BLUE PRINT FOR MODEL QUESTION PAPER – 1

SUBJECT : PHYSICS (33) CLASS : II PUC

Blue Print For Model Question Paper – II PUC-PHYSICS

Topic

10

8

8

3+

3

5

Un

it

Ch

ap

ter

Tea

chin

g H

ou

rs

Mark

s a

llott

ed

1 m

ark

(V

SA

)

2 m

ark

(S

A1

)

3 m

ark

(S

A2

)

5 m

ark

(L

A)

5 m

ark

(N

P)

1 1 Electric Charges and

Fields 9 8

2 2 Electrostatic Potential

and Capacitance 9 8

3 3 Current Electricity 15 13

4 4 Moving Charges and

Magnetism 10 8

5

5 Magnetism and

Matter 8 7

6 Electromagnetic

Induction 7 6

6

7 Alternating Current 8 8

8 Electromagnetic

Waves 2 2

7 9 Ray Optics and

Optical Instruments 9 8

8 10 Wave Optics 9 8

9

11 Dual nature of

Radiation And Matter 6 5

12 Atoms 5 5

10

13 Nuclei 7 6

14 Semiconductor

Electronics 12 10

15 Communication

Systems 4 3

TOTAL 120 105 10 16 24 30 24

MODEL QUESTION PAPER – 1

II P.U.C. PHYSICS (33)

Time: 3 hours 15 min. Max. Marks: 70

General instructions:

a) All parts are compulsory.

b) Answers without relevant diagram/ figure/circuit wherever necessary will not carry any marks.

c) Direct answers to the Numerical problems without detailed solutions will not carry any marks.

PART - A

I. Answer the following 10 x 1 = 10 1. Define the SI unit of charge.

2. Write the colour sequence of the resistance 4.7kΩ 5%. 3. Define dip at a place. 4. Give an example for paramagnetic substance. 5. State Lenz’s law. 6. What is wattles current? 7. Name the experiment which confirms wave nature of electrons. 8. In which region of electromagnetic spectrum does Lyman series of hydrogen spectrum lies? 9. What are isotones? 10. Mention the band width for speech signals?

PART – B

II. Answer any FIVE of the following questions. 5×2=10

11. Write the expression for torque on an electric dipole in uniform electric field and explain the terms.

12. Define drift velocity and mobility

13. Mention the applications of potentiometer.

14. What are permanent magnets and electromagnets?

15. Give any two sources of energy losses in transformer.

16. Mention the expression for Ampere-Maxwell law and explain the terms.

17. Write the logic symbol and truth table of NAND gate.

18. Draw the block diagram of AM transmitter.

PART – C

III. Answer any FIVE of the following questions. 5×3=15

19. What is an equipotential surface? Mention any two of its properties.

20. Derive j E , equivalent form of Ohms law.

21. Mention the expression for force on a moving charge in uniform magnetic field.

When does the force become maxim and minimum?

22. Give any three properties of magnetic field lines.

23. Draw a neat labeled diagram of compound microscope when the final image is at least distance of

distinct vision.

24. Mention any three differences between interference and diffraction.

25. Give any three experimental observations of photoelectric effect.

26. Explain the working of Zener diode as a voltage regulator.

PART – D

IV. Answer any TWO of the following questions. 2×5=10

27. Derive an expression for effective capacitance of two capacitors connected in series.

28. Derive an expression for equivalent emf and equivalent internal resistance of two cells connected in

parallel.

29. Draw a neat labeled of ac generator. Obtain the expression for instantaneous emf.

V. Answer any TWO of the following questions. 2×5=10

30. Derive lens makers formula.

31. Derive an expression for total energy of an electron in a stationary state of hydrogen atom.

32. What is the rectifier? Describe the working of full wave rectifier with circuit diagram and waveform.

VI. Answer any THREE of the following questions. 3×5=15

33. Two point charges 2nC and -2nC are placed at two corners of equilateral triangle of side 5cm. Find the

magnitude of the resultant electric field at the third corner of the triangle.

34. A circular coil of radius 0.1m of number of turns 100, carries a current of 1.2A. Find the magnetic field

a) at the centre of coil, b) at a point distance 0.2m from centre on the axis.

35. A resistance of 100Ω, inductance of 0.2H and capacitance of 10µF are connected in series across an ac

source of 220V, 50Hz. find the average current in the circuit.

36. In Young’s double slit experiment a light of wavelength 5000Ao is used to incident on two slits. If the

separation between the slits is 2mm and distance between slits and screen is 1.5m, find a) distance

of 5th bright fringe from the centre of fringe pattern. b) distance of 4th dark fringe from the

centre of fringe pattern.

