SOLVED Karnataka II PUC PAPER April 2020

SOLVEDPAPER

With Scheme of Valuation

Karnataka II PUCApril 2020Class-XII

PHYSICSCode-33 (NS)

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

Instructions : 1. All parts are compulsory. 2. Answer without relevant diagram/figure/circuit wherever necessary will not carry any marks. 3. Numerical problems solved without writing the relevant formulae carry no marks.

Part-A

I. Answer all the following questions : (10 × 1 = 10)

1. Write the SI unit of electric flux. 2. Graphically represent the variation of resistivity of a semiconductor with absolute temperature. 3. Give any one use of electromagnet. 4. What is the significance of Lenz's law? 5. How does capacitive reactance vary with frequency? 6. Arrange the following electromagnetic waves in ascending order of their wavelength: Radio waves, Gamma rays, Infrared waves, X-rays. 7. Why does sky appear blue? 8. Mention a method to increase the resolving power of a microscope.

9. Write the nuclear reaction equation for alpha decay of 92238U .

10. Draw the logic symbol of NOR gate.

Part-B

II. Answer any five of the following questions : (5 × 2 = 10)

11. State and explain Coulomb's law in electrostatics.

12. A parallel plate capacitor with air between the plates has a capacitance C. What will be the capacitance if

(a) the distance between the plates is doubled?

(b) the space between the plates is filled with a substance of dielectric constant 5?

13. Write two limitations of Ohm's law.

14. In a region, an electric field −= × 3 1ˆ5 10E j NC and a magnetic field of =

ˆ0.1B kT are applied. A beam of charged particles are projected along X-direction.

Find the velocity of charged particles which move undeflected in this crossed fields. 15. Define "retentivity" and "coercivity". 16. Mention two sources of energy loss in transformer. 17. What is displacement current? Give the expression for it.

18. An alpha-particle, a proton and an electron are moving with equal kinetic energy. Which one of these particles has the longest de Broglie wavelength? Give reason.

Part-C

III. Answer any five of the following questions : (5 × 3 = 15)

19. Establish the relation between electric field and electric potential. 20. Derive the expression for the energy stored in a charged capacitor. 21. Give the principle of cyclotron and draw the neat labelled schematic diagram of cyclotron.

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2 ] Oswaal Karnataka PUE, Solved Paper - 2020, Physics, II PUC


22. Mention three properties of diamagnetic materials. 23. Arrive at the relation between focal length and radius of curvature of a spherical concave mirror. 24. Using Huygen's principle, show that the angle of incidence is equal to the angle of reflection, when a plane wave

front is reflected by a plane surface. 25. Define work function. Write Einstein's photoelectric equation and explain the terms. 26. Give three differences between intrinsic and extrinsic semiconductors.

Part-D

IV. Answer any two of the following questions : (2 × 5 = 10)

27. Derive the expression for conductivity of a material : σ τ= nem

2 where the terms have their usual meaning.

28. Obtain the expression for the force between two straight long parallel conductors carrying current. Hence, define "ampere".

29. With the help of a labelled diagram, derive the expression for instantaneous emf induced in an AC generator.

V. Answer any two of the following questions : (2 × 5 = 10)

30. Obtain the expression for the fringe width of interference fringes in Young's Double slit experiment. 31. Using Bohr's postulates, derive the expression for the radius of nth stationary orbit of electron in hydrogen atom.

Hence, write the expression for Bohr radius. 32. What is rectification? Explain the working of a p-n junction diode as a half-wave rectifier. Draw the input and

output wave forms.

VI. Answer any three of the following questions : (3 × 5 = 15)

33. Two point charges qA = 5 µC and qB = –5 µC are located at A and B separated by 0.2 m in vacuum. (a) What is the electric field at the midpoint O of the line joining the charges? (b) If a negative test charge of magnitude 2 nC is placed at O, what is the force experienced by the test charge? 34. (a) Three resistors 3 W, 4 W and 12 W are connected in parallel. What is the effective resistance of the combination? (b) If the combination is connected to a battery of emf 6 V and internal resistance 0.5 W, find the current drawn

from the battery and terminal potential difference across the battery. 35. A series LCR circuit contains a pure inductor of inductance 5.0 H, a capacitor of capacitance 20 µF and a resistor of

resistance 40 W. (a) Find the resonant frequency of the circuit (b) Calculate the Quaity factor (Q-factor) of the circuit. (c) What is the impedance at resonant condition? 36. At what angle should a ray of light be incident on the face of an equilateral prism, so that it just suffers total

internal reflection at the other face? The refractive index of the material of the prism is 1.5.

