cbse 2013 xii physics solution set-ii

7/30/2019 CBSE 2013 XII Physics Solution Set-II

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CBSE Ps-2013 S (Physics-XII) 1

CBSE ParikSha-2013 Solution(PhysicsXii) RSPL/2

1. By enclosing it in the cavity o a metallic shell.

2. As vd

= ,n e A

IwhenIis doubled, v

dis also doubled.

3. H= .(10) 3600

R

Vt

50

2 2#

= = 7200 J

4. Zero, beause F"

is perpendicular to the direction o motion o the charged particle.

5. When magnet is rotated about an axis perpendicular to its length, the orientation o its magneticeld with respect to its coil changes continuously, the magnetic fux linked with the coil changes.Hence an e.m.. is induced at the terminals o the coil.

6. The light rays rom the nearby tiny object spread over the small aperture and as a result thenal image will be highly bright.

7. Remote sensing is the technique o obtaining inormation about an object, area or phenomenon,aquired by a sensor that is not in direct contact with the target o investigation.

8. The AM wave contains three rquencies, viz, fc,f

c+f

mandf

cf

mwheref

cis the requency o

carrier wave andfm is the requency o modulating signal. 9. Initial potential energy o the system

Ui=

r

q q

r

q q

r

q q

41

0 12

1 2

23

2 3

31

3 1

p + +> H

= 9 1091 2 2 3 3 1

1 1 1# # #

+ +; E= 9 109 11 = 9.9 1010 J

Final potential energy o the system

Uf

= 9 109.

1 2.

2 3.

3 10 5 0 5 0 5

# # #

+ +< F

1C

3C2C 1m

0.5m= 9 109 22 = 19.8 1010 J

\ Work done = Uf

Ui= 9.9 1010 J

10. Without the slab the capacity C0 = V

q

0

0

When slab is inserted the p.d., Vbecomes

V= E0 .d

K

E d

4 430+

= E0d K K

E d

41

43

40+ =< F (K+ 3) = V0 K

K

43+c m

\ C = Vq KK340 = +d nC0

11. Let A be the area o cross section o a wire and n be the number o ree electrons per unit volume.Iv

dis the drit velocity o the electron along the wire then the number o electrons passing

in time t through a cross section o the wire will be given by nA Vd. Ie is the charge on each

electron, the charge q passing in time t through the cross section o the wire will be equal to


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2 CBSE Ps-2013 S (Physics-XII)

q = (n Vd

A t)e

As I =t

q;I= n e A V

d ... (i)

Also Vd

= ,m

e E substituting in equation (i)

I = n e Am

e E

m

n e A E2 = ... (ii)

Also E =l

V, substituting in equation (ii)

I =m l

n e A V 2 , on rearranging, we get

V =n e

m

A

l2

d nI=R I, If R the resistance o the wire is constant then V I

This is Ohms law.12. Lenz's law states that the induced e.m.. is developed so as to oppose the cause o its production.

This helps in deciding the direction o induced current when time varying magnetic eld inducese.m.. As the loop PQRS moves, the fux linked with the loop increases. Hence, the induced current

in the loop should fow in such a direction as to oppose this increase o fux. So the current shouldfow anticlockwise i.e., along path SRQP.

OR

Sel induction o a coil is numerically equal to the magnetic fux linkage through the coil whenunit current fows through it.We know that, the magnetic eld inside an air lled long current carrying solenoid is,

B = m0 lN

i, where m0 is the permeability o ree space,N, the total number o turns o the solenoid

and l is its length.

Then, f = BA =l

Ni A0m d n

and total magnetic fux (fux linkage) through the solenoid is equal to

Nf = N lN i A0m d n = lN iA0

2

m Wb-turns

As, Nf = L i,The sel inductance o the solenoid is equal to

L =i

Nf=

l

N iA02m

=l

N r02 2m p

[Q A = pr2]

Hence, LN2.INis doubled,L will become 4 times the original value.

13. (i) When a magniying glass is not used, an objected to be placed at a distance o 25 cm romthe eye. But the use o a magniying glass allows us to place the object much closer to theeye. The closer object has larger angular size. It is in this sense that a magniying lensproduces angular magnication.

