Marking scheme of question paper 2013 phyics set1_dl

Del hi SET I Page 1 of 19 FI NAL Print Draft 11/ 3/ 2013 11: 30 am
Strictly Confi dential (For Internal and Restricted Use onl y)
Seni or School Certificate Exa mi nation
Marki ng Sche me - Physics ( Code 55/ 1/ 1)
1. The marki ng sche me provides general gui delines t o reduce subjectivity i n t he marki ng. The ans wers
gi veni nt he marki ngsche me are suggested ans wers. The content ist hus i ndicated. If a st udent has gi ven
any ot her ans wer, whi ch is different fromt he one gi ven i n t he marki ng sche me, but conveys t he
meani ng correctly, such answers shoul d be given full wei ghtage.
2. In val ue based questions, any ot her i ndi vidual response wit h suitable j ustification shoul d also be
accepted even if there is no reference tothe text.
3. Eval uation is t o be done as per i nstructions provi ded i n t he marki ng sche me. It shoul d not be done
accordi ngt o one' s own i nterpretation or any ot her consideration. Marki ng sche me shoul d be adheredt o
and religiously followed.
4. If a question has parts, please a ward i n t he right hand si de for each part. Marks a warded for different
part of the question shoul d then be totaled up and written inthe left hand margi n and circled.
5. If a question does not have any parts, marks are to be awarded inthe left hand margi n onl y.
6. If a candi date has attempted an extra question, marks obtainedi nt he question atte mpt edfirst shoul d be
retained and the other ans wer shoul d be scored out.
7. No marks aret o be deducted for t he cumul ative effect of an error. The st udent shoul d be penalized onl y
once.
8. Deduct ½ mark for writing wr ong units, missing units, inthe final ans wer to numerical problems.
9. For mul a can be taken as i mplied fromthe calculations even if not explicitly written.
10. In short ans wer t ype question, aski ng for t wo feat ures/ characteristics / properties if a candi date writes
three features, characteristics / properties or more, only the correct t wo shoul d be eval uated.
11. Full marks shoul d be a warded t o a candi dateif his / her ans wer i n a numerical proble mis cl ose t o t he
val ue given inthe sche me.
12. In compliance t ot he j udgement of t he Hon’ble Supreme Court of India, Board has deci ded t o provi de
phot ocopy of t he ans wer book(s) t o t he candi dates who will appl y for it al ong wit h t he requisite fee
from2012 exa mi nation. Therefore, it is all t he more important t hat t he eval uationis done strictly as per
the val ue poi nts gi ven i n the marki ng sche me so t hat t he Board coul d be i n a position t o defend t he
eval uation at any forum.
13. The Exa mi ner shall also have t o certifyi n t he ans wer book t hat t hey have evaluated t he ans wer book
strictly i n accordance witht he val ue poi nts gi ven i n the marki ng sche me and correct set of question
paper.
14. Every Exa mi ner shall also ensure t hat all t he ans wers are eval uated, marks carried over t o t he title
paper, correctlytotaled and writtenin figures and words.
15. In t he past it has been observed t hat t he followi ng are t he common t ypes of errors committed by t he
Exa mi ners
Leavi ng ans wer or part thereof unassessed in an ans wer script.
Gi vi ng more marks for an ans wer than assigned toit or deviation fromthe marki ng sche me.
Wr ong transference of marks fromthe inside pages of the ans wer book tothe title page.
Wr ong question wise totaling on the title page.
Wr ong totaling of marks of the t wo col umns on the title page.
Wr ong grand total.
Mar ks in words and figures not tallying.
Wr ong transference to marks fromthe ans wer book toaward list.
Ans wer marked as correct ( ) but marks not awarded.
Half or part of ans wer marked correct ( ) and the rest as wrong ( ) but no marks awarded.
16. Any unassessed portion, non carrying over of marks tot he title page or t otaling err or detected by t he
candi date shall da mage t he prestige of all t he personnel engaged i n t he eval uation work as also of t he
Board. Hence i n order t o uphol dt he prestige of all concerned, it is agai nreiteratedt hat t he i nstructions
be followed meticulously and judiciously.

