solved ii puc physics subject code march-2018 · what is a rectifier ? with suitable circuit...

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SOLVEDPAPER

II PUCMarch-2018

PhysicsSubject Code

33 (NS)

Time : 3 Hours 15 Min. Max. Marks : 70

General Instructions : 1. All parts are compulsory. 2. Answers 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. What is an equipotential surface ? 2. Define 'drift velocity' of free electrons. 3. Give an application of cyclotron. 4. State Faraday's law of electro magnetic induction. 5. If the peak value of a.c. current is 4.24 A, what is its root mean square value ? 6. Mention one power loss in transformer. 7. Two lenses of power +1.5D and –0.50 are kept in contact on their principal axis. What is the effective power

of the combination ? 8. The decay of proton to neutron is possible only inside the nucleus. Why ? 9. What is 'depletion region' in a semi conductor diode ?

10.

1

0

1

0

What is the output of this combination ?

PART-B II. Answer any five of the following questions : 5 × 2 = 10 11. Mention any two factors on which the capacitance of a parallel plate capacitor depends. 12. State Kirchhoff's laws of electrical network. 13. Define magnetic 'declination' and 'dip' at a place. 14. Write the expression for magnetic potential energy of a magnetic dipole kept in a uniform magnetic field and

explain the terms. 15. Give any two applications of X-rays. 16. What is 'myopia' ? How to rectify it ? 17. Draw the diagram representing the schematic arrangement of Geiger-Marsden experimental set up for the

alpha particle scattering. 18. Write any two characteristics of nuclear forces.

PART-C III. Answer any five of the following questions : 5 × 3 = 15 19. Give three properties of electric charge. 20. State Ampere's circuital law and arrive at the expression for the magnetic field near a straight infinite current

carrying wire. 21. What is hysterisis ? Define the terms 'coercivity' and 'retentivity' of a ferromagnetic material. 22. Arrive at Snell's law of refraction, using Huygen's principle for refraction of a plane wave. 23. Write Bohr's postulates for the hydrogen atom model. 24. Derive the expression of the half-life of a radio active nuclide. 25. Write any three distinctions between p-type and n-type semi conductor. 26. Draw the block diagram of generalised communication system.

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PART-D IV. Answer any two of the following question : 2 × 5 = 10 27. Define electric potential due to a point charge and arrive at the expression for the electric potential at a point

due to a point charge. 28. Obtain the expression for the effective emf and the effective internal resistance of two cells connected in

parallel such that the currents are flowing in the same direction. 29. Derive the expression for the magnetic field on the axis of a circular current loop, using Biot-Savart's law. V. Answer any two of the following questions : 2 × 5 = 10 30. Arrive at the expression of the impedance of a series LCR circuit using phasor diagram method and hence

write the expression for the current through the circuit. 31. Deduce the relation between n, u, v, Q, R for refraction at a spherical surface, where the symbols have their

usual meaning. 32. What is a rectifier ? With suitable circuit describe the action of a full wave rectifier by drawing input and

output waveforms. VI. Answer any three of the following questions : 3 × 5 = 15 33. Three charges each equal to +4nC are placed at the three comers of a square of side 2 cm. Find the electric

field at the fourth corner. 34. 100 mg mass of nichrome metal is drawn into a wire of area of cross-section 0.05 mm2. Calculate the resistance

of this wire. Given density of nichrome 8.4 × 103 kgm–3 and resistivity of the material as 1.2 × 10–6 W m. 35. A circular coil of radius 10 cm and 25 turns is rotated about its vertical diameter with an angular speed of 40

rad S–1, in a uniform horizontal magnetic field of magnitude 5 × 10–2 T. Calculate the emf induced in the coil. Also find the current in the coil if the resistance of the coil is 15 W.

36. In Young's double slit experiment the slits are separated by 0.28 mm and the screen is placed at a distance of 1.4 m away from the slits. The distance between the central bright fringe and the fifth dark fringe is measured to be 1.35 cm. Calculate the wavelength of the light used. Also find the fringe width if the screen is moved 0.4 m towards the silts, for the same experimental set up.