(ii) The magniying power o a compound microsocope is given by

m = m0 me = uv

f

D1

e0

0 +e o =u f

f

f

D1

0 0

0

e

# +e oIu0f0, m0 is large. As u0 is small sof0 is also small and ife is small, me is large. Hencebothf0 andfe should be small or high magnications.


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14. The results o de Broglie concept o matter-waves were conrmed by Davisson and Germeror slow electron and by G.P. Thomson or ast electrons. The only dierence between that thebeam o X-rays was replaced by a beam o electrons which was accelerated by an electric eld.Davisson and Germer in 1927 perormed experiments on the refection o slow electron romthin crystal suraces. The refected beam was received in an ionisation chamber connected to aquadrant electrometer.

They ound that the relected electrons weremore concentrated in certain directions thanothers. They came to the conclusion that a singlecrystal acted as a 3-D diraction grating or thede Broglie electron waves. They observed sharp peaksor certain angles (50) o diraction and calculatedde Broglie wavelength o the electrons. They alsoound that or dierent accelerating potentials thebeam was diracted in dierent directions and thus calculated the wavelength by using Braggslaw which was ound to be in accordance with de Broglie equation. Using Nickel crystal it wasobserved that or electrons accelerated through 54 volts the angle o diraction was ound tobe 50.

Using d sin q = nl and putting d = 2.15 and n = 1,

2.15 sin 50 = l = 1.65 Applying de Broglies equation

l =.

54

12 27= 1.669 , which is very close to the experimental result.

Thus, we nd that electrons behave like a train o waves when passing through space betweenthe atoms o a crystal or through a narrow slit, but when they strike against matter they behavelike particles.

15. As E(eV) =( )A

12420l

=3000

12420= 4.140 eV

\ Kmax=eV

c= 0.59 eV

f = hv Kmax = 4.14 0.59 = 3.55 eV

16. intrn emondutor Extrn emondutor

(a) It is a semiconductor in pure form. (a) It is a doped semiconductor.

(b) No impurities are added. (b) Impurities like In, Ga (Which are trivalent)

or As, Sb(which are pentavalent) are added

in controlled manner.

(c) Electrical conductivity is low. (c) Electrical conductivity is high.

(d) The number of free electrons in conduction

band is equal to number of holes in valence

band i.e., ne

= nh

= ni

(d) Forp type semiconductor

nen

h= n

i2, n

h;N

A>> n

e

wherene

= number density of electrons in C.B,

ni

= number density of intrinsic carriers,

NA

= number of acceptor atoms. Similarly

for n type semiconductor ne

>> nh

(e) Both electrons and holes take part in

conduction.

(e) In p-type, holes are the majority charge

carriers and in n-type, electrons aremajority charge carriers.


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17. (i) When two charged conduction are touched then rearrangement o charge occurs till boththe conduction have equal potential. Then new charge will be

q1 = C1Vand q2 = C1V= C2V

Where C1 and C2 are their capacities and Vis the common potential.

Hence,q

q

2

1

l

l=

C

C

2

1

(ii) Let two charged conductors A and B are having capacities C1 and C2 and their initial chargesand potentials be q1, q2 and V1, V2 respectively. Also assume that their initial energiesbe U1 and U2. As we assume that V1 > V2, on joining the two bodies A and B, the chargewill fow rom body A to bodyB, till their potentials are common. During this process, thesystem loses energy which we have to calculate.

The total charge o the system q1+ q2 can be written asq1+ q2 = C1 V1 + C2 V2

and the total energy o the system beore joining the conductors will be

Ui

=21

C1 V12 +

21

C2 V22 ...(i)

On joining the conductors, the sharing o charge takes place and both the bodies acquirecommon potential V given by

V =arge

Total capacityTotal Ch

=C C

C V C V

1 2

1 1 2 2

++

The nal energy o the system ater joining the conductors will be

Uf =

2

1(C

1+ C

2)V2

or Uf

=( )

C C

C V C V

21

1 2

1 1 2 22

++

...(ii)

on nding the value o Ui

Uf, we have

Ui U

f =

( )( )

C V C V C C

C V C V C C V V

21

21

21 2

1 12

2 22

1 2

12

12

22

22

1 2 1 2++

+ +> H=

( )( 2 )

C C

C V C C V C C V C V C V C V C C V V

2 1 212

12

1 2 12

2 1 22

22

22

12

12

22

22

1 2 1 2

++ + +

= ( )C C

C CV V

21

1 2

1 21 2

2

+

As (V1

V2

)2 will always be positive, Uf

< Ui

. Hence, there is always a loss o energywhen two charged conductors are joined together. The dissipated energy UiUf appearsas heat, light or sound i sparking occurs.


