceptibility

(iv) Magnetic permeability

(v) Relative permeability

� � �

(4)


PHYSICS

SECTION-A

1. According to Gauss Law, the net flux of electric field through any closed surface is equal to the net charge

enclosed by the surface divided by 0.

0

.

qE dS

∫�

S

E

2. Yes, two similar charges can attract, due to induction between them, if one charge is large compared to other, this

is why repulsion is sure test of polarity.

3. When light travels from denser to rarer medium and angle of incidence is greater than critical angle (C) [angle of

incidence when angle of refraction is 90°], ray is reflected back in same medium. This phenomena is called Total

Internal Reflection.

C

i > C

O

Denser

medium

Rarer

medium

4. When a charge particle enters in a region where both electric field and magnetic field exist, force exerted on it is

called Lorentz force.

E MF q E V B F F⎡ ⎤ ⎣ ⎦

��

Model Test Paper-1(for School / Board Exams.)

MM : 70 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

SOLUTIONS

(5)


5. According to Law of radioactivity, the rate of decay is proportional to the number of nuclei N present at that instant.

dNN

dt : decay constant negative shows N is decreasing with time.

dNN

dt

SECTION-B

6. Applying Snell’s Law sin sin since , are small , 1d t t⎡ ⎤ ⎢ ⎥⎣ ⎦

11 1

td t

⎡ ⎤ ⎢ ⎥ ⎣ ⎦.

( – )

t

d

7. Quality factor: The selectivity or sharpness of a resonance circuit is measured in terms of quality factor. It is

defined as voltage magnification in the circuit at resonance.

Q0(Quality factor) =

Voltage across inductor at resonance.

applied voltage

Active power or real power is the power which is dissipated by the resistance in a. c. and reactive power is the

temporarily power stored in inductor and capacitor in a. c.

8. Primary rainbow: It is result of three step process

Refraction with dispersion

Internal reflection

Refraction

R

V

V R

Sunlight

(6)


Secondary rainbow: It is the result of four step process

Refraction with dispersion

Internal reflection

Internal reflection

Refraction

VR

Sunlight

9. 1 & 2 are in series and combination is parallel with 3. Therefore RE = R.

R2R

2R2R

2RR

R

x y

1

3

2

10. Half life: It is the time during which the number of undecayed nuclei reduces to 50% of its initial value or time

during which 50% of the nuclei have decayed.

12

Ln 2T

Mean Life: It is the time during which the number of undecayed nuclei reduces to 37% of its initial value.

Avg0

0

1T Ndt

N

∫

OR

1

1 0

tA A

�

2

2 0

tA A

�

2 11

2

t tA

A

�

1

2 1

2

LnA

t tA

2 1

1

2

1

Ln

t tT

A

A

(7)


SECTION-C

11. (i) It can be operated at 120 V by using step up transformer

V1 = 120, V

2 = 240

2 2

1 1

2402

120

N V

N V

Secondary coil must contain double turns that of primary.

(ii) P = V1I1

1

1

9608

120

PI A

V

(iii)

2

2

2 1

1

NR R

N

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

1

1

12015

8

VR

I

222 15 60R

12.1 2

1 2

1 11 1S t t⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

(Apparent shift)

1 18 1 4.5 1

4 33 2

S

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

S = 2 + 1.5 = 3.5 cm

Apparent depth = 8 + 4.5 – 3.5 = 9 cm

Effective 1 2

real depth 12.5R.I. = 1.39

app. depth 9

t t

t

13. Moving coil Galvanometer: When a current carrying coil is placed in a magnetic field, torque acts on it. In

moving coil galvanometer radial field is used which is obtained from magnet having concave shape poles.

