PAPER CODE : A Class : XII & XIII
Time : 3 hour Max. Marks : 210
INSTRUCTIONS
PART-B
For example if Correct match
for (A) is P, Q; for (B) is P, R;
for (C) is P and for (D) is S
then the correct method for
filling the bubble is
P Q R S
(A)
(B)
(C)
(D)
PART-C
Ensure that all columns
(4 before decimal and 2 after
decimal) are filled. Answer
having blank column will be
treated as incorrect. Insert
leadingzero(s)ifrequiredafter
rounding the result to
2 decimal places.
e.g. 86 should be filled as
0086.00
.
.
.
.
.
.
.
.
.
.
PART-D
Ensure that all columns
{1 before decimal and 2 after
decimal with proper sign (+)
or (–)} arefilledand columns
after 'E'usedfor fillingpower
of 10 with proper sign (+) or
(–). Answer having blank
column will be treated as
incorrect.
e.g. – 4.19 × 1027 should be
filled as – 4.19 E + 27
P
A
P
E
R
-
USEFULDATA: Atomic Mass:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O = 16, H = 1, P = 31,
Ag = 108, N = 14, Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40,
Zn = 65.4, Ba = 137, Co = 59, Hg = 200, Pb = 207, He = 4, F=19.
Radius of nucleus =10–14 m; h = 6.626 ×10–34 Js; me = 9.1 ×10–31 kg, R = 109637 cm–1.
Take g = 10 m/s2 where ever required.
1. The question paper contain 7 pages and 2-parts. Part-B contains 6 questions of "Match the Column" type and Part-C
contains 18. All questions are compulsory.
Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some
mistake like missing questions or pages then contact immediately to the Invigilator.
PART-B
(i) Q.1 to Q.6 are "Match the Column" type which may have one or more than one matching options and carry 8 marks
for each question. 2 marks will be awarded for each correct match within a question.
ThereisNEGATIVE marking. 0.5 Marks will be deducted for each wrong match. Marks will be awarded onlyif all the
correct alternatives are selected.
PART-C
(ii) Q.1 to Q.17 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded onlyifall the correct
bubbles are filled in your OMR sheet.
PART-D
(iii) Q.1 to Q.2 are "Subjective" questions. There is NO NEGATIVE marking. Marks will be awarded only if all the correct
bubbles are filled in your OMR sheet.
2. Indicate the correct answer for each question by filling appropriate bubble(s) in your answer sheet.
3. Use only HB pencil for darkening the bubble(s).
4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.
5. The answer(s) of the questions must be marked by shading the circles against the question by dark HB pencil only.

PART-B
MATCH THE COLUMN [2 × 8 = 16]
There is NEGATIVE marking. 0.5 Marks will be deducted for each wrongmatch.
INSTRUCTIONS:
Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q.1 On a capacitorof capacitanceC0 followingsteps are performed in the order as given in column I.
(a) Capacitor is charged byconnecting it across a batteryof EMF V0.
(b) Dielectricofdielectricconstant k and thicknessd is inserted
(c) Capacitor is disconnected from battery
(d) Separation betweenplates is doubled
ColumnI ColumnII
(Steps performed) Final value of Quantity (Symbols have usual meaning)
(A) (a) (d) (c) (b) (P) Q =
2
V
C 0
0
(B) (d) (a) (c) (b) (Q) Q =
1
k
V
kC 0
0

(C) (b) (a) (c) (d) (R) C =
1
k
kC0

(D) (a) (b) (d) (c) (S) V =
k
2
)
1
k
(
V0 
[Ans.(A) P, R, S; (B) P, R, S; (C) R; (D) Q, R]
[Sol. Q C V
(A) C0V0 C0 V0
2
V
C 0
0
2
C0
V0
2
V
C 0
0
2
C0
V0
2
V
C 0
0
1
K
KC0
 K
2
)
1
K
(
V0 
(B) 0
d
2
C0
0
V0 d
2
C0
d
2
C0
V0
V0
d
2
C0
d
2
C0
V0
2
V
C 0
0
1
K
KC0
 K
2
)
1
K
(
V0 

(C) 0
d
KA 0

0
KC0V0 KC0 V0
KC0V0 KC0 V0
KC0V0
1
K
KC0

(K+1)V0
(D) C0V0 C0 V0
KC0V0 KC0 V0
1
K
V
KC 0
0
 1
K
KC0

V0
1
K
V
KC 0
0
 1
K
KC0

V0 ]
Q.2 ColumnIhas someconductoraccrosswhich batteryis connected asshown.Variationofresisitivityis
alsoindicated.WhichofthequantitiesincolumnIIremainconstantthroughout thevolumeofconductor.
ColumnI ColumnII
(A) (P)Magnitudeofelectricfield
(B) (Q) Magnitudeofcurrent density
(C) (R) Powerdissipated per unit volume
(D) (S) Drift speed of free electron
[Ans.Bonus]

PART-C
SUBJECTIVE
ThereisNO NEGATIVE marking.
Q.1 Findthekineticenergyofpulsetravelling inatautstring.Given T=10N&=0.1 kg/m.[Giveanswer
inmillijoules]. [Ans. 0.15 mJ]
[Sol. at giventimey=m1x 0 < x < 0.1 m
= m2 x + C 0.1 m < x < 0.15 m
m1 =
1
.
0
10 3

= 10–2











T
V
m2 =
05
.
0
10 3

= 2 × 10–2
dK =
2
1
dm 2
P
V =
2
1
dx
2
x
y
V 







 
K
d
 =
2
1
 2
V 








 