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OR

1 mF

3 mF

2 mF

A B

6 mF

U=21

CV2

U6

=21

6 106 V2 = 27 J (Given)

\ V2 =6 10

27 26

#

#

= 9 106

Similarly, U2

=21

2 106 9 106 = 9 J

and U3

=21

3 106 9 106 = 13.5 J

\ Total energy in parallel combination

Up

= U2

+ U3

+ U6

= 49.5 J

Effective capacity of parallel circuit = 2 + 3 + 6 = 11 mF

\ 49.5 =11 10

q

21

6

2

#

UC

q

2

2

== GAs parallel circuit is in series with 1 mF capacitor, it has the same charge q

\ U1

=C

q

21

1

2

=2 1 10

49.5 2 11 10

6

6

# #

# # #

= 544.5 J

\ Total energy in the arrangement

U= 544.5 + 49.5 = 594.0 J

18. Resistance o wireAJ= 1010040

# = 4 W

Current through wireAJ,

G

40 cm

100 cm

10 mV

J

+

+

R

A B

I

2 V

I=RV=

10 104

3#

= 2.5 103 A

In principal circuit

I=R 10

2+

R + 10 =2.5 10

23

#

= 800

\ R = 790 W.

19. (i) The electrons revolve round the nucleus in an atom. The orbits o electrons may be thoughto as the small current loops. As a result, the atom possess magnetic dipole moment andhence behaves like a magnetic dipole.

Let an electron o chargee and mass m be revolving round

the nucleus o an atom in an orbit o radius r. The angularmomentum due to orbital motion o the electron is given by

L = mvr

where v is the linear velocity o the electron. (angularmomentum is deined as the moment o linearmomentum).


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The angular momentum vector will be perpendicular to the plane o the orbit and upwardsor the given direction o rotation o the electron.

Let Tbe the time period o rotation o the electron. Then, T=v

r2p

and i =T

e

r

ev

2p=

The areaA o the orbit is pr2. \ The magnetic dipole moment o the atom,

M = iA =r

evr

evr

2 22

#

pp =

As we know that, vr =m

L

\ M =m

eL

2e o

In vector notation,

M = m

eL

2e o

As electrons are negatively charged particles, the negative sign is used in the above

expression. In case o an electron, the magnetic dipole moment will be in opposite directiono the angular momentum.

In Bohrs theory o hydrogen like atoms, the angular momentum o an electron in a stationary

orbit is equal to integral multiples oh

2porL =

nh

2p

\ M =m

e nh

2 2pe o = n

m

eh

4pd n

(ii) Retentvt, coervt and htereWhen a specimen o erromagnetic material is placed in amagnetic eld, the specimen is magnetised by induction.

As the magnetic intensityHis varied, the fux densityB in the (erromagnetic) material does not vary linearly

withH(Intensity o magnetisation). In other words, thepermeability.

m is not a constant, but varies withHand also dependsupon the past history o the material.

The variation inB with variation inHis shown in thegure. The point O represents an initially unmagnetisedspecimen and a zero magnetic intensity. AsHis increased,

B also increases, but not uniormly, and a point such as a is reached.

IHis now decreased,B also decreases but ollowing a path ab (and not the original pathaO). Thus,B lags behindH. WhenHbecomes zero,B still has a value equal to Ob. Thismagnetic fux density remaining in the specimen in the absence o any external eld is called

the residual magnetism. The power o retaining this magnetism is called the retentivityor remanence o the specimen. Thus, the retentivity o a specimen is a measure o themagnetic fux remaining in the specimen when the magnetising orce (magnetic intensity)is removed.


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When the magnetic intensity is now increased in reverse direction the value oB decreases,and still lags behindH. It becomes zero whenHhas a value equal to Oc. This value o themagnetic intensity is called the Coercive force o the specimen.