37. Light of frequency 8.41 × 1014 Hz is incident on a metal surface. Electrons with their maximum speed of 7.5 × 105 ms–1 are ejected from the surface. Calculate the threshold frequency for photo emission of electrons. Also find the work function of the metal in electron volt (eV). Given Planck's constant h = 6.625 × 10–34 Js and mass of the electron 9.1 × 10–31 Kg.

qqq

SOLUTIONSAs per Scheme of Valuation

(Issued by Department of PUE, Karnataka)

PART - A I. 1. An equipotential surface is a surface with constant value of potential at all points on the surface. 1 2. The average velocity with which the free electrons are drifted in a direction opposite to the applied field is

called drift velocity. 1 3. Cyclotron is used to accelerate the charged particles or ions to high energies. 1 4. The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the

circuit. 1

5. Irms = Io

2 =

4 241 41..

= 2.9981 A ≈ 3A ( Io = maximum or peak current) 1

6. Flux leakage / resistance of the windings / eddy currents / hysteresis loss (Any one) 1 7. P = P1 + P2 = 1.5 – 0.5 = 1 D 1 8. The decay of proton to neutron is possible only inside the nucleus because proton has smaller mass than

neutron. 1 9. The space charge region on either side of the p-n junction together is known as depletion region. 1 10. 1 1

PART - B II. 11. Area of the plates / distance between the plates / permittivity of the medium between the plates. (Any two) 2 12. Junction rule : At any junction, the sum of the currents entering the junction is equal to the sum of the

currents leaving the junction. 1 Loop rule : The algebraic sum of the changes in potential around any closed loop involving the resistors and

cells in the loop is zero. 1


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13. Declination : The angle between the true geographic north and the north shown by the compass needle is called declination. 1

Dip : The angle between the direction of total intensity of magnetic field of earth and a horizontal line in the magnetic meridian. [Scheme of Valuation, 2018] 1

Detailed Answer : Angle of Declination : At any place on earth, the acute angle between magnetic meridian and the geographical

meridian is called the angle of declination. Angle of Dip : The angle of dip at any place is the angle between earth's magnetic field intensity B with horizontal

in the magnetic meridian at that place.

14. Um = – m B→ →

. 1

Um = magnetic potential energy, m→

= magnetic dipole moment, B = uniform magnetic field. 1

15. (i) X-rays are used as diagnostic tool in medicine. 1 (ii) To treat certain forms of cancer. 1 16. Myopia is a defect in human eye in which light from the distant objects arriving at eye lens get converged at

a point before to retina. Using concave lens to rectify this defect. 1

17.

θ

2 18. (i) Strongest of all fundamental forces in nature. (ii) Charge independent force. (iii) Short range force. (iv) Saturated force (v) Spin dependent force. (vi) Non-central force. (Any two) 1 × 2 = 2

PART - C III. 19. (i) Additivity of charges : Charges can be added algebraically. 1 (ii) Charge is conserved : The total charge of the isolated system is always conserved. 1 (iii) Quantisation of charge : All free charges are integral multiples of a basic unit of charge 'e'. 1 20. Ampere's circuital law : The integral of the product of magnetic field and length element is equal to µ0 times

the total current passing through the surface.

B dl.→ →

= µ0I. 1

r

P

I

Consider a point at distance 'r' from a straight infinite current carrying wire.

From Ampere's circuital law B dl.

→ →

= B. 2pr = µ0I.⇔ 1

B =

��0

2Ir

1

21. Hysteresis : The phenomenon of lagging behind of magnetic induction (B) with respect to the magnetizing field (H) is called hysteresis. 1

Coercivity : It's a phenomenon of completely demagnetizing the magnetic material by applying the magnetizing field in the opposite direction. 1

Retentivity : The property of the magnetic material to retain magnetism even in the absence of the magnetizing field is known as retentivity. 1




22.

Incident wavefront

B

v1�

Refracted

wavefront

v2�

Ev <v2 1

Medium 2

P

Medium 1

v1

A'

i

i

P'

rCR

AB = plane wave front A'A = direction of propagation of wave front PP' = surface of separation of medium 1 and 2 v1, v2 = speeds of light in medium 1 and 2 respectively i = angle of incidence r = angle of refraction τ = time taken by the wave front to travel a distance BC 1 To determine the shape of the refracted wave front a sphere of radius, v2 τ is drawn and CE = tangent plane

if the refracted wave front.

