Answers to CBSE Pariksha-2015 (Physics-XII).pdf

Answers to CBSE Pariksha-2015 (Physics-XII) 1

Answers to CBSE Pariksha-2015 (Physics-XII) RSPL/1Section A

1. Potential is same at every point on the surface of the conducting sphere.

\ VA = VB = VC = V So the surface will be a equipotential surface.

Hence, work done in moving a charge of + 1mC on the equipotential surface W = q (VA VC) = q (V V) = 0

2. Copper.

3. The element may be pure inductor or capacitor.

4.

5. Space wave propagation or line of sight communication.

Section B

6. The distribution of current is as shown in the figure.

(i) Applying Kirchhoffs law, we get

VB VA = 1 + 4 = 3,

where VA and VB are potential at A and B

VB VA = 3 Now, A is directly connected to B and is connected to C

\ VA = VD and VB = VC

\ VD VC = 3 = VF VE \ Potential difference across R is 3 volt.

2 Answers to CBSE Pariksha-2015 (Physics-XII)

(ii) Also, VF VE = 3i + 6

or 3 = 3i + 6

3i = 3

i = 1 A \ Current through EF is 1 A.

7. (a) Fringe width in YDSE is given by b = d

Dl i.e., b l so fringe width decreases when

monochromatic source of short wavelength is used. (b) The central bright fringe due to zero path difference for different colours is white. As the

violet colour has the lowest l, the fringe closest on either side of the central white fringe is violet, the farthest is red. After a few fringes, no clear fringe pattern is obtained due to large overlapping of different colours.

OR

(a) It is so because (i) grinding lens of very small focal length is difficult. (ii) on decreasing focal length, aberrations (both spherical and chromatic) become more

pronounced. (b) Angular magnification of eye-piece is

me = f25

1e

+ It increases if fe is taken smaller. Magnification of objective is

mo = uv

As object is kept close to the focus of objective lens, u fo. Thus to increase magnification fo should be smaller.

8. As total E is constant, let n1 and n2 be the number of photons of X-rays and visible region, we have

n1E1 = n2E2

nhc

11l

= nhc

22l

nn

2

1 = 2

1

ll

nn

2

1 = 5001

9. (a) Negative sign signifies that the electron is bound to the nucleus by an electrostatic force of attraction.

(b) Energy in ground state of Hydrogen atom E1 = 13.6 eV Energy in the second excited state (n = 3)

E3 = . . 1.51 eV

n

13 6

3

13 62 2

= =

\ Energy required to take an electron of this atom from ground state to second excited state is

DE = E3 E1 = 1.51 (13.6) = 12.09 eV


10. Radio waves of frequency upto 30 MHz can be reflected by the ionosphere but frequency of 100 to 220 MHz used by T.V. signals cannot be reflected by the ionosphere. Hence, transmission of T.V. signal via sky wave is not possible.

Range of transmission of T.V. signal can be increased by using (i) tall antenna (ii) geostationary satellite.

Section c

11. (a) When Q = 360 mC, then voltage (V) is 360 = CV ...(i)

When V = (V 120), then Q = 120 mC 120 = C (V 120) ...(ii)

From equations (i) and (ii),

C = 2mF and V = 180 V. (b) When voltage increased by 120, V = (V + 120)

Q = CV = 2 106 (V + 120)

= 2 106 (180 + 120)

= 600 mC 12. (a) The path of a free electron in a conductor (i) In the presence of electric field curved

(ii) In the absence of electric field straight line between two successive collision.

(b) The average velocity of a free electron in the presence of an electric field = m

eEt

where, t = average time between two successive collision of an electron. No all the electrons do not have the same average velocity. Different electrons have different

velocity. (c) When the temperature of the conductor increases

The thermal speed of electrons increases

The amplitude of vibration of metal ions also increases.

So the collision of free electrons with positive metal ions occurs more frequently. Hence due to decrease in mean free path between the two successive collision, the relaxation time t decreases. Therefore, average velocity of free electrons in the presence of electric field decreases.

13. (a) Magnetic field at the centre O of a circular coil X carrying current in clockwise sense,

Bx = RI

2om (Vertically downward)

Total magnetic field at the common centre of two coils X and Y to be zero only when

Bx + By = 0

or By = Bx


So the current in coil Y must flow anticlockwise as shown below.

Also, By = Bx

O

I

I

RR/2

Y

X

( / )R

I2 2

om l = RI

2om

I = I2

(b) When the coil Y is lifted vertically upwards through a distance R, centre of coil Y lies on the axial line of coil X.

So, Bx = ( )R R

IR

2 /o

2 2 3 2

2m+

= R

I

4 2om (Vertically downward)

and By = ( / )( / )RI

RI

2 22

2o om m= (Vertically upward)

R

O X

Y

I

R

R/2

I = I/2

Net magnetic field at the centre of coil Y.

Bl = B yl + B xl

= RI

R

I2 4 2

o om m

= RI

21

2 2

1om e o acting vertically upwards.oR

(a) (i) For 1 c < 0 Diamagnetic material Required range 0 mr < 1 (ii) For 0 < c < e Paramagnetic material Required range 1 < mr < 1 + e (b) (i) Figure (a) indicates diamagnetic material

(ii) Figure (b) indicates ferromagnetic material

(iii) Figure (c) indicates paramagnetic material.

14. tan3 =

RLw , and tan

3 = /

RC1 w

\ wL = C1

w XL = XC

This is resonance condition of series LCR circuit.