Coercivity can be dened as the measure of the magnetic intensity required to destroy theresidual magnetism of the specimen.

When H is increased beyond Oc, the specimen is increasingly magnetised in theopposite direction and a point such as d is reached where the substance is magneticallysaturated.

By takingHback rom its negative maximum value, through zero, to its original positivemaximum value, a symmetrical curve is obtained. At point b ande where the substance ismagnetised in the absence o any external magnetic intensity, it is said to be a permanentmagnet.

It is ound that the fux density B always lags behind the magnetic intensity H, whenHchanges. The lagging oB behind H is called Hysteresis. The closed curve abcdefarepresents a cycle of magnetisation. The curve is called a hysteresis curve.

20. GivenBy

= (8 106) sin (2 1011 t + 300px) Tesla

Oncomparing with standard equation. B

y=B0 sin T

t x2p

l+d n< F or By =B0 sin T t x

2 2pl

p+< F, we get

B0 = 8 106 T,

T2p = 2 1011

2l

p= 300p

(i) l, wavelength =300

2p

p=

1501

m = 0.67 cm

(ii) E0 = cB0 = 3 108 8 106 = 2400 V m1

(iii) Ex

=E0 sin Tt x

2pl

+d n< F = 2400 sin [2 1011t + 300px] V m1 21. (i) Let the incident ray meet reracting aceAB o the prism (at angle o incidence i) at point

P. RayPQ is the reracted ray inside the prism and 1 and r1 are the angles o deviationand reraction at interaceAB.At interace AC the ray goes out o the prism. Let e be the angle o emergence. Theangle o deviation at point Q is

2as shown in the gure.

Using geometry, we see that at point P, i =

1+ r1 \ 1 = i r1

and at point Q, e = 2

+ r2\

2= e r2

The total deviation , suered by the incident ray is equal to 1

+ 2

or =

1+

2= (i r1) + (e r2)

= (i + e) (r1 + r2) ...(i)

In quadrilateral POQA, the sum o all our angles is 360.


8/18


Principal section of a

equilateral glass prism

Emergent

ray

Incident

ray 2

1

iP

B C

O

Q

A

A

r1

r2 e

or P+ O + Q + A = 360

as Pand Q both are right angles

P+ Q = 180

\ O + A = 180 ...(ii)

In triangle POQ,

O + r1

+ r2

= 180 ...(iii)

Comparing equations (ii) and (iii), we have A = r

1+ r

2...(iv)

Substituting this value in equation (i)

= i + e A

or +A = i +e ...(v)

(ii) = (i +e) AOnly at minimum deviation i =e and we have single value o

m. For all other values same

is obtained or corresponding values oi ande respectively.

m

i

22. Consider a beam o parallel rays betweenLO andPD incidenton the plane surace o a water medium rom air in thedirection shown, and suppose that a plane waveront hasreached the position OA at a certain instant, see Fig.Each point between O and A becomes a new centre odisturbance as the waveront advances to the surace othe water, and the waveront changes in direction whenthe disturbance enters the liquid.Suppose that, t is the time taken by the light to travelrom A to D. The disturbance rom O travels a distanceOB, or vt, in water in a time t, where v is the velocity olight in water. At the end o the time t, the waveronts in


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the water rom the other secondary centres between O and D reach the suraces o spheresto each o which DB is a tangent. Thus, DB is the new waveront in the water, and the rayOB which is normal to the waveront is consequently the reracted ray.Since, c is the velocity o light in air,

AD = ct. Now,

\ sin

sin

r

i

= sin

sin

BOM

LON

= sin

sin

ODB

AOD

sinsin

r

i=

//

OB OD

AD OD=

OB

AD=

vt

ct=

v

c... (i)

But c, v are constants or the given media,

\sinsin

r

iis a constant,

which is Snells law o reraction.

It can now be seen rom (i) that the reractive index m o a medium is given by m =vc

where,c is the velocity o light in air and v is the velocity o light in the medium.

23. There are 26 protons and 30 neutrons in 26Fe56

\ Z mp + (A Z) mn = 56.463400 amu M26F

56 = 55.934939 amu.