MARKI NG SCHEME
SET 55/ 1/ 1
Q. No. Expected Ans wer / Value Poi nts Marks Total
Marks
1. Substances, which at roomte mperat ure, retaintheir ferromagnetic property
for a long period of ti me are called per manent magnets. Al nico, cobalt, steel
and ticonal(any one)
½ +½ 1
2. Spherical . 1 1
3. Heat waves, as they are transverse/electromagnetic in nat ure ½ +½ 1
4. Magnitude of conduction &displace ment currents are zero. 1 1
5. 2mA i 1 1
6. ( 1, 3) and ( 2, 4) ½ +½ 1
7. 190
5
38
V
i A
R
Award full 1 mark if student calculates current directly
½ +½ 1
8. Because t he cell has some fi nitei nternal resistance./ Emf is deter mi ned when
the cell is in open circuit and no current is drawn.
1 1
9.
(a) i) Ray of light shoul d travel fromdenser torarer medi um.
ii) Angle of inci dence shoul d be more than the critical angle.
(b)
1
sin ci
where ic is the critical angle
½
½
1 2
10.
The polarity of induced emf is such that it tends to produce a current which
opposes the change in magnetic flux that produced it.
Yes, as t he magnetic flux due t o vertical component of Earth’s magnetic
keeps on changi ng as the metallic rod falls down.
1
½ +½ 2
11.
Power of convex lens,
½
Conditions ½+½
Rel ation 1
St atement of lenz law 1
Emf and justification ½ +½
Det er mi nation of power 1½
Nat ure ½

Rt ot al
Power of concave lens,
P2 =
Power of the combi nation P= P1 +P2 =-1D
Nat ure: Di vergi ng
½
½
½ 2
12.
(i) Shunt
A g
g
R i
S
i i
=
0.8 1.0
0.2
5.0 1.0
(ii) Co mbi ned resistance of ammeter and shunt
1 1 1
1 1
0.8 0.2
0.8
5
0.16
total A
total
R R S
R
½
½
½
½ 2
13.
(i) Increases.
As t he val ue of t he base current i ncreases, t he collector current
will increase proportionately.
(ii) Increases.
Due t oi ncrease i n collector current, voltage drop across la mp will
increase.
½
½
½
½ 2
14.
(i) Effect on Bri ght ness of the bul b and reason ½ + ½
(ii) Effect on volt meter reading and reason ½ + ½
(a) Sketch of propagation 1 ½
(b) Rel ation ½
(i) Val ue of Shunt Resistance 1
(ii) Combi ned resistance 1

(a)
[ NOTE: Accept the alternative choices indicating the correct directions of the
oscillating components of Eand B]
(b) 0
0
E
c
B
1 ½
½
2
15.
X: IF stage
Y: Amplifier
The carrier frequency is changed to a lower frequency by inter mediate
frequency (IF) stage precedi ng the detection.
It increases the strengt h of detected signal
½
½
½
½ 2
16.
Circuit diagramof an illumi nated phot odi ode:
½
Circuit diagramand worki ng 1½
Its use to detect the optical signal ½
1
Identification of Xand Y ½ +½
Function of Xand Y ½ +½

When t he phot odi ode is illumi nated wit h radiations (phot ons) wit h energy
(hν) greater t hant he energy gap( Eg) of t he se mi conduct or, t hen electron-hol e
pairs are generated due to the absorption of phot ons.
The j unctionfieldsends the electrons t o n-side and holes t o p-side t o produce
the emf. Hence current flows through the load when connected.
It is easier to observe the change inthe current with change inthe radiation
intensity, if a reverse bias is applied. Thus phot odiode can be used as a
phot odetect or to detect optical signals.
OR
I mportant considerations 1
Or der of band gap 1
1. It is a heavily doped p-n junction.
2. The reverse breakdown voltages of LEDs are very low.
3. The se mi conduct or used for fabrication of visible LEDs must
atleast have a band gap of 1. 8 eV
( Any t wo of the above)
Or der of band gap is about 3 eVto 1. 8 eV
1
½
½ + ½
1 2
17.
1. Practical Size of the antenna or aerial
2. Effective power radiated by an antenna
3. Mi xi ng up of signals fromdifferent trans mitters
½
½
½
½
½
½ 3
18.
I mportant factors justifying the need of modulation 1½
Di agra mshowi ng, howAM wave is obtained 1½
(i) Calculation of potential Vand unknown capacitance C 2
(ii) Calculation of charge stored Q’ 1
+1/ 2