\ Impedance of the circuit, Z = R. The current in the circuit is

Irms = Z Rrms rmse e= = 200 V

100 = 2A

O


The power dissipated in the circuit is

P = erms Irms cos f where cos f is the power factor

Now, tan f = R

X XL C = 0 (At resonance)

\ cos f = 1 and P = 200 2 1 = 400 W.

15. (a) With modern electronic circuits, we hardly get a frequency of approximate 1011 Hz. To produce an em wave in the visible region, we need to set up an oscillatory circuit of frequency of order of 1014 Hz which is not possible with present electronic device.

(b) In AM wave, amplitude of modulated wave varies as the amplitude of the modulating wave. On transmission noise is also added. So AM signal is more susceptible to noise.

However in FM, carrier wave frequency is changed according to the instantaneous voltage of modulating waves. It is done at the modulating stage not during the transmission through channel. As amplitude remains constant in FM, the FM signal is less susceptible to noise than AM signal.

(c) Every e-mail ID has two parts separated by a sign @ :

(i) Part before @ sign: Personal information part

(ii) Part after @ sign: Domain name which provide the information about the server that provide this e-mail facility.

16. Using lens formula, Focal length of lens in air

f1

a

= ( ) R R

1 1 1a g1 2

m e o ...(1) when lens is placed in the liquid, its focal length is

f1

l

= ( ) R R

1 1 1l g1 2

m e o ...(2) On dividing (1) by (2), we get

ff

l

a =

1

1a g

l g

mm

= 1

1

a g

a l

a g

mmm

f

20

l

= . ..

1 6 11 31 6 1

= . .

..1 3 0 6

0 31 3 2

1# #

=

fl = 20 1.3 2 = 52 cm.

17. (a) The size reduces by half according to the relation, size ~ dl .

Intensity increases four times. (b) Let a and a be the width of the slits in the two cases

q = al and qq =

apll

aal =

Pq

Yes, ratio will be equal.


18. (a) l = cn

As (n0)x < (n0)y

(l0)x > (l0)y

Metal X has larger wavelength.

(b) We know hcl

= hc

0l + K.E.

As the L.H.S. is constant, for lesser value of work function K.E. of photoelectron will be more. So, metal X will give out electrons of larger kinetic energy.

(c) Kinetic energy will not change. On reducing the distance only intensity of light changes, frequency remains same, K.E. of emitted photo electrons depend on frequency.

19. Tn = vr2

n

n vr

n

n

But rn n2 and vn n

1

\ Tn vr

n

n n3

So, TT

2

1 = nn

2

1 3c m 8 =

nn

2

1 3c m

nn

2

1 = 12

\ Required ratio = 2 : 1. Proved.

20. (a) AND gate

Q Y = A B+ = A B

(b) Truth table

A B A B A B+ = AB0 0 1 1 0

0 1 1 0 0

1 0 0 1 0

1 1 0 0 1

It represents AND gate.


(c) Output wave form of a given AND gate for the given input is shown below:

A

t1 t2 t3 t4 t5 t6

B

Y

21. (a) Circuit diagram of an illuminated photodiode in reverse bias:

R

+

p-side n-side

hv

mA

Photoelectric current is proportional to the intensity of incident light. In reverse bias change in photoelectric current is prominantly large for a slight change of intensity of incident light. Thus photodiodes can be used to measure light intensity.

(b) Voltage gain for 1st amplifier , VG1 = 10

Voltage gain for 2nd amplifier, VG2 = 20

Vi = 102 V, V0 = ?

Total gain, VG = 10 20 = 200

But VG = VV

i

0

V0 = Vi VG = 102 200

V0 = 2 V.


22. Drawbacks from this arrangement suffer: (i) Signals cannot be sent very far without employing large amount of power. (ii) Bandwidth is very short.

Alternative arrangement (a) For transmitter

(b) For receiver

Loudspeaker

We attach modulator in the transmitter. It superimposes baseband signals on carrier waves generated by radio frequency oscillator. It is done so because audio signals cannot cover a large distance. Signals get attenuated before they reach the receiving end. Carrier waves have high frequency. They can cover large distances without being attenuated. They act as a means to transport baseband signals over a large distance. On the receiving side demodulator takes out audio signals back from modulated signals. Antenna is used on both sides to radiate and pick up signals respectively.

Section D

23. (a) Research mind, responsibility, curiosity and nature of appreciation.

(b) The horizontal component of earth's magnetic field is zero at the pole. This make the compass needle pointed in any direction at the geomagnetic north or south pole of the earth.

(c) The radius of circular path described by the charged particle is given by

r = qBmv =

.1 6 10 6 10

9 10 3 1019 4

31 7

# # #

# # #

= 26 102 m = 26 cm.


\ n = r

v2

= .2 3 14 26 10

3 102

7

# # #

#

= 2 106 Hz = 2 MHz

Energy, E = mv21 2

= 21 9 10 9 1031 14# # # #

= 40.5 1017 J

= 4.05 1016 J

= 2.5 KeV.

Section e

24. (a) Force on +q charge = qE+

Force on q charge = qE

As forces are equal in magnitude and opposite in direction net force = 0, i.e., no translatory force acts on the dipole.

(b) In a non-uniform electric field, the net torque on the dipole is zero because of linear force on charge q. However, there is a net force on the dipole as shown below:

(i) When P || E

qFq = qE

Less force More force

direction of increasing field

F+q = +qE+ q

p

E

Direction of net force is in the direction of increasing electric field.

(ii) When P is anti parallel to E .

+ q

Fq = qELess force on +q More force on q

Direction of increasing field

F+q = +qE

qP

So, net force acts in the direction of decreasing electric field.