37. A copper coin has a mass of 63.0 g. Calculate the nuclear energy that would be required to separate all the neutrons

and protons from each other. The coin is entirely made of 2963Cu atoms.

Mass of 2963Cu atom = 62.92960 u

Mass of proton = 1.00727 u Mass of neutron = 1.00866 u Avogadro's number = 6.022 × 1023.


Oswaal Karnataka PUE, Solved Paper - 2020, Physics, II PUC [ 3


SOLUTIONSAs Per Scheme of Valuation

(Issued by Department of PUE, Karnataka)

Part-A

1. NC–1 m2 OR V m OR J C–1 m 1

2.

T

� 1

3. In electric bells / In loudspeakers (speakers)/ In telephone diaphragms/ In cranes/ In motors/ In generators/ In MRI machines/ In hard disks.

(Any one) 1 4. •Conservationofenergy 1 •Itgivespolarityofinducedemf •Itgivesdirectionofinducedcurrent. (Any one) 5. Capacitive reactance varies inversely as the

frequency. X XC C∝ ∝

1 1υ ωOR 1

OR When frequency increases capacitive reactance decreases and vice versa.

6. Gamma rays, X-rays, Infrared waves, Radio waves.1

7. Due to scattering of light OR Rayleigh scattering. 1 8. • Using a medium of higher refractive index

between object and objective. • Using light of shorter wavelength. (Any one) 1

9. 238 234 4

92 90 2U Th + He→

OR

92238

90234

24U X+ He→

OR

92238 -decay

90234U Thα → 1

10. A

B

Y 1

Scheme of Valuation, 2020

Detailed Answer :

10.

C = A + B

A

B

C

Part-B

11. Statement : The electrostatic (or electric) force between two point charges is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between them. 1

Explanation: Let q1, q2 be the two point charges separated by a distance r in vacuum, then magnitude of electrostatic force (F) between the charges is

F kq qr

= | |1 22 ; where k = 1

4 0πε is a constant. 1

12. Capacitance: CK Ad

Cd

= ⇒ ∝ε0 1 and C ∝ K

(a) C/2 OR capacitance is halved. 1 (b) 5C OR capacitance becomes 5 times the initial

values. 1Scheme of Valuation, 2020

Detailed Answer :

e

= 0K AC

d

d = distance between the plates

A = area of the plates.

K = dielectric constant

(a) e= =0 ( 1 for air)

AC K

d If the distance between the plates is doubled, then

e′ = =0 1

2 2A

C Cd

So, the capacitance will be halved.

(b) If dielectric of dielectric constant 5 is inserted then

05 5A

C Cd

e′ = × =

13. • Ohm’s law fails, if the current varies non-linearly with voltage (potential difference). 1

• Ohm’s law fails, if the relation between voltage and current is not unique (In GaAs). 1

• Ohm’s law is not applicable at very low temperature and very high temperature.

• Ohm’s law fails, if the relation between voltage (V) and current (I) depends on the sign of voltage. OR Ohm’s law is not applicable to




non-ohmic devices (In diode/ triode etc.)(Any two)

14. Velocity: vEB

= 1

v = × = × −5 100 1

5 103

4 1

.ms 1


Detailed Answer :

Charged particles move undeflected in crossed electric and magnetic field, if

eE = evB

where, E = electric field, B = magnetic field, v = velocity

e = charge of the charged particles

∴ v = EB

v = 5 × 103 / 0.1 = 5 × 104 m s–1

15. • The value of magnetic field B when magnetic intensity H = 0 is called retentivity. 1

• The value of magnetic intensity H when magnetic field B = 0 is called coercivity. 1

(Any other correct definition/s should be considered)


Detailed Answer :

Retentivity: The ability of a substance to retain magnetization after removal of the inducing magnetic field.