\ Dm = [Zmp

+ (A Z)mn] M

= 56.463400 55.934939

Dm = 0.528461 amu

\ B.E. o 26F56 = 0.528461 931 = 491.99 MeV 492 MeV

\ B.E. per nucleon =56492

= 8.79 MeV/amu.

Doing similar calculation or 83Bi209, Dm = 210.741265 208.980388 = 1.760877 amu.

\ B.E. o83

Bi209 = 1.760877 931 = 1639.37 MeV

\ B.E. per nucleon =.

2091639 37

= 7.84 MeV/amu

Hence, Fe has greater B.E./Nucleon than Bi.

24. (i) In hydrogen like atom, there is only one electron in outer orbit. For example singly ionizedHe atom or doubly ionized Li atom may be considered as single electron system. We know

that or the moving electrons, centripetal orce, r

mv

n

n2

is equal to Coulombian attraction

between nucleus and the electron,( )

r

kZe e

n2

.

r

mv

n

n2

=r

kZe

n2

2

or mvn2

= rkZe

n

2

(i) \Also the angular momentum o revolving electrons is quantised.

\ mvnr

n =

nh

2p

or mvn

=r

nh

2 np(ii)


10/18


Multiplying equation (i) by m and squaring equation (ii), we get let hand side equalto m2v2. Hence equating the R.H.S, we have.

r

kmZe

n

2

=r

n h

4 n2 2

2 2

p

\ rn =

kmZe

n h

4 2 2

2 2

p ...(iii)

This is the radius o the nth preerential orbit o a hydrogen like atom.

Putting the value o k =4

10p, we have

\ rn

=4

mZe

n h

4 2 2

2 20#

p

p=

mZe

n h2

2 20#

p

; r

n n2.

The radius o rst orbit o hydrogen, (Z = 1) will be

r1 =me

h2

20

p

= a0.

This is called the Bohr radius, represented by a0. Its numerical value comes out to be5.29 1011 m or 0.53 .

(ii) As n = 3 and h = 6.6 1034 Js

L = nh2p

=2 3.14

3 6.6 1034#

# # = 3.07 1034 Js.

25. A radio wave directed towards the sky and refected by ionosphere towards the desired locationo the earth is called a sky wave propagation. Radiowaves o requencies between 2 MHz to 30 40 MHz, can be refected by ionosphere. This region o AM band is called short wave band.These waves radiated rom a certain point and ater being refected by the ionosphere can bereceived at another point on the surace. This is known as sky wave or ionospheric propogation.The radiowaves can travel very large distances in this way.

The reractive index o ionosphere is less than its ree space value thus the ionosphere behavesas a rarer medium. Consequently, the EM wave gradually turns away rom the normal as itpenetrates the ionosphere. The reraction or bending o the beam continues till it reaches critical

angle and ater which, it gets refected back.

RReceiver

R'R''

IONOSPHERE

Transmitt

er

I the requency o transmission is too high, then ater a certain value the electron density in theionosphere may never be so high as to produce enough bending or critical angle or the conditionor refection. This value is called critical requency. I maximum electron density o ionosphereisNmax per m

3, then critical requencyfc 9(Nmax)

1/2

The value ofcranges rom about 5 to 10 MHz. Frequencies higher thanf

c, cross ionosphere and

do not return back to the earth.


11/18


26. (a) Enterprising, witty and has thorough knowledge o the concept o conversion o galvanometerinto voltmeter.

(b) I G be the resistance o the galvanometer o 3 V range. I a high resistance R is joined inseries than

GRi

3 V

i =G R

3+

VG

= 5 mV= i G =G R

3+

d nG \ G +R =

5 10

G36

#

or R = 3 10G G5

6#

-c mI G = 100 W (suppose) then

R =

5

300 106 100 6 107W

This is the required resistance or conversion o voltmeter into given range.

27. (a) R =IV W

XL

, inductive reactance = wL = 2pfLW

XC

= Capacitative reactance =wC

1=

2 fC1p

W

Z = ( )R X XL C2 2+ is called impedance.