(c) W = pE(cos q1 cos q2)

= pE(cos 0 cos 180)

= pE[1 (1)]

= 2pE.

OR

(a) Suppose n identical cells, each of emf e and internal resistance r, are connected in series in each now and m such rows are connected in parallel across the external resistance R as shown below:

Total number of cells in the combination = mn Net each series emf = e + e + e + ... n times = ne \ Net emf across parallel = ne Internal resistance of each row = nr Total internal resistance of mn cells in parallel

r1l = ... times

nr nr nrm

1 1 1+ + +

r1l =

nrm

r = mnr

\ Current through R, I = cetanTotal resis

Total emf = R r

ne+ l

= R

mnr

nmR nr

mne e

+=

+

For maximum I, denominator should be minimum, i.e. mR + nr is minimum.

Now, mR + nr = ( ) ( ) mR nr mR nr22 2 $+ + mR nr2 $

= imumin( ) 2 MmR nr mnRr2 + = .

It will be so of ( )mR nr 2 = 0

or mR = nr


or mR = nr

or R = mnr

or External resistance = Total internal resistance of all cell. Hence, the current in the circuit of mixed grouping of cells will be maximum when

R = mnr . Hence proved.

(b) Ratio of heat produced in both the wires when same voltage is applied across each is

PP

B

A = /

/

V R

V RRR

Al

lA

B

A

A

B

B

B B

A A

A2

2

#r

r= =

Given lA = lB = l; rA = rB

\ PP

B

A = :AA

41 1 4

B

A = =

25. (a) Force experienced by the sides PQ and RS of the loop are

Fb = IlB sin q Fb = IlB sin (180 q) = IbB sin q

We observe that

Fb = bFl (acting along the axis of the loop) These forces do not provide torque because their lines of action coincide. Forces acting on the sides QR and SP of the loop are

Fl = IlB

and Fl = IlB


These forces provide a torque because

(i) Fi = Fll

(ii) Lines of action of the forces do not coincide.

Now, torque experienced, by the loop is

t = IlB (b sin f)

t = IAB sin f (... lb = A)

t = mB sin f (... IA = m)

t = m B# (b) Current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer

per unit current.

If a current i produces a deflection q in the galvanometer, then current sensitivity Si is given by

Si = iq

Voltage sensitivity of a galvanometer is defined as the deflection produced per unit voltage i.e.,

SV = V iR RSiq q= = (Q V = IR)

If the current sensitivity is doubled say by doubling the number of turns, then voltage sensitivity may be not be increased because it will increase the resistance of the galvanometer and voltage sensitivity may remain the same.

(c) Consider mass of proton = m and mass of a-particle = 4m charge on proton = e and charge on a-particle = 2e (i) Cyclotron frequency is given by,

nc = mqB

2

For proton, nP = meB

2 ...(i)

For a-particle, na = ( )meB

meB

2 42

4 = ...(ii)

Hence, a-particle and proton will not accelerate at the same cyclotron frequency. From equations (i) and (ii), we get

Pnn

a = 2 or nP = 2na

(ii) Maximum velocity by the accelerated particle is given by

r

mv

max

max2

= qv Bmax

or vmax = mqBrmax


For proton, (vmax)P = meBrmax

For a-particle, (vmax)a = ( )

me B r

meBr

42

2max max$ =

\ ( )( )vv

max

max P

a = 2

Hence, at the exist slit of the dees, when they are accelerated in turn, proton will have the higher velocity which is twice than the velocity of a-particles.

OR (a) Applied a.c. voltage is E = E0 sin wt ...(i) Let q be the charge on the capacitor at any time t. The instantaneous voltage across the

capacitor is

E = Cq

E0 sin wt = Cq

q = CE0 sin wt

dtdq = CE0w cos wt

I = C

E1

0

w sin (wt + /2) =

XE

C

0 sin (wt + /2)

I = I0 sin (wt + /2) (ii) From equations (i) and (ii), we conclude that current leads the voltage by a phase angle /2. Capacitive reactance is the opposition offered by a capacitor towards the flow of current

passing through it.

XC = nC21


(b)

(i) E = BHLv

(ii) e.m.f reduces from east to west

(iii) end of the wire towards east.

26. (a)

(b) f0 = 4 cm

fe = 10 cm

u0 = 6 cm

(i) m = uu

feD1

0

0 +e o

f1

0 =

v u1 1

0 0

v0 = 12 cm

So, m = 6

12 11025

+c m = 6

121035 7# =

(ii) f1

e

= v u1 1

e e

fe = 10 cm

ve = 25 cm

ue = 7 cm

Length of the compound microscope

L = v0 + ue

L = 12 + 7 = 19 cm


OR (a) According to question T2P = D + x, T1 P = D x

S1P = ( ) ( )S T PT1 12

12+

= [D2 + (D x)2]1/2

S2P = [D2 + (D + x)2]1/2

Minima will occur when

[D2 + (D + x)2]1/2 [D2 + (D x)2]1/2

= 2l

In case x = D

(D2 + 4D2)1/2 = 2l

(5D2)1/2 = ,2l

Thus D = 2 5

l

(b) Brewster angle: The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of completely plane polarised light.

According to Brewster law, m = tan iB = cot i1

B

Also, m = sin i

1

C

, where iC = critical angle

So, sin i

1

C

= cot i

1

B

sin iC = cot iB

iC = sin1 (cot iB).


Answers to CBSE Pariksha-2015 (PhysicsXII) RSPL/2Section A

1. In series combination, equivalent resistance of the circuit is greater than that of parallel combination. So terminal potential difference across series combination will have higher value.