Scale

SN

Pointer

Spring

Coil

Soft iron core

= NIAB = K [Where is angle formed by the pointer and C is restoring torque/ twist]

KI

NBA

⎡ ⎤⎢ ⎥⎣ ⎦

(8)


14. Wheatstone bridge: For balanced wheatstone bridge

I2 = I

4

I1 = I

3

Ig = 0

Applying KVL (ABDA) I2R2 + 0 – I

1R1 = 0

Applying KVL (BCDB) I4R4 – I

3R3 + 0 = 0

A

B

C

E

DI3

I2

I

I1

I4

I

R1

R4R2

R3

G

Ig

1 4

2 3

I R

I R and

1 2

2 1

I R

I R

2 4

1 3

R R

R R

Condition of balanced wheatstone bridge.

15. Considering small element ‘dx’ .V B dx x Bdx ��

2

0

1

2

l

E Bxdx Bl ∫

B

O

A

dx

x

The end A is positive and the end O is negative.

16. According to the classical theory of EMW, light is a varying electromagnetic field and the strength of the electric

and magnetic forces in this field increases, with increasing intensity. If the execution of electrons from metal

surface is due to the electric forces of the incident waves, then the energy with which the electrons are ejected

should increase with increasing light intensity. But in the photoelectric effect, the intensity of radiations affect only

the number of effected electrons and does not effect their velocity or energy.

Further, according to classical theory, the energy of light waves, so the amount of energy and electron could

absorb from one tiny spot on such a spreading wave front would be negligible. Hence classical picture of light

waves does not represent what actually happens.

17. Working of semi-conductor diode as full wave rectifier: During positive half cycle of the input sine wave, the

diode D1 is forward biased while diode D

2 is reverse biased, while vice-versa in the negative half cycle. Hence in

both time current flows via R0 in the same direction. Thus d.c. voltage shape is obtained in output. This process is

known as full wave rectification.

–

+

D1

D1

D2

R0

a. c. v

D2

D1

t

(9)


18. NOR gate: Is called universal gate as all the fundamental gates i.e., AND, OR, NOT can be obtained from the

combination of NOR gates.

x

y

z1

zz2

x y z1

z2 z

0011

0101

1100

1010

0001

Truth Table

Thus, AND gate can be obtained for NOR gates.

19. Young’s double slit experiment: Here parallel wave fronts from a distant monochromatic source are incident

normally on two closely spaced silts S1 & S

2. The slits are rectangular in shape with their length normal to the

plane of paper. The two slits behave as two coherent sources, with zero initial phase difference [ = 0]. Light waves

from the two slits spread out and fall on the screen S at a distance D from the slits. [D >> d]. Alternate bright and

dark bands are obtained on screen due to interference.

sin tann

y

D

2 1sin

ny

p S P S P d dD

since form nth maxima p = n

Ddsin

d

S1

S2

Screen

yn

P

ndy

nD

n

n DY

d

where [n = 0, 1, 2, .....]

Position of bright fringe.

for nth minima:

1

2p n

⎡ ⎤ ⎢ ⎥⎣ ⎦

1

2

ndy

nD

⎡ ⎤ ⎢ ⎥⎣ ⎦

1

2n

Dy n

d

⎡ ⎤ ⎢ ⎥⎣ ⎦

[n = 1, 2, 3, .....]

Position of dark fringe

At n = 0, central bright fringe is obtained.

20.3Li6 +

0n1

2He4 +

1H3

Mass of 3Li6 = 6.015126 amu

Mass of 1H3 = 3.016049 amu

Mass of 2He4 = 4.002604 amu

Mass of 0n1 = 1.008665 amu

Total mass of reactant = 7.023791 amu

Total mass of product = 7.01865 amu

Mass difference (m) = .005138 amu

Energy released = .005138 × 931 MeV = 4.783 MeV

(10)


21. Lens Maker’s formula

Let R1 and R

2 be radius of curvature.