(b) 100 mF 40 W

100 V

60 Hz

(i) R = 40 W

(ii) XC

=2 fC

1p

=2 60 100 10

16

# # #p= 26.5 W

Z = R XC2 2+ = 1600 702+ = 2302 48 W

(iii) \imax = RV

40100

= = 2.5 A

OR

AlternatingCurrent Generator (A.C. Dynamo).A device used to convert mechanical energy into electrical energy is called an electric

generator.Principle. An alternating current generator, designed by Nikola Tesla, is based upon theprinciple o electromagnetic induction. When a closed coil is rotated rapidly in a strongmagnetic eld, the number of magnetic ux lines (magnetic ux) passing through the coilchanges continuously. Hence, an e.m.f. is induced in the coil and a current ows in it in a

direction given by Flemings right-hand rule. In act, the mechanical energy expanded inrotating the coil appears as electrical energy (current) in the coil.


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Construction. An a.c. generator consists o the ollowing parts :

(i) Armature. It is a rectangular coil ABCD (Fig.) having a large number o turns oinsulated copper wire wound over a sot-iron core. The core increases the magneticfux linked with the armature. (not shown in the diagram)

Armature

Field

magnet

Slip rings

Brushes

Ac Generator

(ii) Field magnet. It is a powerul permanent magnet having concave pole-pieces N andS. The armature is rotated (say, by a water turbine) between these pole-pieces aboutan axis perpendicular to the magnetic eld lines.

(iii) Slip rings. The leads rom the armature coil ABCD are connected to two copperrings R1 and R2 called the slip rings. These rings are concentric with the axis o thearmature-coil and rotate with it.

(iv) Brushes. These are two carbon piecesB1 andB2 called brushes which remain stationary,pressing against the slip-rings R1 and R2 respectively. The brushes are connected tothe external circuit in which current is to be supplied by the generator.

Working. As the armature-coil ABCD rotates, the magnetic fux linked with it changes.Hence, an e.m.. is induced in the coil and current fows in it. Suppose, the coil is rotatingclockwise and is horizontal at some instant o time. At this instant, the arm AB o the coilis moving up and the arm CD is moving down. By Flemings righthand rule, the current

fows rom A to B in the arm AB and rom C to D in the arm CD, as shown. The brushB1 at which the current is entering the external circuit is positive relative to the brush B2at which the current is leaving the external circuit. During the time the coil rotates to thevertical position, the current in the armature will fow in the direction ABCD.When the plane o the coil reaches the vertical position, the arms AD and CD move momentarilyparallel to the eld. Thereore, at this instant, the induced e.m.. and hence, the current is zero.As the coil rotates rom its vertical position, the arm AB moves down and the arm CD movesup. Hence, the direction o current in the coil reverses. The brush B2 is now positive relativeto B1 and remains so long as the coil reaches the vertical position again. Ater this, B1 isagain positive relative to B2.Thus, an alternating potential dierence is developed between B1 and B2 generating analternating current in the circuit.

Expression for the Induced e.m.f. In Fig. (a) is shown a coil rotating clockwise abouta horizontal axis perpendicular to a magnetic eld B . Suppose, we start timing rom theinstant when the plane o the coil is perpendicular to the eld B . In this position themagnetic fux linked with the coil is maximum.


13/18


As the coil rotates, the magnetic fux linked with it changes. Suppose, in t second, the coilrotates through an angle q. At this instant, the normal to the plane o the coil makes anangle q with the direction o the eld B (Fig. b) and the component oB perpendicular to theplane o the coil is B cos q. IA is the area o the plane o the coil, then the instantaneousmagnetic fux linked with the coil is

f = (B cos q) AI is the angular velocity o the coil, then q = t\ f = BA cos tThe rate o change o magnetic fux is

dt

df= BA sin t.

By Faradays law o electromagnetic induction, the e.m.. induced in each turn o the coil is df/dt. I the coil has Nturns, the e.m.. induced in the coil is given by

e = Ndt

df= NBA sint.

This shows that the magnitude of the e.m.f. induced in a coil rotating in a magnetic eldchanges continuously with time. The maximum value o sin t is 1 and so the maximumvalue oe is NBA. Let it be e0 that is,

e0 = NBA

Thus, e = e0 sint.

In the upper part o Fig. shows the positions o the coil at dierent times and in the lowerpart is shown the time-variation o the e.m.. induced in the coil rotating with a constantangular velocity in a uniorm magnetic eld B

.