2. Atmagneticequatorverticalcomponentofearth'smagneticfieldBv is zero.

3. Iftheresistanceofthecircuitshowninfigurea is decreased, then current increases. Due to changeinthecurrent,aninducedemfissetupinthesecondarycircuit[figureb], which opposes thegrowthofcurrentincircuita.Soinducedcurrentinthecircuitbmustflowclockwise.

4. For large aperture mirror such as 5 cm.

Onlytheparaxialraysarefocussedattheprincipalfocus.

Marginalraysmeettheprincipalaxisatapointnearbythepolenotatthefocus.Thisresultin blurred image of the object.

5. TheworkfunctionofX is higher than that of Y.Asworkfunction,W0 = hv0 so metal X has higher threshold frequency.

Section B

6. (i) Here XR =

ll

100

As on doubling the value of R and X, its ratio does not change. Hence, the balance point on meter-bridge remains same.

(ii) Oninterchangingthepositionofgalvanometerandbattery,theratioofR : X remains same. Thebalancepointagainremainssame.

7. AccordingtotheAmperescircuitallawthelineintegralofthemagneticfield B around any closed loop is equal to 0m times the net current i flowing through the area enclosed by theloop.Mathematically

.dlByo = 0i ... (i) Where 0 is the permeability of free space.

Maxwell in1864showedthe logical inconsistencyofequation (i), when applied to the charg-ing or discharging of a capacitor. He suggested that during charging of capacitor, a changing electricfieldexistsintheregionbetweentheplatesofthecapacitor,whichproducesthesamemagnetic effect as does the conduction current, called displacement current. Hence, during charging of a capacitor, conduction current flows through the wires and a displacment cur-rent, id, exists in the region between the plates of the capacitor, fulfilling the condition ofcontinuity of current.

Let q be the charge on capacitor plate at any instant. Then the instantaneous electric fieldbetween the plates will be

E = Aq

0, where A is area of each plate.


By differentiating, we get

dtdE =

A dtdq1

a

= A

i1

0

i = AdtdE

0 ;

This current is definedasdisplacement current id = dt

dE0 (where E = EA is the electricflux.)

Thus themodifiedAmperes circuital law canbe statedas

.dlByo = 0(i + id)

= m idt

d E0 0+c m

The displacement current between the plates is exactly equal to the conduction current inthe connecting wires. Conduction current and displacement current individually are discon-

tinuous, but their sum is continuous. Further since 00 = c

12

1.1 1017 S2m2, is a very

small quantity, the second term will not contribute unless dt

d E is very large i.e., the electric

fieldmustbe changingvery rapidly.

8. (a) As dtdN = lN, graph will be a straight line as shown.

dtdN

N

(b) Weknowthat

RR

0 =

21 nc m

or 161 =

21 nc m

or 21 4c m =

21 nc m

or n = 4

But n = Tt

t = 4T


9.

YAB

X

A

1 1

1 1

1

B

X

truth table

A B X = A + B0011

0101

0111

It is equivalent to (Y = A B+ = A + B)

A

BX

Thus,ORlogicoperationisperformedbythiscircuit.

10. (i) Amplitude modulation: It is a process in which original signal is superimposed over a high frequency carrier wave in such a way that amplitude of modulated wave varies with the amplitude of the modulating signal whereas frequency modulated wave is same as that of the carrier wave.

(ii) Frequency modulation: It is a process in which frequency of carrier wave varies in accordance with the instantaneous value of modulating signal whereas its amplitude remains same as that of the carrier wave.

(a) Frequency modulation gives better quality transmission. (b) Amplitude modulation has a larger coverage.

Modulation of a carrier wave: (a) a sinusoidal carrier wave; (b) a modulating signal; (c) amplitude modulation (d) frequency modulation.


or

10. (a) All information related to a localnetwork is stored in server computer ofLAN.

(b) With increasing use of mobile phones and advancement of technology, it is pertinenttomake themobile phonenetworksmore efficient. The efficiency ofmobile networks ismentioned by word 'Generation' and abbreviation 'G'

1G were first generation of mobile networks, which were based on analogue radiosignals.

2G were narrow band digital signal based networks with good quality of calls. Theyprovided world over connectivity.

3G networks increased the data transfer speed for efficient use of Internet onmobilephone.

4G networks are going to provide a high-speed internet facility onmobile phones forsurfingnet, chattingviewing, television, listeningmusic etc.

Sectionc

11. (a) (i) TheelectricfieldatOiszerobysymmetry.Thisisbecause,thefieldatthecentreduetochargesatA,B,CandDisequalandoppositetothefieldduetothechargesqatEalone.

(ii) By symmetry, E E E E EA B C D E+ + + + = 0

or E E E E EB C D E A+ + + =

Thus,thefieldatthecentreduetochargeatB,C,DandEareequalandoppositetothefieldduetochargeq at 'A' alone.

Thefieldatthecentreduetocharge'q' at A is

.p

aloner

kq

r

qE AO

41

A 20

2= =

Thus,thefieldatOduetochargeB,C,DandEis

.along directionr

kqE OA

2=

(iii) TheelectricfieldatO,ifthecharge'q' at 'A' is replaced by q,thenthenetfieldatOwill be

4p

alongr

qE OA

1 2

02

=

(b) Bysymmetry, theelectricfieldat thecentreofn-sides regular polygon with charge q at each of its corner is zero.

12. (a) Equipotential surfaces for a system of two identical positive point charges placed a distance d apart.


(b) Workdoneinbringingthechargeq1frominfinitytor1.Againsttheexternalelectricfield

W1 = q1V(r1)

Workdoneinbringingthechargeq2frominfinitytor2.