Coercivity: Coercivity is defined as the ability of a ferromagnetic material to withstand an external magnetic field without becoming demagnetized.

16. • Loss due to flux leakage/Magnetic loss. 1 • Ohmic loss (heating) due to the resistance of

the windings (wires)/Copper loss. 1 • Loss due to eddy current. • Loss due to hysteresis/Iron loss. (Any two) 17. The current due to changing electric field/flux is

called displacement current. 1

Displacement current: iddtdE=

ε φ0 1

18. de Broglie wavelength: λ = hmK2

⇒ ∝λ 1m

for equal kinetic energy

Electron has the longest de Broglie wavelength because it has least mass. 1+1

Part-C

19.

Equipotentials

A

VB

P

E

V + V�

�l

1 Work done to move unit positive charge from P to A

is W = F dl For unit positive charge, q = 1C, force F = E W = E dl 1 Also, work done W = q(VA – VB) = [V–(V+dV)] = –dV Thus, Edl = –dV

∴ E = − δδVl

1

(Any other correct diagram/method should be considered)


Detailed Answer :

Relation between electric field and electric potential.

Say, A and B are two point in an uniform electric field. Potential at point B is greater than the potential at point A.

B A

+Q0

E

If +Q0 amount of charged is moved between the points and WAB is the work done, then work done per unit charge

= WAB / Q0 = Potential difference between the points = (VB – VA) ...(i)

A charge +Q0 in an electric field E will experience a force Q0E in the direction of the field.

An external agent must apply same amount of force to move the charge from A to B.

∴ FEXT = Q0E Work done = WAB = FEXT × AB ∴ WAB = Q0E × AB ...(ii)

Comparing equation (i) and (ii)

Q0E × AB = (VB – VA)Q0

or, E = V VAB

B A−( )




The above results shows that if the potential difference between two plates separated by a distance l, then assuming an uniform electric field E between the two

plates, EVl

= .

20. Let C be the capacitance of the capacitor. At intermediate situation, the plate-1 and plate-2

have charges Q’ and –Q’ respectively.

The potential difference between plates is VQC

''= 1

Work done to move a charge dQ’ from plate-2 to

plate-1 is δ δ δW V QQC

Q= =

' ''

'

The total work done in building the charge from

Q’ = 0 to Q’ = Q is WQCdQ

Q

= ∫'

'0

1

WQC

=2

2

This work done is stored as potential energy in the

capacitor: UQC

=2

2 1

(Any other correct method should be considered)

21. PRINCIPLE : 1 • The frequency/time period of revolution of

a charged particle in the magnetic field is independent of its velocity/ energy/ radius. 1

• The charged particle is made to move in the perpendicular magnetic and electric fields (crossed fields), the electric field accelerates the charged particle and the magnetic field deflects the charged particle. (Any one)

Magnetic field outof the paper Deflection plate

Chargedparticle

D1D2

Exit Port

OSCILLATOR

2

22. • Diamagnetic materials are repelled by a magnet. 1

OR They tend to move from a region of stronger magnetic frield to the weaker magnetic field.

• Magnetic susceptibility of the substance is negative (c < 0). 1

OR Relative permemability of is less than 1 (µr < 1) OR Permeability of the substance µ < µ0. 1

• In a diamagnetic material, the magnetic dipole moment of an atom is zero. 1

OR Atoms of diamagnetic material have no un-paired electrons.

• They develop a net magnetic moment in a direction opposite to the applied field.

OR They are magnetised in a direction opposite to the applied field.

• The magnetic field lines are repelled/expelled/flow out of the material when place in an external field.

• The susceptibility/permeability/magnetization is independent of temperature.

OR They do not obey Curie’s law. 23.