(i)
Q=CV
360
120 = CV – 120 C
= 360
120 = 360
= 240
 Capacitance C = 2
Substituting the val ue of C
Potential V= 180V
(ii) Charge stored when voltage is increased by 120 V
Q’ = 2
= 600
OR
(i) Cal culation of net electric flux 2
(ii) Calculation of charge 1
(i)The magnitude of the electric field at the left face is E= 50 NC- 1
Therefore flux through this face
=
The magnitude of the electric field at the right face is E= 100 NC- 1
Therefore flux through this face
(ii) Charge encl osed by the cylinder
C
½
½
½
½
½
½
½
½
½
½
½
½
3
3

19.
(a) Si nce t he t otal i nitial mass of nuclei on t he left si de of reaction is
greater t hant he t otal final mass of nucleus on t he right hand si de, t his
difference of mass appears as the energy released.
(m= mass of each nucleon)
As R= R0 A⅓
Therefore N
=
1
½
½
½
½ 3
20.
(a) Reasons of failure of wave theory to explain Photoelectric effect. 1 ½
(b) Basic features of Phot on picture 1 ½
(a) Accordi ng to wave theory
(i) The maxi mu m ki netic energy of t he e mitted electron shoul d be
directly proportional t othe i ntensity of i nci dent radiations but it is
not observed experi mentally. Also maxi mumki netic energy of t he
e mitted electrons should not depend upon i nci dent frequency
accordi ng to wave theory, but it is not so.
(ii) El ectron e mmi ssion shoul d take place at all frequencies of
radiations i.e. t here should not exist t he t hreshol dfrequency. This
fact contradicts experi mental observation
(iii) There shoul d be a time l ag i n phot oelectric e mmi ssion but
accordi ng to observation phot oelectric emmi ssion is instantaneous
(b) Accordi ng to phot on picture
(i) Each quant umof radiation has energy h
½
½
½
½
(a) Cause of release of energy 1
(b) Proof for independence of nuclear density on mass number 2

(ii) In phot oelectric effect the electrons i n t he metal absorbs t his
quant umof energy (h )
(iii) When t his energy exceeds t he mi ni mu m energy needed for t he
[ Award 1 mark of this part if student writes h = 0 + Kmax ]
½
½ 3
21.
As t he rod is rotated, free electrons i n t he rod move t owards t he out er end
due t o Lorentz force and get distributed over t he ri ng. Thus, t he resulting
separation of charges produces as emf across the ends of the rod.
The magnitude of t he emf generated across t he lengt h ‘ dr’ of t he rod as it
moves at right angle tothe magnetic fieldis given by
½
1
½
½
½ 3
Explanation, howe mf is induced 1 ½
Deri vation of the expression 1 ½

Ib
22.
(1) Dyna mi c out put resistance
C
CE
I
V
r0
Fromt he graph = (12 - 8) =4V
=( 3. 7 - 3. 5) mA
= 0.2 mA
r0 = = 20 K
(2) dc current gai n, at 10 V, CI = 3. 6 mA
=
b
c
I
I
= 6
3
1030
106.3
x
x
=120
(3) ac current gai n
bI = 40 µA – 30 µA = 10 µA
cI = 4. 7 mA – 3. 6 mA = 1. 1 mA
ac =
b
c
I
I
= 6
3
1010
101.1
x
x
=110
[ NOTE: Credit shoul d also be gi ven to candi date who uses the right
procedure, but consi ders the val ues slightly different fromt hose used above]
½
½
½ + ½
½
½ 3
23.
½
Det er mi nation of
(i) Dyna mi c out put resistance ½ + ½
(ii) d.c current gai n ½ + ½
(iii) a.c current gai n ½ + ½
Deri vation of expression for total energy of the electron 2
Energy level diagra mfor Bal mer series 1