W2 = q2V(r2)

Workdoneonq2againstthefieldduetoq1

W3 = rq q

4 o 121 2

p

Potential energy of the system = W1 + W2 + W3

= ( ) ( )q V r q V rr

q q41 1 2 2 0 12

1 2

p+ +

13. Potential difference across 20 W and 30 W will be same as they are connected in parallel combination.

V20 = V30 I3 20 = I2 30

2.420 = I2 30

I2 = .

3048 0

=1.6A and the current through ammeter A3. I1 = I3 + I2=2.4+1.6=4.0A

14. Potentiometer is preferred over a voltmeter: Potentiometer is a null method device. It measures the emf of cell when cell is in open circuit which is equal to the actual emf of cell.

Whenemfofacellismeasuredwiththehelpofavoltmeter,terminalpotentialdifferenceofthecellisobtainedbecauseofclosedcircuitasvoltmeterdrawsomecurrentfromthecell.Terminalpotential difference is always less than the emf of a cell. So potentiometer is preferred over a voltmeter.

Let kbethepotentialgradient(V/cm).Then

1 + 2 = 351 K

and 1 2 = 70.2 K

1= 210.6K and 2=140.4K.

Hence, 2

1

= ..

KK

140 4210 6 =

23

or

Capacitoroffersinfiniteresistancetotheflowofd.c.throughitinitssteadystate.So,nocurrentwillflowthrough4W resistor. Hence, it is ineffective.

Totalresistanceofthecircuit, R = . .2 82 32 3

4 0# W++

=


Current drawn from the battery

I = 1.5 AR 4

6= =

Potential difference between A and B is

VAB = 1.5 1.8 VIR 56

AB #= =

Current through 2 W resistor, I = . 0.9 ARV

21 8

= =

15. (a) Gausss law in magnetism: Itstatesthatthesurfaceintegralofamagneticfieldsovera closed surface is always zero.

Mathematically, dsBs

$y = 0. Significance:

(i) Magneticfieldlinesformacontinuousclosedcurve,i.e., if a certain number of magnetic fieldlinesenterinclosedsurface,samenumberoffieldlinesmustleavethatsurface.

(ii) Thereisnopointatwhichthefieldlinesstartandthereisnopointatwhichthesefieldlinesterminated.Hence,isolatedmagneticpoleormonopoledonotexist.

(iii) Thereforemagneticpolealwaysexistasunlikepairsofequalstrength.

(b) Difference between paramagnetic and ferromagnetic material

Paramagnetic Material Ferromagnetic Material

(i) They are feebly attracted by amagnet.

(ii) Inanon-uniformmagneticfield,theyhave a tendency to move slowly from weakertostrongerpartofthefield.

(iii) As soon as magnetising field is removed, magnetisation is lost

(i) They are strongly attracted by amagnet.

(ii) When these are placed in a non-uniform magnetic field, it moves quicklyfromweakertostrongerpartofthefield.

(iii) Theyretainthemagnetismevenaftertheremovingofmagnetisationfield.

16. Let XX and YY be two long parallel straight conductors carrying current i1 and i2 respectively in the same direction and placed in vacuum or air at a distance R metre apart.

ThemagnitudeofmagneticfieldatanypointonY due to current 1 in XX is given by

B1 = Ri

2o 1$pm

AccordingtoRightHandPalmRule,thedirectionofB1 is perpendicular to the plane of paper, directed inward.

TheconductorYY, carrying current i2 is situated in B1magneticfieldproducedbythecurrenti1 in XX.ThereforelengthofYYexperiencesaforcewhichisgivenby

F = i B l iRi

l2

o2 1 2

1$ $pm

=


Force per unit length of YY is therefore

lF =

Ri i

2o 1 2$pm

So attract each other.

FlemingsLeftHandRuleshowsthatthedirectionofthisforceistowardsXX. Similarly, force per unit length of XX due to current in YY is

lF =

Ri i

2o 1 2$pm

AccordingtoFlemingsLeftHandRule,thedirectionofforcewillbetowardsYY.Thuswiresattract.

Thedirectionoftheforcewhentheconductorscarrycurrentinthesamedirectionistowardseachother,whiletheforceisawayfromeachotherasthecurrentflowsinoppositedirection.Thusattractionandrepulsionareproved.

(b) Accordingtotherighthandthumbrule,themagneticfieldduetoboththewiresatP is directed normally into the plane of paper.

I IP

q

r/2 r/2

Alsodirectionofmotionofthechargeparticleisperpendiculartotheplane.Thuswefind

v || B , so q = 0. F = qvB sin q = qvB sin q = 0. Themagnitudeoftheforceduetothemagneticfieldactingonthechargeatthegiven

instant is zero.


17. (a) Self-inductance is that property of a coil by virtue of which it opposes any change in the magnitude of current passing through it by inducing an emf in itself.

Consideran inductor of inductance L, carrying alternating current through it.

Suppose at any instant of time e.m.f. induced in the inductor is

= LdtdI

Tomaintain thegrowthof current through the inductor,powerhas tobe supplied fromexternal source.

dt

dW = I LdtdI

I =

dW = LI dI

Total amount ofworkdone to buildup current fromzero to I is

W = L I dI LI21I 2

0

=y

Thisworkdoneget stored in the inductor in the formofmagnetic energy.

Thus, U = W

U = 21

LI2

(b) Step-up transformer is based on the principle of mutual induction.