F D

��

�C P

2�

M

R

(Any three)1

C – Centre of curvature F – Principal Focus R – Radius of curvature f – focal length of the mirror ∠MCP = q and ∠MFP = 2q

Now, tanθ = MDCD

and tan2θ = MDFD

For small q, tan q ≈ q and tan2q ≈ 2q

Then, MDCD

= θ and MDFD

= 2θ 1

i.e., MDFD

MDCD

FDCD=

⇒ =22

For small q, the point D is very close to P. Applying sign conventions, FD = –f and CD = –R

Thus, f = R2

1




Note : Full marks should be awarded for substituting, FD = +f and CD = +R also.

24.

MA

N

Incidentwavefront

Reflectedwavefront

B

C

i

i r

E

1 MN → Reflecting plane surface (mirror) i → angle of incidence r → angle of reflection v → speed of light wave Let t be the time taken by the wave front to travel

from the point B to C, then distance BC = vt. In order the construct the reflected wave front,

a sphere of radius = vt, is drawn from the point A as shown in the above figure. Let CE represent the tangent plane drawn from the point C to this sphere. Then AE = BC = vt, ABC = CEA = 90o and AC is common. 1

The triangle ABC and CEA are congruent. Thus i = r1


Detailed Answer : AB is the wavefront incident on a reflecting surface

XY with an angle of incidence ∠i. According to Huygen’s principle, every point on AB acts as a source of secondary wavelets.

I

XA D E

Y

R

BN

ri

i r

irC1

D1

N1A1

At first, wave incidents at point A and then to points C, D and E.

A1E represents the tangential envelope of the secondary wavelets in forward direction.

InΔABE andΔAA1E,

∠ABE = ∠AA1E = 90°

Also, side AE is common.

AA1 = BE = distance travelled by wave in same time

So, these triangles are congruent.

So, ∠BAE = ∠A1EA

So, ∠i = ∠r

Hence, angle of incidence = angle of reflection. 25. Work function: The minimum energy required to

remove an electron from the metal surface. 1 Einstein’s photoelectric equation: Kmax = hn – f0

OR Kmax = hn – hn0 OR eVo = hn – f0 1 Kmax – Maximum kinetic energy of photoelectrons. n – Frequency of incident light. 1 h – Planck’s constant. f0 – Work function of the metal.

(Any other equivalent correct equation should be considered)

26.

Intrinsic semiconductor

Extrinsic semiconductor

1. Semiconductor without doping.(It is pure semiconductor).

Semiconductor doped with impurity atoms.(It is impure semiconductor).

1

2. Conductivity is low. Conductivity is high. 1

3. Number of free electrons (ne) is equal to the number of holes (nh).OR ne = nh

Number of free electrons (ne) and number of holes (nh) are unequal.OR ne ≠ nh

1

4. Conductivity depends only on temperature.

Conductivity depends both on temperature and impurity added.

5. Conductivity due to electrons and holes is nearly equal.

Conductivity is mainly due to majority charge carriers.

6. It does not conduct at zero kelvin.

It conducts at zero kelvin.

(Any three) 3

Part-D

27.

� �x v t=d

�

1

Volume = A(Dx) = A vd(Dt)

Let ‘n’ be the number of free electrons per unit volume (number density of free electrons) in the material, then there are n vd(Dt)A electrons; A-area of cross section and vd-drift velocity.

The amount of charge crossing the area A to the left time Dt is,

I (Dt) = neAvd(Dt) 1

Current : I = neAvd

Magnitude of drift velocity of electrons is

vd = emτ

E ...(1) 1




e – magnitude of electron charge t – relaxation time

E – electric field m – mass of electron

Substituting vd from eq. (1)

I = neA em

Ene Am

Eτ τ

=

2

Thus, current density; j = IA

nem

E=

2τ ...(2) 1

But j = sE ...(3) 1

Thus, from eq. (2) and eq. (3), conductivity: στ

=nem

2

1

28.