The e.m.. is zero when the coil is perpendicular to the eld (q= t = 0, 180, 360) and is

maximum when the coil is parallel to the eld (q = t = 90, 270).


14/18


28. Letf,flandFbe this ocal lengths o lens, liquid lens and combination o the two. Then

F

1=

f f

1 1

l

+ orf

1l

= f F

1 1e oAlso

f

1l

=( 1)

R

m

\

1R

m

= f F1 1

Alsof

1=

R23

12d dn n =

R

1 \ f=R

Substituting

f

1m=

f F

1 1

f f

1m=

f F

1 1

f

m=

f F

2 1or m = 2

F

f

\ m = 2 4530

= 2 32

=34

OR

(a) compound Mroope :

Formaton of mage : Suppose AB is a small object placed slightly away rom therst ocusF0 o the objective O (Fig.) which orms a real, inverted and magnied image

AB. This image lies between the eyepieceE and its rst ocusFe

and acts as an objector the eyepiece which orms a magnied, virtual nal imageAB. To nd the positionoB, two dotted rays ( ) are taken rom B. One ray, which is parallel to theprincipal axis passes, ater reraction, through the second ocus Fe oE. The other raywhich passes through the optical centre oE travels straight. Both the reracted rayswhen produced backward meet at B. The imageAB is generally ormed at the leastdistance o distinct vision although it can be ormed anywhere between this position

and innity. The rays by which the eye sees the image are clearly shown in the Fig.


15/18


Magnfng Power : Suppose, the nal imageABsubtends an angle at the eyepieceE. Since, eye is very near to the eyepiece, the angle can also be taken as subtendedby AB at the eye. Suppose, when the object AB is at the least distance o distinctvision D, then it subtends an angle at the eye. The magniying power o microscopeis

M =

ce

l

tanngle subtended by the object at the eye whenplaced directly at least o distinct vision

nglesubtended by the ina image at the eye

A

A

dis

=

Since, and are very small, we may write, = tan and = tan . Thus,

M =tantan

.

From the geometry o the gure, tan = AB/EAand also, tan = AB/D

\ M =//

AB D

A B EA

AB

A B

EA

D=

l l l l l

ld n.

I the distances o the object AB and the image AB rom the objective O be uo and vorespectively, then rom the magnication ormula we have (taking proper signs).

AB

A Bl l=

u

v

0

0+

Similarly, i the distance oAB rom the eyepiece be ue, then EA= ue Also, D is tobe replaced by D. Then, rom the above ormula, we have

M =

u

v

u

D

e0

0 d n = u

v

u

D

e0

0 d n ...(i)Now, there are two possibilities.

(i) The nal image is formed at the least distanceD of dtnt von : Ithe distance o the nal image ABrom the eyepiece be D, then in applying the

lens ormula, v u

1 1=

f

1or the eyepiece,

we shall have

v = D, u = ue

and f= + fe

where fe

is the ocal length o the eyepiece. Now, we get

or

D u

1 1e

=f

1e

...(i)

oru

1e

=D f

1 1e

+

u

D

e

=f

D1

e

+

Substituting this value oD/ue in eqn. (i), we get

M = u

v

f

D1

e0

0 +e o ...(ii)In this position, the length o the microscope will be v

0+ u

e.


16/18


(ii) Thenalimageisformedat innity: To see with relaxed eye, the nal imageAB should be ormed at innity (Fig.). In this case the image AB will be atthe ocus F

eo the eyepiece E, i.e., u

e= f

e. Substituting this value in equation (i),

we get the magniying power o the relaxed eye, which is given by

M = u

v

f

D

e0

0 e o ...(iii)In this position, the length o the microscope will be v0 + feIn ormulae (ii) and (iii), we shall substitute only the numerical values o v0, u0,

fe

and D. Negative sign shows that the nal image is inverted.

It is clear rom these ormulae that in order to increase the magniying power omicroscope.