An alternating potential (Vp) when applied to the primary coil induced an e.m.f. in it.

p = N dtd

p

If resistance of primary coil is low Vp = p.

i.e. Vp = N dtd

p

As same flux is linked with the secondary coil with the help of soft iron core due tomutual induction e.m.f. is induced in it.

s = N dtd

s

If output circuit is open Vs = s

Vs = N dtd

s

ThusVV

p

s = NN

p

s


For step-up transformer NN

p

s > 1

Incaseofd.c.voltagefluxdoesnotchange.Thusnoe.m.f.isinducedinthecircuit.

18. (a) Onlyanaccelerated charge canact as a source of electromagneticwave.

(b) Direction of propagation of em wave = Direction of E B vector.

(c) (i) Visible lighthas shorterwavelengthnext to infrared radiations.

(ii) Microwaveshave longerwavelengthnext to infrared radiations.

19. (a) TheresultantintensitytotwowaveshavingintensitiesI1 and I2 with a phase difference is

I = 2 cosI I I I1 2 1 2+ + But given I = I1 + I2 2 cosI I1 2 = 0 or cos = 0 i.e. average value of cos over one complete cycle (from 0 to 2p) is zero. It means that

the phase difference between two waves emitting from the two source is not constant. So, both the source are incoherent source of light.

(b) When thewaveundergoes refraction,nophase change occurs.

Change in phase = 0.

20. (a) Given l1 = 6000,b1=0.8mm,d1 = d

l2 = 7500,d2 = 2d, b2 = ?

Fringe width in YDSE is given by

b = d

Dl

1

2

bb

= d

DDd

dd

122

1

1

2

1 2# #l

l ll

=

0.8 mm

2b =

dd

2 6000

750085

# =

or b2 = 0.8 0.5 .mm85# =

(b) (i) Magnifyingpowerofarefractingtelescopewhenthefinalimageisformedatinfinityisgiven by

m = ff

e

0 .

andwhenthefinalimageisformedattheleastdistanceofdistinctvision(D)is

m = ff

Df

1e

e0 +c m. Clearly magnifying power of a telescope depends on focal length of objective and its

eyepiece.


(ii) Resolvingpowerofaastronomicaltelescopeisgivenby

R P = .D

1 22 l.

Clearly it depends on

Aperture(D) of the objective lens Wavelength(l) of light.

21. Bohrs Postulate Quantum condition: Electron revolves around the nucleus in those orbits in which the angular momentum of an electron is an integral multiple of h

2p, h is plank's

constant.

i.e., L = mvr = nh2p

, n = 1, 2, 3, ...

h=6.61034 Js.

K.E. of electron = r

kZe2

2

P.E. of electron = r

kZe2

(a) Inthe1stexcitedstate,totalenergy=3.4eV

As K.E. = T.E.

K.E. = 3.4eV

(b) P.E.ofelectroninthe1stexcitedstate

= 2 K.E.

P.E. = 23.4

= 6.8eV

(c) K.E. will not change on changing the reference of P.E. It is the P.E. and total energy of the state that would change.

22. (a)

ORgategivesoutputonthebasisofthefollowingtruthtable.

A B C

0011

0101

0111


(b) IB =

.R

V VBB BE1

If R1 is increased, IB will decrease.

Because IC = bIB, it will result in decrease in IC i.e. decrease in ammeter and voltmeter readings.

23. (a) Keen observer, curiosityandquest forknowledge. (b) Photoelectric effect: Aphotocell connected to amotorwhich opens the door.When a

person interrupts a beam of invisible ultraviolet light falling on a photo cell installed in front of the doorway, a photoelectric current is suddenly changed. This starts amotorwhich open the door.

(c) Photoelectric equation is

Given l1= 400nm l2= 400+200=600nm According to Einstein's photoelectric equation, the maximum kinetic energy of emitted

photoelectron is

kmax = hn wo or eVo = hn wo

Vo = ne

he

wo

= le

hce

wo

Hence, change in stopping potential

DVo = Vo2 Vo1

= l lehc

ew

ehc

ewo o

2 1e eo o

= l le

hc 1 1

2 1e o

= .

.

1 6 10

6 6 10 3 10

6 10

1

4 10

119

34 8

7 7

= G So, Vo2 Vo1 = 1.03 volt

or Vo2 = Vo1 1.03=61.03

= 4.97V.

Section D

24. (a) Charged particle in a magnetic field: LettherebeauniformfieldofmagneticinductionB at right angles to the plane of the paper and directed into the page, indicated by symbols x.Supposeaparticlewithacharge+q is introduced at point Ointhefieldwithvelocity v in a direction at right angles to B .Therelation qvF B#= and the right hand screw rule show that an upward force F , of magnitude qv B and lying in the plane of the paper, is exertedontheparticleatthepointO.


If m is the mass of the particle and r is the radius of the circle then, we have

F = qvBr

mv2=

r = qBmv

That is, the radius r of the path is proportional to the momentum mv of the charged particle.

Iftheparticlewerenegativelycharged,theforceatOwouldhavebeenadownwardforceandtheparticlewouldhavedescribedaclockwisecircle.

As p = mv = ,mk2 where K.E. of the particle is k = qV, where V is the accelerating potential.

r v V

Thetimeperiod, T = vr

v qBmv2 2

$p p

=

T = qB

m2p and frequency f = .T m

qB12p

=

Both values are independent of radius of the orbit, speed or of the particle and depend only onBandspecificcharge(q/m).