Ia

Ib

L

d

ba

�abF

�baF

�aB

�bB

1

The magnetic field at the location of ‘b’ produced by

the conductor ‘a’ is Ba = 02π

aµ Id

....(1) 1

The magnetic force on a segment L of the conductor ‘b’ due to ‘a’ is Fba = Ba Ib L sin 90o

Using eq. (1), Fba = 02π

a bµ I I Ld

. This force Fba is towards

conductor ‘a’. 1 Similarly, we can find the magnetic force Fab on a

segment L of the conductor ’a’ due to ‘b’. The force Fab is equal in magnitude to Fba, and directed

towards ‘b’. Thus = −

ba abF F . 1

Definition of ampere: One ampere is that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section and placed one metre apart in vacuum (free space), would produce a force of 2 × 10–7 newton per metre of length on each other. 1

(Note : Any other equivalent correct definition with required key terms should be considered)

29.

Coil

SN

Alternating emf

�Axle

1

Let the coil having N turns be rotated with a constant angular speed w, the angle q between the magnetic field

B and the area vector

A of the coil at any instant of time t is q = wt

The magnetic flux at any time t is

ΦB B A BA BA t= = =

. cos cosθ ω 1

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

e = −

NddtBΦ

1

= –NBAddt

(cos wt)

= NBAw sin wt 1

Thus, the instantaneous value of the emf e = e0 sin wt 1

where e0 = NBAw is the amplitude (peak/maximum value) of the emf.

30.

F

O

E

P

d

2

� ��

d

2

� ��

D

x

d

S2

S1

Double slits screen

2

� �� d

x

2

� �� d

x

1

1

S1 & S2 → Slits

d → Slit separation

D → Distance of screen from the plane of slits

From the figure,

S2P2 – S1P2 = D xd

D xd

xd22

22

2 22+ +

− + −

=

1

⇒ − =

+S P S P

xdS P S P2 12 1

2

Since P is very close to O and d << D, (S2P + S1P) ≈ 2D

∴ Path difference: S2P – S1P = 22xdD

xdD

= ...(1)

Let l be the wavelength of light, for a bright fringe at P.

Path difference: S2P – S1P = nl ...(2) 1

Thus, from (1) and (2), xdD

n= λ




⇒ For nth bright fringe: x nDdn =

λ 1

For (n+1)th bright fringe: xn+1=(n+1)x nDdn =

λ

Fringe width : b = xn+1 – xn = λDd

1

Note : Alternate method using sinq and tanq should be considered.

31. Considered an electron revolving around the nucleus of hydrogen atom. The electrostatic force of attraction, between the revolving electron and the nucleus provides the required centripetal force to keep the electron in its orbits.

i.e., mvr

2 =

14 0

2

2πεer

1

⇒ mv2r = e2

04πε

For an electron in the nth orbit,

mvn2rn =

e2

04πε ...(1)

From Bohr’s II postulate, angular momentum:

Ln = mvnrn = 2πnh

1

Squaring, 2 2 2n nm v r =

2 2

24π

n h ...(2)

EqEq

21

( )( ) ⇒ m rn =

n he

2 2

20

244

ππε× 1

Radius of nth stationary orbit of electron:

rn = επ0

2 2

2n hme

1

The Bohr radius (n = 1),

r1 = επ0

2

2hme

1

32. Rectification: The process of conversion of ac into dc. 1

Primary

Transformer

Secondary

OUTPUT VOLTAGE

INPUT acCIRCUITDIAGRAM

WAVE FORM

Volta

ge a

cros

s RL

Volta

ge a

t A

B Y

A X

RL

t

t

1

1

Working: During the positive half cycle of the input ac signal, the diode is forward biased and hence it conducts and a current flows through it. There is a potential drop/voltage across RL. 1

During the negative half cycle of the input ac signal, the diode is reverse biased. It does not conduct. Hence no current flows through it and no voltage across the load resistor RL. 1

33. q = 5µC

Aq = –5µC

Bq

A O B

F�

E�

A BE E��

r = 0.1 m r = 0.1 m

(a) Electric field at O due to qA :