37. Find the binding energy per nucleon of the nuclei 6C12 and 8O16 having rest masses 12.0000u

and 16.00744u respectively. (given mP =1.00727u and mn = 1.00866u)

SCHEME OF EVALUATION

SCHEME OF EVALUATION FOR PHYSICS MODEL QUESTION PAPER – 2

Q. No ANSWERS Marks

PART – A 1. One coulomb of charge is the charge that when placed away from a similar charge at a distance

1m in vacuum exerts a force 9x109 N.

1

2. Yellow – violet – red – gold 1

3. The angle between the direction of the magnetic field at a place and the horizontal at the same

place.

1

4. Chromium 1

5. The induced emf in an EMI opposes the creator of itself. 1

6. In a pure inductor or a pure capacitor when current flows no power is dissipated, this current is

called as wattles current.

1

7. Davison and Germer experiment. 1

8. Ultraviolet region 1

9. The atoms having the same number of neutrons. 1

10. Band width of speech signals 2800 Hz. 1

II. PART – B

11. = p E sin

Labeling

1+1

12. The average velocity with which the electrons drift opposite to the direction of applied electric

field is known as drift velocity.

The drift velocity acquired per unit electric is known as mobility

1

1

13. To compare the emf’s of two cells and to determine the internal resistance of a cell. 1+1

14. Magnets which have permanent magnetic moment are known as permanent magnets. Magnets

which gets magnetic moment on application of electric field is known as electromagnets

1

1

15. Eddy currents and hysteresis. 1+1

16. ∮

and Explanation of the terms 1 + 1

17. Logic symbol and Truth table:

Input Output

A B BAY

0 0 1

1 0 1

0 1 1

1 1 0

1+1

18.

2

III PART – C

19. A surface in which all the point on it are at same electric potential is called equipotential surface.

The electric field must, be normal (perpendicular) to the equipotential surface at every point.

Any equipotential surface, there is no potential difference (p.d. =0) between any two points on

the surface and no work is required to move a test charge on the surface (W=0).

1

1

1

20. I = neA vd and vd =

E =

Arriving at j = E

1

1

1

21. Force: F = qv B sin

Case-1: = 90 F = maximum

Case-2: = 0 F = 0.

1

1

1

22. Any three properties 1 each

23. Diagram of Compound microscope, 3

24. Any three differences 1 each

25. Any three observations 1 each

26.

WORKING: The unregulated (fluctuating) dc voltage (filtered output of a rectifier) is

connected to the reverse biased Zener diode through a series resistance Rs. Regulated

output is taken across the load resistor RL .

Any increases or decreases in the input voltage results in the increase or decrease in the

voltage drop across the resistance SR but the voltage across the Zener diode remains

constant. Therefore a constant output voltage acts across the load .RL

1

1

1

IV. PART – D

27. Figure shows capacitors C1 and C2 combined in series.

In the series combination, charges on the two plates (± Q) are

the same on each capacitor. The total potential drop V across

the combination is the sum of the potential drops V1 and V2

across C1 and C2, respectively.

Then V = V1 + V2 = 1 2 1 2

Q Q 1 1 + =Q +

C C C C

;

( since Q = C1 V1 = C2 V2 V1 = Q /C1 and V2 = Q /C2 )

The combination has an effective capacitance with charge Q & potential difference V.

The effective capacitance of the combination is C = Q/V V = Q/C

We compare Eq. (7) with Eq. (6), we obtain 1 2

Q 1 1 = Q +

C C C

1 2

1 1 1 = +

C C C

1

1

1

1

1

28. CIRCUIT DIAGRAM for two cells.

The potential difference across the first cell: 1 1 1V = ε r I

Similarly, p.d. across second cell: 2 2 2

V = ε r I

1 2

1 2

1 2

ε V ε V= & =

r rI I

But I = I1 + I2

I = 1 2

1 2 1 2

ε ε 1 1+ V

r r r r

….. (1) and P.d. across

equivalent cell:eq

eq eq

eq eq

ε VV = ε r

r r I I = ..... (2)

Comparing (1) and (2), Equivalent emf:

eq eq1 2 1 2

eq 1 2 eq 1 2 n

ε εε ε ε ε ε = + and for n cells = + +r r r r r r r

n............

Equivalent internal resistance of the combination:

1 2

eq

1 2 eq 1 2 n

r r 1 1 1 1= and for n cells : = + +

r r r r r rr

........