(b) ThecyclotronwasdevisedbyLawrencein1932,foracceleratingpositivelychargedparticlessuch as protons and deutrons, to very high energies so that they can be used in disintegration experiments.

construction: It consists of two horizontal D-shaped hollow metal segments D1 and D2 (called dees) (Fig.) An alternating p.d. of the order of 105 volts at a frequency of 10 to 15 megacycles issuppliedacrossthedees.AnintensemagneticfieldBofabout1.6Weber/meter2 is set up perpendicular to the plane of dees by a large electromagnet.


cyclotron

Thewholespaceinsidethedeesisevacuatedtoapressureofabout106 mm of Hg. Anion source is located at the centre S is the gap between the dees. It consists of a small chamber containingaheatedfilamentandagassuchashydrogen(forprotons)ordeuterium(fordeutrons).Thethermionsgivenoutbythefilamentproducepositiveionsbyionisationofthegas.Theionscomeoutthroughasmallholeintheionsourceandareavailabletobeaccelerated.

theory and Working: Suppose an ion of mass m and charge + q emerges from the ion source at an instant when D2 is at a negative potential. It will be accelerated towards D2 by the electricfieldinthegapbetweenthedeesandenterD2 with a velocity v(say).Onceinside,itisscreenedfromtheelectricfieldbythemetalwallsofthedees.Now,undertheactionofthemagneticfield,whichisperpendiculartotheplaneofdees,theionsadoptacircular

path with a constant speed v and of radius r given by, r = qBmv , where Bisthemagneticfield

induction.Thetimet required by the ion to complete a semi-circle is

t = vrp =

qBmp

Thisshowsthatthe time of passage of the ion through the dees is independent of the speed of the ion and of the radius of the circle, it depends only on the magnetic induction B and the charge to mass ratio (q/m) of the ion. (Greater the speed of the ion, larger will be the circle in which it travels, the period of motion remaining the same.)

Let us assume that the frequency of the applied p.d. has been so adjusted that during the one-half cycle of p.d. the ion completesa semi-circle.Then the ionwill emerge fromD2 into the gap at the instant when D1isatanegativepotential.Theionisthereforefurtheraccelerated while crossing the gap between D1 and D2 and enters D1.Onaccount of itsincreased velocity, its semi-circular path in D1isnowofgreaterradius.Thetimeofpassaget through D1,however,isstillthesame.Theprocessisrepeatedaftereveryhalf-cycleofp.d.and the ion gains in speed each time it passes from one dee to the other. Finally, the ion becoming enough energetic reaches the outer edge of one dee where it is pulled out of the systembyanegativelychargeddeflectorplate.

Achievement of resonance condition:Wehaveseenthatthecyclotronoperatesunderthe condition that the frequency n0 of the applied p.d. must be equal to the frequency n of thecircularrevolutionoftheion.Thatis,

n0 = n


But, n = t m

qB21

2p=

Therefore, n0 = mqB

2p

In practice, the frequency n0oftheelectricoscillatoriskeptfixedandthemagneticinductionBisvarieduntiltheaboveconditionissatisfied.

or

(a) Moving coil Galvanometer: It is an instrument used for the detection and measurement of current. Its action is based on the torque acting on a current-carrying coil placed in a magneticfield.

construction:Itconsistsofarectangularcoilmadeupofalargenumberofturnsoffineinsulatedcopperwirewoundonalightnon-magneticmetallicframe.Thecoilissuspendedbetweenthecylindricalpole-piecesofapermanenthorse-shoemagnetNSbymeansofathinphosphor-bronzestrip,theupperendofwhichisattachedtoatorsionhead.Thelowerendofthecoilisattachedtoalooselywoundspringofveryfinephosphor-bronzewire.Asoftironcylindrical core C is placed symmetrically within the coil without touching it. It concentrates thelinesofforceandthusmakesthemagneticfieldbetweenthepole-piecesstrong.

ThecurrenttobemeasuredentersatoneterminalT1 and passes through the suspension, coilandspringandfinallyleavesatthesecondterminalT2.

Moving coil galvanometer

theory:Whenacurrentflowsinthecoil,horizontalandoppositelydirectedforcesareexertedontheverticalsides,producingacoupleaboutaverticalaxisthroughthecentre.Thecoilrotates in the direction of the couple until the rotating couple is balanced by the restoring elasticcoupledevelopedinthesuspension.Thesteadyangulardeflectionisobservedbymeansofabeamoflightreflectedfromthemirrorandproducingaspotonascaleatadistanceofone meter from the galvanometer.


Let Bbethemagnitudeofthemagneticfieldinductionduetothepermanentmagnet,Athearea of the coil, and Nthenumberofturnsinit.Themagnitudeofthetorque(momentofthe couple) acting on the entire coil when carrying a current i is given by

t = NiBA sin q,

where qistheanglewhichthenormaltotheplaneofthecoilmakeswiththedirectionofB .

Now,inthegalvanometer,themagneticfieldB ismaderadialbymakingthepole-pieces N and S cylindrical and placing a soft-iron cylindrical core within the coil. In this case, the normaltotheplaneofthecoilisalwaysatrightanglestothemagneticfieldsothatq = 90. Hence, the torque acting on the coil is

t = NiBA sin 90 = NiBA

If be thesteadyangulardeflectionof thecoil, thentheelasticcoupledeveloped inthesuspension is c, where cisthetorsionalconstantofthesuspension.Therefore,wehave,

t = NiBA = c

i = NBA

C

i = k,

As i .Thisistheprincipleofmovingcoilgalvanometer

where k = NBA

c=theconstantoftheinstrument.Thus,thecurrentpassedthroughthe

galvanometerisproportionaltothedeflectionproduced.