EA = 20

14πe

Aqr

1

Magnitude of electric field at O due to each charge is

AE =

BE

= 9 × 1096

25 10

0.1

− ×

= 45 × 105 NC–1 1

Net electric field at O is

E = EA + EB

= 90 × 105 NC–1 OR 9 × 106 NC–1

1

(b) Magnitude of force experienced by the test charge

F = qE

F = 2 × 10–9 × 90 × 105 1

= 180 × 10–4 N OR 1.8 × 10–2N 1

34. (a) Effective resistance RP is given by

1RP

= 1 1 1

1 2 3R R R+ + 1

Substituting, 1RP

= 13

14

112

+ +

1RP

= 4 3 112

812

23

+ + = =

Effective resistance of the combination:




RP = 3/2W = 1.5W 1

(b) Current drawn from the battery is,

I = ε

R rP + 1

I = 6

1 5 0 53

. .+= A 1

Terminal potential difference:

V = e – I r = 6 – 3(0.5) = 4.5V 1

OR Terminal potential difference:

V = IRP = 3(1.5) = 4.5V

35. (a) Resonant frequency: 01

w =LC

16

1100 rad s

5 20 10

−−

= =× ×

OR

fo or uo = 1

2π LC 1

( ) 6

115.9 Hz

2 3.14 5 20 10−= =

× ×

1

(b) Quality factor: 0w=

LQ

R OR

0

1=

wQ

RC

100 512.5

40×

= =Q

OR

1=

LQ

R C 1

6

1 512.5

40 20 10−= =×

Q 1

(c) Impedance at resonance: Z = R = 40W 1

36.

n = 1.5

90o

60o

ir

1r

2

Critical angle: ic = sin–1 1 n 1

ic = sin–1 11.5

= sin–1 (0.6667)

= 41o49’ 1

Angle of incidence on the second face:

r2 = 41o49’

Angle of refraction at the first face:

r1 = A – r2

= 60o – 41o49’ = 18o11’ 1

Applying Snell’s law to the first face:

n = sinsin

ir1

[OR using n1 sin i = n2 sin r1; n1 = 1, n2 = 1.5] 1

⇒ sin i = n sin r1

⇒ sin i = 1.5 sin 18o11’

= 1.5(0.3121) = 0.4682

Angle of incidence on the first face:

i = sin–1 (0.4682) = 27o55’ 1

(Award full marks for the answers rounded off to 2 significant figures: 1/1.5 = 0.67, ic = 42o, i = 28o)


Detailed Answer :

Refractive index = m = 1.5

Critical angle = qC

∴ sin qC = 1µ

or sin qC = 23

∴ qC = 42o

∴ ∠ADP = 90° – 42° = 48°

∠BAC = 60°

∴ ∠APD = 180° – 60° – 48° = 72°

∴ ∠r = 90° – 72° = 18°

sinsin

ir

= m

Or, sin

sini

18o = 1.5

Or, sin i = 1.5 × sin 18° Or, sin i = 1.5 × 0.31 Or, sin i = 0.465




∴ i = 27.7° The angle of incidence should be 27.7°

37. Mass defect per copper nucleus:

Dm = Zmp + (A – Z) mn – MCu 1

Dm = (29 × 1.00727) +

(34 × 1.00866) – 62.92960

Dm = 0.5757 u 1

Energy required to separate neutrons and protons in one nucleus (is equal to binding energy)

Eb = Dm × 931.5 MeV

Eb = 0.5757 × 931.5 = 536.3 MeV 1

Number of atoms in the copper coin (mass = 63 g) is N = 6.022 × 1023 1

Total energy required to separate all neutrons and protons in the coin = Eb × N

E = 536.3 × 6.022 × 1023

= 3230 × 1023 MeV

OR 3.23 × 1026 MeV 1

OR E = 5.17 × 1013 J

Note : (1) Full marks should be awarded if 931 MeV or 932 MeV is used while finding binding energy.

(2) Full marks should be awarded if MCu = [Mass of copper atom – 29 × mass of electron] is used.

i.e., Mass of copper nucleus:

M or MCu = 62.92960 – (29 × 0.00055) = 62.9316 u

Mass defect: Dm = 0.59167 u, B.E.: Eb = 551.1 MeV and Total energy: E = 3319 × 1023 MeV

qqq


SOLVED Karnataka II PUC PAPER April 2020

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