Consider triangles ABC and AEC sin i = BCAC

= v1τAC

, sin r = AEAC

= v2τAC

sinsin

ir

= vv

1

2

ττ =

vv

1

2 .........(1) 1

If c = speed of light in vacuum then, n1 = cv1

, n2 =

cv2

\ Equation (1) ⇒ sinsin

ir

=

cn

nc1

2× =

nn

2

1

n1 sin i = n2 sin r This is the required Snell's Law 1 23. (i) An electron in an atom could revolve in certain stable orbits without the emission of radiant energy. 1 (ii) Electron revolves around the nucleus only in those orbits for which the angular momentum in some

integral multiple of h

2p where h = Planck's constant.

L = n

h.2p

. 1

(iii) When an electron make a transition from one non- radiating orbit to another of lower energy a photon of energy is equal to the energy difference between the two states is emitted. i.e. hn = Ei – Ef. 1

24. We have N = N0e– λt 1

when t = T12

, N = N0

2

\ N0

2 =

N e0

12

��T

\ eλT1

2 = 2 1

λT12

= ln2 = loge2

T12

= log . loge2

1022 303

� �� = 0 693.

λ. 1

25.

n – type p – type

(i) Obtained by doping pure semiconductor by pentavalent element like As, Sb, P.

Obtained by doping pure semiconductor by trivalent element like Al, B, In.

(ii) Majority charge carriers are electrons and minority charge carriers are holes.

Majority charge carriers are holes and minority charge carriers are electrons.




(iii) Donor energy level lies close to the conduction band.

Acceptor energy level lies close to the valence band.

(iv) ne >>>nh nh >>>ne

(Any three) 3

26.

3

PART - D 27. Electric potential is defined as the work done in bringing a unit positive test charge from infinity to that point

against the direction of the field. 1 Suppose a point-charge of +q coulomb is situated at a point O in a medium of dielectric constant K. Let P be

a point, distant r from O, at which the electric potential is to be determined. For this, we must calculate the work done in bringing a test-charge from infinity to P. Suppose, a test-charge + q0 is placed at point A, distant

x from O, and away from P. By Coulomb's law, the magnitude of the electric force F→

acting on q0 is given by

F =

14 0

02�� K

qqx

newton ...(i)

xr

PO+q +q

0

AB

dx

F

––

–

–––

The direction of F

→

is away from O, If the test charge + q0 is displaced by dx to the point B, the work dopn

dW = F · (– dx) Negative sign is due to the reason that displacement dx is opposite to the direction of force. Hence work done

in bringing + q0 from infinity (x = ∞) to the point P(x = r) is

W =– Fdxr

�� = � �

q qxdx

r0

0204

1�� K

= � ��

��

��

q qK x

r0

041

�� =

� � �

��

��

q qK r

0

041 1

��

W = 1

4 0

0

�� Kqqr

. .

Thus, the potential at P V = Wq0

= 1

4 0�� Kqr

.

For vaccum (or air) K = 1

Vqr

�1

4 0��volt.

28. Consider two cells of emf ε1 & ε2 and corresponding internal resistances r1, r2 are connected in parallel. Since as much charge flows in as out, we have I = I1 + I2 1 Both the cells are at same potential since they are in parallel. V = ε1 – I1r1 For first cell V = ε2 – I2r2 For second cell 1




I1 I1

I2I2

I IB2B1A C

r2

r1

ε1

ε2

A I I C

———

So, I1 = �1

1

�Vr

, I1 = �2

2

�Vr

I = � �1

1

2

2

��

�V Vr r

=

� �1

1

2

2 1 2

1 1r r r r� � �

�

��

�

��V

I = � �1 2 2 1

1 2

r rr r�

–

V

r rr r

2 1

1 2

��

��

�

��

⇒ I (r1r2) = ε1r1 + ε2r2 – V (r2 + r1)

⇒ I ( )

( )r r

r r1 2

1 2+ =

� �1 2 2 1

1 2

r rr r��

�( )

V

\ V = � �1 2 2 1

1 2

r rr r��

�( )

V ( )( )

r rr r

1 2

1 2+ 1

V = εequivalent – Irequivalent

\ εeq = � �1 2 2 1

1 2

r rr r

req��( )

and =

( . )( )r rr r req

1 1

1 2

1+

or = 1 1

1 2r r+ . 1

29. Consider a current loop of radius R carrying a steady current I. let 'P' be a point at a distance 'x' from the center of the current loop on its axis.