Measurement of current:Theconstantkofthegalvanometerisobtainedbypassingknowncurrentsthroughit.Thentheunknowncurrentispassedandthecorrespondingdeflectionismeasured.Thestrengthofthecurrentisobtainedbytheaboverelation.

In portable galvanometer, a pointer is attached to the coil which moves over a scale graduated directly in amperes and attached to the galvanometer itself.

(b) (i) As the iron core has high value of relative permittivity, mr,ithelpsinmakingthemaximumfluxconcentratedinthiscore.HighvalueofB is available due to this core.

(ii) current sensitivity ofagalvanometer isdefinedas thedeflectionproduced in thegalvanometer per unit current.

If a current Iproducesadeflectionq in the galvanometer, then current sensitivity SI is given by


SI = Ii

Voltage sensitivityofagalvanometerisdefinedasthedeflectionproducedperunitvoltage i.e.,

SV = V IR R

SIi i= = (QV=IR) If the current sensitivity is doubled say by doubling the number of turns, then voltage

sensitivity may not be increased because it will increase the resistance of the galvanometer and voltage sensitivity may remain the same.

25. n1 < n2

In DAOC, i = yMOAM

MCAM + = +

In DNCI, r = MCAM

MIAM b =

WhenanglesaresmallM approaches P.

i.e., i = POAP

PCAP

+ ...(i)

r = PCAP

PIAP ...(ii)

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

For small angles n1 i = n2r ...(iii)

From (i), (ii) and (iii) nPO PC1 1

1 +c m = n PC PI1 1

2 c m

POn1 +

PIn2 =

PC

n n2 1

Now PO = u

PI = + v

PC = + R

vn

un2 1 =

R

n n2 1


Sign conventions

(i) Allthedistancesaretobetakenfromopticalcentre(P).

(ii) Distancestakeninthedirectionofrayoflightaretaken+veandvice-versa.

Weknow,f1 = ( )

R R1 1 1

1 2m = G

Whenlens(convex)isimmersedinwater,m decreases. From the above formula it is clear that focal length will increase.

(a) Diffraction of light: Phenomenon of bending of light round the corners of an obstacles or aperture is called diffraction of light.

Graphshowingthevariationofintensitywithangleinsingleslitdiffractionexperimentis

In diffraction pattern, the brightness of successive bright fringes from the centre goes on decreasing whereas in interference pattern all bright fringes are equally bright and have the same width.

Widthofcentralmaxima=dD2l

(i) Whenthewidthofslit(d) is decreased, angular width increases.

(ii) Whenthemonochromaticsourceoflightisreplacedbysourceofwhitelight,thediffractionpatterniscoloured.Thecentralmaximaisbrightbutotherbandsarecoloured.

Since band width l and lred > lviolet

So, bred > bviolet

(b)

If I0 is intensity of unpolarized light, the intensity of polarized light would be l20 ,

i.e., IA = l20

SupposepassaxisofpolaroidCmakesangleqwithpassaxisofA.Wehave Ic = IA cos

2q [LawofMalus]

Ic = I20 cos2q


As the polaroids A and BarecrossedanglebetweenthepassaxesofC and B would be 90o q.

AccordingtolawofMalusintensityoflighttransmittedbypolaroidB would be

IB = Ic cos2 (90o q)

= Ic sin2q

IB = I2

o cos2q sin2q

IB = I8o (sin 2q)2

Intensity transmitted by polaroid B would be I0/8if sin 2q = 1

2q = 2p

q = 4p

Thus,polaroidCshouldbeplacedmakingangle4pwiththepassaxisofpolaroidA.

26. Zener diode is a special purpose semiconductor diode because it is designed to operate under reversebiasinthebreakdownregion.

IV characteristics of a Zener diode:

Use of a Zener diode as a voltage regulator : Voltage regulation is a measure of a circuits ability to maintain a constant output voltage even when either input voltage or load current varies.

A zener diodewhenworking in the breakdown region can serve as a voltage regulator. Inthefigure,Vin is the input d.c.voltagewhosevariationsaretoberegulated.Thezenerdiodeis reverse connected-across Vin. When potential difference across the diode is greater thanV, it conducts and draws relatively large current through the series resistance R.The loadresistance RL across which a constant voltage Vout is required, is connected in parallel with the diode. The total current I passing through R equals the sum of diode current and load current.

i.e., I = Id + IL


It will be seen under all conditions Vout = Vz Hence, Vin = IR + Vout Vin = IR + Vz Consider, say, the case of n-type semiconductor.On illumination excessholes and electrons

generated would be Dp and Dn. n = n + Dn p = p + Dp Here Dn = Dp

So, Dnn > p Sowefindthatfractionalchangeinthemajoritycarriers(Dn/n) would be much less than that

in the minority carriers (Dp/p). As change in reverse saturation current is more pronounced than that current in forward biased, photodiodes are used in reverse bias.

or

(a) (i) Forward biasing of a p-n junction. When forward biased, majority charge carriers in both the regions are pushed through the junction.

The depletion regions width decreases and the junction offers low resistance. (ii) Reverse biasing of a p-n junction. When reverse biased, majority charge carriers in both the regions are pushed away from the junction.

The depletion regions width increases. The minority charge carriers, however, are pushed through the junction thereby causing a little current in the reverse biased p-n junction. The p-n junction thus offers a very high resistance when reverse biased.

(b)

(c) Since IC increases with IB almost linearly and IC = 0, when IB = 0 the value of the small signal current gain (bac)ofatransistorcanbetakenasnearlyequaltoitsd.c.currentamplificationfactor (bdc).

Answers to CBSE Pariksha-2015 (Physics-XII).pdf

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