Also =
 (ii)
From(i) and (ii)

Ki netic energy
KE =
=
Pot ential energy
Substituting the expression for
U = -
Tot al Energy
TE = KE + U
EnergyLevel Diagra m of Bal mer Series
½
½
½
1 3

24.
a) In double slit experi ment, t he pattern on t he screen is act ually a super
position of si ngle slit defraction from each slit and double slit
interference pattern. As a result, t here appears a broader diffraction peak
in whi cht here occur several fringes of s maller widths due t o double slit
interference.
b) Di stance of first secondary maxi mu mfromcentre of the screen
x =
a
D
2
3
Therefore spaci ng bet ween first secondary maxima on t he screen for two
gi ven wavelengt hs
x = 12
2
3
a
D
= 4
1022
5.13
590596 × 10 -9
=
4
1065.4 5
=6. 75× 10- 5
m
1
½
½
½
½ 3
25.
Condition for resonance XL = XC
1
½
(a) Relationshi p bet ween interference pattern and diffraction fromeach slit 1
(b) Calculation of separation bet ween the position of first maxi ma of t wo
wavelengt hs 2
Pl ot of variation of current with angular frequency 1
Condition for resonance ½
Val ue of resistance for sharper resonance ½
Defi nition of Q– factor and its significance ½+½

26.
a) Because during thunder stor mcar woul d act as an electrostatic shield
b) Dr. Pat hak displayed values of safety of human life, helpful ness,
e mpat hy and scientific temper. ( or any ot her t wo relevant val ues)
c) Gr ateful ness , indebtedness ( or any ot her relevant val ue)
d) Exa mpl e of any si milar action
1
½ + ½
1
1 4
27.
Magnification of objective
m0 =
h
h'
=
0f
L
Angular magnification due to eyepiece
Tot al magnification when i mage is for med at infinity
1
½
½
Alternativel y, =
LC
I
(or any ot her correct
Resonance will be sharper for resistance R2
Q fact or is defined as
Si gnificance of Qfact or
For large Qfact or, resonance will be sharper and t herefore circuit will be
more selective
½
½
½
3
Four parts 1 mark for each part
(a) Ray diagra mshowi ng i mage for mation 1
Derivation of expression for magnification 2
(b) Distinction bet ween myopia and hyper metropia 1
Correction of defects by diagra m. 1

m= m0 me
=
(b)
Myopi a Hyper metropi a
1. Distant object arriving at the eye
lens get converged at a poi nt infront
of the retina
1. Eyelens focuses the incomi ng
light behi nd retina
2. The eye ball is elongated 2. The eye ball is shortened
3. Person cannot see distant objects
clearly.
3. Person cannot see nearby objects
clearly.
( Any t wo or any ot her correct ans wer)
Myopi a can be corrected by
interposi ng a concave lens bet ween
eye and object
Hyper metropia can be corrected by
interposi ng a convex lens bet ween
eye and object
[ Award onl y half mark if diagra ms not drawn, award full mark even if
explanation is not written]
OR
(a) Accordi ngt o Huygens princi ple, each poi nt of t he wavefront is t he source
of a secondary disturbance and t he wavelets emanating fromt hese points
spread out i n all directions wit ht he speed of t he wave. Acommon t angent t o
all these wavelets, gives the new position of the wavefront at a later ti me.
½
½
½ + ½
½ + ½
1
5
(a) Statement of Huygen’s princi ple 1
Di agra m 1
Verification of Snell’s law 1
(b) Explanation of (i) and (ii) 1+1