For first surface ADB

2 1 2 1

V U R

2 1 2 1

1 1V U R

...(i)

C1C

2P

A

B

D

E

IOO1

G

H

1

1

[Image][Object]For second surface

1 2 1 2

1 2V V R

...(ii)

(i) + (ii)

2

1

1 1 2

1 1 1 11

V U R R

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1 2

1 11

R R

⎡ ⎤ ⎢ ⎥

⎣ ⎦

2

1

⎧ ⎫ ⎨ ⎬⎩ ⎭

According to equation

f; U = ; V = f

1 2

1 1 11

f R R

⎡ ⎤ ⎢ ⎥

⎣ ⎦Lens Maker’s formula

22. Essential elements of communication system:

InformationSource

Transmitter Channel Receiver

Userof

Information

Noise

Communicationsystem

(i) Transmitter: It processes the incoming message signal so as to make it suitable for transmission through a

channel and subsequent receiver.

(ii) Channel: It is the range provided by the government to the service provider. Noise insertion takes place during

this phenomenon.

(iii) Receiver: It extracts the desired message signals from the received signals at the channel output.

OR

Length of antenna should have a size comparable to the wavelength of the signals [at least 4

in size]

e.g., f = 15 KH3;

8

3

3 1020000 m

15 10

C

f

Length of antenna 20000

500 m4 4

This height of antenna is practically impossible therefore, there is an urgent need of converting the information low

frequency signal in to high frequencies before transmission.

(11)


23. If an object has to coincide with its image then the rays have to retrace its path. It happens when rays meet the

reflecting surface perpendicularly. That means object must be placed at the focus of the equivalent combination of

two lenses (glass lens + liquid lens).

1 1 1

g eF f f

1 1 1 11.5 1

gf R R R

⎡ ⎤ ⎢ ⎥⎣ ⎦

11 1 11

ef R R

⎡ ⎤ ⎢ ⎥⎣ ⎦

D

O I

Liquid

Plane mirror

11 1 2

F R R R

since F = D

2R

D

SECTION-D

24. Van de Graaff generator

Supply circuit

Motorrotating belt

Chargetaken on belt

Insulator++++++

Hollow metal sphere

+++

++ + + + +

+++++

An insulated belt continuously brings charge to the inside of hollow conductor which then moves to the outside

surface of the conductor. The electric potential on the spherical conducting surface increases as charge flows to

its surface 0

4

qV

R

. An ion source produces charged atoms whose sign is such as to be repelled from the

region of high potential and thus accelerated. Such devices are called Van de Graaff generator.

OR

The cyclotron is a machine to accelerate charged particles or ions to high energies. The cyclotron uses both

electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpen-

dicular to each other they are called crossed fields. Cyclotron used the fact that the frequency of revolution of the

charged particle in a magnetic field is independent of its energy and speed. The particles move most of the time

inside two semicircular disc like metal containers, D1 and D

2, which are called dees as they look like the letter D.

(12)


Figure shows a schematic view of the cyclotron. Inside the metal boxes the particle is shielded and is not acted

on by the electric field. The magnetic field, however, acts on the particle and makes it go round in a circular path

inside a dee. Every time the particle moves from one dee to another it is acted upon by the electric field. The sign

of the electric field is changed alternately in tune with the circular motion of the particle. This ensures that the

particle is always accelerated by the electric field. Each time the charge enters the gap, the electric field

increases the energy of the particle. As energy increases, the radius of the circular path increases. So the path is

a spiral one. Two whole assembly is evacuated to minimize collisions between the ions and the air molecules. A

high frequency alternating voltage is applied to the dees. In the sketch shown in figure, positive ions or positively

charged particles (e.g., protons) are released at the centre P.

Charged Particle

D2

Exit port

Deflection plate

Magnetic field outof the paper

D1

P

OSCILLATOR

They move in a semicircular path in one of the dees and arrive in the gap between the dees in a time interval

T/2; where T is the period of revolution, given by

1 2

c

mT

f qB

2c

qBf

m

This frequency is called the cyclotron frequency for obvious reasons and is denoted by f

c.

The frequency fa of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that

it takes the ions to complete one half of the revolution. The requirement fa = f

c is called the resonance condition.