15
.
0
1
.
0
2
2
1
.
0
0
2
1 dx
m
dx
m
K
2
1
 
)
5
.
0
(
m
)
1
.
0
(
m 2
2
2
1 
T = K + U =  
)
5
.
0
(
10
4
)
1
.
0
(
10
T
2
1 4
4 



=
2
10
[10–4] (0.3) = 0.15 mJ Ans. ]
Q.2 A sampleofideal gas is taken through the cyclic process shown in the
figure.The temperature of the gas at stateAis TA=200 K.At states B
and C the temperature of the gas is the same.
Whatisthegreatesttemperatureofthegasinkelvinsduringthecyclicprocess?
[Sol. At B and C (TB = TC)
B
B
B
T
V
P
=
C
C
C
T
V
P
B
A
B
T
V
P
=
C
A
C
T
V
3
P
PB = 3PC
for lineAC PV (straightlinethroughorigin)
so
A
C
P
P
=
A
C
V
V
=
A
A
V
V
3

 PC = 3PA
Thus PC = 3PA ; PB = 3PC = 9PA .....(I)
VC= 3VA ; VB = VA
TA =
nR
V
P A
A
.....(II)
fromAto B ; sochoric P  T 
so TB > TA
for C toA; both (P, V)  so T 
Thus from BtoC (we couldhavemaximum temperature)
P = aV + b
 P = 








A
A
V
2
P
6
+ 12 PA  P = –
A
A
V
V
P
3
+ 12 PA
PV = nRT
V
P
12
V
V
P
3
A
A
A









 = nRT
T
for Tmax dV
dT
= 0
A
A
V
P
6

V + 12PA = 0
V = 2VA  P = 6PA
T =
nR
)
V
2
(
P
6 A
A
=
nR
V
P
12 A
A
Tmax = 12 TA = 2400 K ]
Q.3fl TwoobjectsofequalvolumeV=1m3 anddifferent densities d1=500kg/m3and
d2 = 1000 kg/m3 areglued to eachother so that theircontact surface is flat and
has an areaA= 0.1 m2. When the objects are submerged in a certain liquid,
theyfloatinstableequilibrium, thecontact surface beingparallelto thesurface
oftheliquid(seethediagram). How deep (Hin meters) canthecontact surface
be inthe liquid so that the objects are not torn apart? The maximum force that
the glue can withstand is F=250 N.
[Ans H =3m]
[Sol. Atthe"critical"depthH,theequilibrium conditionforthecompound object suggests thatthedensityof
theliquidisd=(d1 +d2)/2.Forthetopandthebottomobjecttakenseparately,theequilibriumconditions
are,respectively:
d1Vg + F – (dVg – dgAH) = 0
d2Vg – F – (dVg + dgAH) = 0
(For eachobject, the term inparentheses indicates the "effective" buoyancyforce.)
Combining the last two equations yields the answer : H = [(d2–d1)Vg – 2F] / [(d1 + d2) gA] ]

Q.4 An inclined plane is placed on a horizontal smooth surface. The plane is
struckbyanelasticballwhosevelocityishorizontaljustbeforetheimpact.
The ball bounces off the inclined plane and then lands on it again at the
pointoffirstimpact.Findtheratioofthemassesoftheballandtheinclined
plane. (Angle  = 30°)
[Sol. Thishappensof after collision both ball and inclined planehavesamehorizontal velocities say V2 say
V1 beinitialvelocityofballVy beverticalvelocityofballaftercollision,m1 –mass ofball,m2 –mass of
inclinedplane
Conservationoflinearmomentuminhorizontaldirection
2
2
1
1
1 V
)
m
m
(
V
m 
 ....(1)
coffeicientofrestitution=1
1 = – 






sin
V
0
)]
cos
V
sin
V
(
sin
V
[
1
y
2
2


 sin
V
cos
V 1
y ....(2)
Componentofvelocityofballalongtheinclinedplaneremainsame




 sin
V
cos
V
cos
V y
2
1 ....(3)
Vy sin  = (V1–V2) cos 
Vy cos  = V1 sin  
(3)
&
(2)
using



tan2  =
1
2
1
V
V
V 
V2 = V1(1– tan2) ....(4)
(m1 + m2) V2 = m1V1 ....(1)
Divide
m1 + m2 =

 2
1
tan
1
m
2
1
m
m
= cot2 –1; cot2 30° –1 = 2 Ans. ]
Q.5 A radioreceiveris set upon a mast in themiddle of acalm lake to track the radio signal froma satellite
orbitingtheearth.As the satelliterises abovethehorizon,theintensityof thesignal varies periodically.
Theintensityisat a maximum when thesatellite is 1 =0.01 radian abovethe horizon and then again at
2 = 0.03 radian above the horizon. What is the wavelength  (in meters) of the satellite signal? The
receiver is h = 4.0 m above the lake surface. [Ans.0.16 m]
[Sol. PR =

sin
h
; QR = PR cos 2 =


sin
2
cos
h
X = PR – RQ +
2
h
X =

sin
h
(1–cos2) +
2
h
= 2 h sin +
2
h
X  2 h +
2
h
sin  =

Maximumtomaximum
 = 2 h (2–1) + 0
 = 2 (4) [0.03 – 0.01]
= 2 (4) (0.02)
= 0.16 m ]
Q.6 Initiallyswitch S was closed for long time. Switch is opened at t = 0. Find the current (in amperes)
through battery at time RC ln2. Take V = 12 V, R = 4 ohm, C = 1.0 F. [Ans.1.5A]
[Sol. ]