Verification of Snell’s law
Fromfigure
sini =
AC
BC
=
AC
tv1
sin r =
AC
AE
=
AC
tv2
r
i
sin
sin
=
2
1
v
v
=
(b) Yes,
(i) Reflection and refraction arise through interaction of incident light
wit h t he at omi c constituents of matter. At oms may be viewed as oscillators,
whi ch take up t he frequency of t he external agency (light) causi ng forced
oscillations. The frequency of light e mitted by a charged oscillator equals its
frequency of oscillation. Thus, t he frequency of scattered light equals the
frequency of incident light.[ Any ot her correct explanation]
(ii) No. Energy carried by a wave depends on the a mplitude of the wave, not
on the speed of wave propagation.
1
½
½
1
1 5
28.
(a) Princi ple : When a constant current flows through a wire of unifor marea of
cross section then potential difference bet ween t wo poi nts on the wire is
directly proportional tolengt h of this section of wire.
Vα l
1
(a) Worki ng princi ple of potentiometer 1
Di agra m 1
Expression 1
(b) Two possible causes for one sided deflection 1+1

2 = l2
=>
2
1
=
2
1
l
l
(b) (i) When the driver cell/ source cell has emf less than the emf of the cells
to be compared.
(ii) When the positive end of the potentiometer wire is connected to
negative ter mi nal of the cell whose emf is to be compared / deter mi ned
OR
(a)(i) Al gebraic sumof the currents entering the junction is equal tothe sum of
currents leavi ng the junction.
(ii)The Al gebraic sumof the changes in potential around any closed loop
invol vi ng resistors and cells is zero.
[ Alternativel y accept the mat he matical for mof the Kirchhoff’s rule]
1
½
½
1
1
½
½
5
(a) Statement of Kirchhoff’s rule ½ + ½
Obt ai ni ng the balance condition in Wheatstone Bridge 2
(b) Calculation of val ues of R1 and R2 2

Inloop ADBA
-I1 R1 + 0 +I2 R2 = 0
=> I1 R1 =I2 R2
Inloop CBDC
I2 R4 + 0 – I1 R3 = 0
=> I2 R4 =I1 R3
=>
2
1
R
R
=
4
3
R
R
(b)
2
1
R
R
=
60
40
=
3
2
2
1 10
R
R
=
40
60
=
2
3
2
1
R
R
+
2
10
R
=
2
3
=>
3
2
+
2
10
R
=
2
3
=>R2 =12
Substituting for R2 and findi ng the val ue of R1
R1 =8
½
½
½
½
½
½
½
½ 5
29.
(a) Derivation of the expression for the torque with diagra m 3
(b) Depiction of the traject ories 2

(a)
The magnetic field exerts no force on the t wo ar ms ADand BC of the loop.
Force F1 acts on ar m AB directing intothe plane.
F1 =IbB
Force F2 acts on ar m CD directing out of the plane.
F2 =IbB = F1
Hence there is a torque on the loop due to forces F1 and F2
2
a
2
a
=IbB
2
a
+ IbB
2
a
=I(ab) B= I AB where A=ab is the area of the loop
(b)
1
½
½
½
½
1
1

Here r1 =r2
( Si nce the momenta of charged particles are equal and they have equal
charge, therefore they will describe circular trajectories of same radi us)
[If the candi date onl y mentions that they describe circular trajectories without
the diagra m, one mark shoul d be awarded]
OR
(a) Torque acting on the compass needle suspended freelyin a unifor m
magnetic field
In magnitude = MBsin
It will be balanced by the restoring torque
=- MB sin
For s mall angle sin
=- MB
In equilibirum, the resulting equation of motion
=> 2
2
dt
d
= -
I
MB
2
d
[If the student just writes that the needle ,
(i) When slightly disturbed fromits stable position experiences a
torque due tothe magnetic field and
(ii) writes the expression for this torque ,
½
½
½
½
½
½
5
(a) Execution of SHMof compass needle in magnetic field 2
Deri vation of its ti me period 1
(b) Fi ndi ng (i) horizontal component of earth’s magnetic field (ii) angle
of dip 1+1

Award (1 + 1 =2 ) marks ]
(b) (i) Horizontal component of Earth’s magnetic field =0
(ii) The val ue of angle of dip at that place =900
1
1 5

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