Let 'dl' be the current element as shown. The magnetic field due to it is given by Biot- savart's law.

d B

→

=

��0

4

I| |dl rr

� ��3

But r2 = x2 + R2

The displacement vector r→

from dl→

to the axial point P is in the X-Y plane. Hence

| |dl r� �

� = r dl 1Y

dl

R

OI

dl

Z

x

rdB

PX

�

�

\ dB = ��0

34I dl r

r| |� ��

=

��0

34I r dlr.

=

��0

24I . dlr

=

��0

2 24I dl

x R� The component of magnetic field along X- direction is dBx = dB cosθ

cosθ =

Rr

=

R

x R( ) /2 2 1 2+ 1 \

dB = ��0

2 2 24 12

I dlr

R

x R( ) /� =

��0

2 2 2 24 12

I dlx R

R

x R( ) ( ) /� �




\ dB =

µπ

02 24

3 2

IR

x R( ) /+

dl

Therefore the total magnetic field at P due to the loop 1

B = ��

02 24 3 2

.( ) /

IRdlx R�

\ As ∫ dl = 2pR

B =

�02

2 223 2

.

( ) /

IR

x R�

= �02

2 22 3 2

IR

x R( ) /�

1

30. Consider a voltage source V = Vm sin wt connected to a series LCR with inductance L, capacitance C and resistance R.

If q is the charge on the capacitor and 'I' the current at any time 't' then from Kirchhoff's rule we have, VL + VR + VC = 0

L

C

R

L IRdIdt

qC

+ + = 0

I

VR

� �t +

VC

VL

V- VC

Lm

m

VC

VL

+

� t

�

VR

Vm

V

VR

m

From the diagram I = Im sin (wt + φ) where φ = phase difference between voltage across the source and the current in the circuit.

1

V→

= V V VL C R� � �

� �

VC→

and VL→

are along same line but in opposite direction.

\ the magnitude |VCm – VLm| 1 \ Vm

2 = VRm2 + [VCm – VLm]2 1

Vm2 = (Im R)2 + (ImXC – ImXL)2

Vm2 = i2m [R2+ (XC – XL)2]

Z = Vm

mI = R X XC L

2 2� �( )

\ Z = R X XC L2 2� �( ) 1

Im = V

R X XC L

m2 2� �( )

or, Im = V

R X Xm

L C2 2� �( )

1

31. OM = u = object distance MI = v = image distance MC = R = radius of curvature Angle i = angle of incidence Angle r = angle of refraction ON = incident ray NI = refracted ray




O

M R

N

In1

n2

C

r

vu

NC = normal & n1, n2 are the refractive indices

1

From the figure for small angles

tan ∠ NOM =

MNOM

, tan ∠ NCM = MNMC

, tan ∠ NCM = MNMI

1

In the triangle NOC, i� = exterior angle = sum of the interior opposite angles

∠ i = ∠ NOC + ∠ NCO = ∠ NOM + ∠ NCM = MNOM

MNMC

+

Similarly ∠ NCM = ∠ CNI + ∠ NIC = ∠ CNI + ∠ NIM

∠ r = ∠ CNI = ∠ MCN – ∠ NIM = MNMC

MNMI

− 1

From Snell's law n1 sin i = n2 sin r

For small angles n1i = n2r 1

Substituting the values of i and r we get

n1

MNOM

MNMC

��

��

=

n2

MNMC

MNMI

-��

��

n n1 2

OM MI+

=

n n2 1−MC

Applying sign convention, OM = – u, MI = v, MC = R

1

nv

nu

1 2− =

n n2 1−R

. 1

This is the required expression.