The phase of the supply is adjusted so that when the positive ions arrive at the edge of D1, then D

2 is at a lower

potential and the ions are accelerated across the gap. Inside the dees the particles travel in a region free from the

electric field. The increase in their kinetic energy is qV each time they move from one dee to another (V refers to

the voltage across the dees at that time). As radius is directly proportional to speed, the radius of their path goes

on increasing each time their kinetic energy increases. The ions are repeatedly accelerated across the dees until

they have the required energy to have a radius approximately that of the dees. They are then deflected by a

magnetic field and leave the system via an exit slit. We have

;qBR

vm

where R is radius of the trajectory at exit and equals the radius of a dee.

Hence, kinetic energy of the ions is,

2 2 2

21

2 2

q B Rmv

m

The operation of a cyclotron is based on the fact that the time for one revolution of an ion is independent of its

speed or radius of its orbit. The cyclotron is used to bombard nuclei with energetic particles, (accelerated by it),

and study the resulting nuclear reactions. It is also used to implant ions into solid and modify their properties or

even synthesis new materials. It is used in hospitals to produce radioactive substances which can be used in

diagnosis and treatment.

(13)


25. VA = 10 V

VC = 0

Applying KCL at B

(10 – VB)6 + (V

D – V

B)3 + (0 – V

B)3 = 0

12VB – 3V

D = 66 ...(i)

3 F

3 F

3 F6 F

6 FCA

D

B

1 2

3 4

10 V

Applying KCL at D

(10 – VD)3 + (V

B – V

D)3 + (0 – V

D)6 = 0

–3VB + 12V

D = 36 ...(ii)

After solving (i) and (ii)

VB = 6 V; V

D = 4 V

q1 = (10 – 6)6 = 24 C

q2 = (6 – 0)3 = 18 C

q3 = (10 – 4)3 = 18 C

q4 = (4 – 0)6 = 24 C

OR

Let q2 and q

3 be the charges on the 2F and 3F capacitors, then

(a)3 2 3 02

2 3 2 3 5

q q q qq

q0 is the initial charge on the 2F capacitor.

q0 = CV = 2 × 100 = 200 C

Thus 2

200 280 C

5q

3 F2 F+

–100 V

3

200 3120 C

5q

(b) Fraction ‘f’ lost = loss 2

1 2I

U C

U C C

3.6

2 3f

26. Diamagnetic material

(a) They have tendency to move from stronger to weaker part of external magnetic field.

(b) Magnetic field inside material is less than free space.

1R

(c) 1 ;R m m

is negative

(d) m

is independent of temperature

Paramagnetic substance

(a) These are feebly magnetised in the direction of external field.

(b) Magnetic field lines tend to pass through these substances.

(14)


(c) 0 0

0

; 1,B

B BB

0

1; 1R

(d) 1 ;R m m

is positive

(e)1

Tempm

T

m

Ferro-magnetic

(a) They get strongly magnetized in the direction of external field

(b) R >> 1

(c) m is very large

(d) m

C

C

T T

; TC: critical temp.

T

m

OR

(a) Intensity of magnetising field (H): The ratio of magnetising field to the permeability of free space.

0

0

BH

��

��

unit: L–1 A (A/m)

(b) Intensity of magnetization (I): When a magnetic material is kept inside a magnetizing field, the dipole

moment is induced. This induced dipole moment per unit volume.

MI

V��

�

unit: A/m

(c) Magnetic susceptibility (m): It is the ratio magnitude of intensity of magnetization to that of magnetizing

field (H).

m

I

H Unitless and dimensionless

(d) Magnetic permeability (): Ratio of total magnetic field (B) inside a material to that of magnitude of mag-

netic intensity.

B

H

⎡ ⎤ ⎢ ⎥⎣ ⎦

unit: b/A-m

(e) Relative permeability (r): Ratio of permeability of a medium to that of permeability of free space.

0

R

⎡ ⎤ ⎢ ⎥⎣ ⎦ dimensionless / unitless.

� � �

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PHYSICS Complete Syllabus of Class XII