1

1

1

1

1

29. Labelled diagram

DERIVATION OF EXPRESSION FOR SINUSOIDAL EMF:

Consider a rectangular coil of N turns and area A.

The coil is free to rotate about an axis passing through its

centre and perpendicular to the direction of the magnetic

field.

Let the coil be rotated in a uniform magnetic field B.

Let the coil is rotated with a constant angular speed ω, the

angle θ between the magnetic field vector and the area

vector of the coil at any instant t is θ = ωt

(assuming θ = 0o at t = 0).

Due to rotation, the effective area of the coil exposed to the magnetic field lines changes

with time, and the magnetic flux at any time t is θ ω

From Faraday’s law, the induced emf in the rotating coil of N turns,

(

)

( ω ) ω ω , since

( ω ) ω ω ,

Thus the instantaneous value of the emf ;

where 0 = NBAω is the maximum value of the emf.

1

1

1

1

1

IV.

30. RAY DIAGRAMS:

The image formation can be seen in terms of two

steps:

The first refracting surface forms the image I1 of the

object O.

The image I1 acts as a virtual object for the second

surface that forms the image at I.

Applying refraction at spherical surface to the first

interface ABC,

1

we get, 1 2 2 1

1 1

n n n n + =

OB BI BC

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

Applying refraction at spherical surface to the 2nd

interface ADC gives, 2 1 2 1

1 2

n n n n + =

DI DI DC

….....(2)

For a thin lens, BI1 = DI1.

Adding Equations (1) and (2),

To get 1 12 1

1 2

n n 1 1 + = n n

OB DI BC DC

……(3)

Suppose the object is at infinity, i.e., OB → ∞ and DI = f,

Equation (3) gives, 12 1

1 2

n 1 1 = n n

f BC DC

….…(4)

By the sign convention,BC1 = + R1 and DC2= – R2

Eqn.(4) becomes, 2

1 1 2

n1 1 1 = 1

f n R R

2

21

1

n Using n =

n

Lens Makers formula: 21

1 2

1 1 1 = n 1

f R R

1

1

1

1

31. Consider an electron revolving around the nucleus of hydrogen atom. The electrostatic force

of attraction, Fe between the revolving electron and the nucleus provides the requisite

centripetal force (Fc) to keep them in their orbits. Thus, for a dynamically stable orbit in a

hydrogen atom. Fc = Fe 2 2

20

mv 1 e =

r 4πε r

22

0

1 emv =

4πε r.

The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom

are 2 2 2

2

0 0 0 0

1 1 1 e e 1 (e)( ) eK = mv = = and U

2 2 4πε r 8πε r 4πε r 4πε r

e

Thus the total energy E of the electron in a hydrogen atom is

E = K + U = 2 2

0 0

e e

8πε r 4πε r =

2

0

e

8πε r

The total energy of the electron in the stationary states of the hydrogen atom can be obtained

by substituting the value of orbital radius from, 2 2

0n 2

ε n hr

π m e

Total energy: 2

n 2 20 0

2

e 1E =

8 π ε ε n h

π m e

4

n 2 2 20

m eE =

8 n h ε

1

1

1

1

1

32. Circuit diagram

Waveform

Rectifier is a circuit which

converts ac in to dc.

Working: During the positive half cycle of the input ac signal, the diode D1 is forward biased

and hence it conducts, the diode D2 is reverse biased and hence it does not conduct.

During the negative half cycle of the input AC signal, the diode D1 is reverse biased and hence

it does not conduct, the diode D2 is forward biased and hence it does conduct.

Hence the output is as shown in the diagram.

1

1

1

1

1

VI.

33. Diagram

Substitution

Arriving at 0.8446

Writing the unit (N/C)

1

1

1

1

1

34. Formula B =

( )

⁄

Substitution

Answer

B = 0.754 mT

1

1

1

1

1

35. Z = √ ( )

Substitution

Z = 274.53

I = V/Z

I = 220/274.53= 0.801 ampere

1

1

1

1

1

36.

Substitution

( )

31 mm

1

1

1

1

1

37. Mass of the constituents of carbon nucleus mC =6 x mn + 6 x mp = 12.09558 u

Mass defect in carbon nucleus: 0.09558u

BE of carbon nucleus: Eb = 0.09558 x 931.5 MeV = 89.033 MeV

Binding energy per nucleon: Ebn = 89.033/12 = 7.42 MeV

NOTE: Any other alternate correct method should be considered.

1

1

1 + 1

1