(i) u0 should be small, i.e., the objectAB should be placed quite close to the objectiveO. But, to obtain a real and magnied image o the object, the object should beplaced beyond the ocal length f0 o the objective. Hence, for greater magnifying

power of the microscope, the focal length of the objective should be small. (ii) The distance v0 o the image AB rom the objective O should be large. For this,

the object should be placed near the rst ocus o the objective. (iii) The ocal length,fe o the eyepiece should be small.

Thus, it is clear that the magniying power o the microscope depends upon the ocallength o both the lenses. Hence, by taking proper focal lengths the magnication canbe increased.

(b) As m = u

v

f

D

e0

0# =

f

L

f

25e0

# =.

2

156 2525

# = 30 Dioptre

29. I a donor impurity, which is pentavalent atom like phosphorus, antimony or arsenic is diusedinto one side o a crystal and an acceptor impurity, which is trivalent atom like indium, boron,aluminium or gallium is diused into the other, one region o the crystal will have excess oree electrons and exhibit n-type conduction and the other region will have excess o holes andexhibitp-type conduction. The boundary between these regions is called ap-njunction.Near the junction, due to the diusion ree electrons o the n-section combine with the holesop-section leaving unneutralised positive donor atoms and negative acceptor atoms in n and

p-sections respectively.This produces an electric eld across the junction as i a ctitious battery were connectedwith its positive terminal to n and negative to p section. It prevents the electrons and holesrom crossing the junction against the electrostatic orce so developed.The small region in the vicinity op-n-junction, where electrons and holes meet and combineand thus ceases to exit as mobile charge carriers, is called depletion layer. It is a zone oonly a ew micron in width.


17/18


Fg. (a)

P N

Fg. (b)

SemiconductorDiode as a FullWave Rectier

The two junction diodes D1 and D2 alternately supply rectied current to the band duringboth halves o the alternating input voltage and are always in the same direction. During thesuccessive hal cycles o the alternating input voltage, junction diodes D1 and D2 continue toconduct alternately, each permitting current to fow during one hal cycle whenever its p-terminal

is positive with respect to the n-terminal.

A AB

Output (pulsating d.c.)

A A

B

Input (a.c.)

Input

Central

tapping

RL Output

D1 +ve

ve

D2

The resulting output current is a series o unidirectional pulses.

The output requency called ripple requency is twice that o the alternating supply requency.The current is much less discontinuous than that o the hal wave rectier and the eciency isalso ar better than that o the hal wave rectier.

OR

In the common emitter or grounded emitter conguration, the input signal is applied across the baseand the emitter, while the amplied output signal is taken across the collector and the emitter.

Input

Output

R1

R2

B

C

ib

iC

iE

V0

Vi

E

~

~

A simple circuit of a CE transistor amplier


18/18


This is the most ecient amplier circuit. Its input resistance is some what higher and the outputresistance is lower than those o the common base circuit. It provides the highest voltage andpower gains. It however, reverses the phase o the output signal with respect to the input. Thed.c. current gain is dened as the ratio of the collector-current to the base current and is denotedby (d.c.).

\ (d.c.) =

i

i

b

c

In a typical transistor, a small base current ( 10 mA) can produce a large collector-current( 500 mA)

\ (d.c.) =10500

= 50

The value o lies between 50 and 200. The (a.c.) current gain is dened as the ratio o thechange in the collector-current to the change in the base-current at a constant collector-to-emittervoltage.

\ (a.c.) = ii

Vb

cce

D

Dd n = constant.Trans-conductance is dened as the ratio o the change in the collector current to the change inthe base to emitter voltage at constant collector to emitter voltage.

\ gm

=V

iV

be

cce

D

Dd n = constant.We see that,

gm

=i

i

V

i

b

c

be

b#

D

D

D

D= a.c. R

11

The unit ogm

is W1 or S (siemen)The a.c. voltage gain is dened as the ratio o the change in the output voltage to the change inthe input voltage and is denoted by A.

As AV

V

input

output=e o

\ A =i R

i R

b

c

1

2

#

#

D

D=

i

i

R

R

b

c

1

2#

D

D=

R

R

1

2

The a.c. power gain is dened as the product o a.c. current gain and a.c. voltage gain.

\ Power gain = a.c. A = R

R

1

2 = 2 R

R

1

2

This value is extremely large as compared to the value in common base amplier.

cbse 2013 xii physics solution set-ii

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