32. Rectifier is a device which converts alternating current (ac) into direct current (dc) 1

For full-wave rectifier :

Centre TapTransformer

CentreTap

A

B

Diode 2(D )2

Diode 2(D )1

R OutputL

(a)Y

X




Due toD1

Due toD2

Due toD1

t

t

t

Due toD2

(II)

(I)

(b)

Ou

tpu

tw

avef

orm

(acr

oss

R)

L

wav

efo

rmat

Bw

avef

orm

atA

(c)

du

eto

D1

du

eto

D2

1Input and Output Waveforms

The circuit connections are as shown in the figure. In the positive half cycle of input AC diode D1 is forward biased and conducts. In the negative half cycle of the input AC diode D2 is forward biased and D1 is reverse biased. D2 gives output across RL. The input and output waveforms are as shown. The output is a pulsating DC. 2

33. The elecrtic field due to charges at A, B & C can be calculated at point D as. 1

EA = Q

4A

02�� r

= 9 10 4 102 10

9 9

2 2� � �

�

�

�( ) = 9 × 104 NC–1 along AD

EC = Q

4C

02�� r

= 9 10 4 102 10

9 9

2 2� � �

�

�

�( ) = 9 × 104 NC–1 along CD 1

EB = Q

4B

02�� r

= 9 10 4 102 2 10

9 9

2 2� � �

�

�

�( ) = 4.5 × 104 NC–1 along BD 1

A+4 Cn B+4 Cn

C+4 Cn

D

EA

EB

EC

E+

EA

C

2

2

2x

10- m2

2

2 x 10- m1

Resultant of EA and EC at D ED = E EA C2 2+

= 162 108×

= 12.73 × 104 NC–1 1

Total field at D due to all the charges ED + EB = 12.73 × 104 + 4.5 ×104 = 17.23 × 104 NC–1. 1




34. Since, R = ρlA

...(i) 1

Density (D) = mass (m)volume

=mass (m)

length × area (A), length (l) =

mA × D

1

R = �m l

A × D A�

= ρ

mA D2 1

Substituting the values given, R = 1 2 10 100 100 05 10 8 4 10

6 6

6 2 3.

( . ) .� � ��

�

�

-

W =

R = 5.71W 1 + 1

35. Emf induced in the coil = ε = nABw sin (wt) = ε0 sin wt 1 and Area = pr2

ε = (25 × 3.14 × 10 × 10–2 × 10 × 10–2 × 5 × 10–2 × 40) sin (40t) 1 ε = 1.57 sin 40 t ε0 = 1.57 Volt 1

I = εR

=

1 57 4015

. (sin )t V, Imax =

ε0

R =

1 5715. V

W = 0.1047 A = 0.11A 1

Current Iinst = 0.1047 sin 40t A OR Imax = 0.11 A 1

{Note : since it is not mentioned either instantaneous or maximum or average value of induced emf in the question, if a student calculates by mentioning any of the above values marks can be awarded, also for the other values of θ = wt}.

36. Fringe width b = λDd

1

Given The distance of the fifth dark fringe from the central bright fringe = 1.35 cm

b = md

��

��

12

D�

1

Wavelength of light used λ = 600 × 10–9 m, m = 5 1

Taking distance between the slits and the screen as 1m ⇒ b =

600 10 9 12 1 35 10 0 28 10

9

2 3× × ×

× × × ×

-

- -. . 1 Calculation of fringe width = 7.142 × 10–3 m 1

37. Kmax = hn – w0 OR hn = hn0 + 12

mv2 1

hn0 = (6.625 × 10–34 × 8.41 × 1014) – 12

× 9.1 × 10–31 × (7.5 × 105)2

1

Calculation of threhold frequently n0 = 4.54 × 1014 Hz 1 Work function, w0 = hn0 1

Work function, w0 = 6 625 10 4 54 10

1 6 10

34 14

19. .

.� � �

�

�

�

Work function, w0 = 1.88 eV. 1qqq


solved ii puc physics subject code march-2018 · what is a rectifier ? with suitable circuit...

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