PAPER CODE : A
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 - 2
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.
Class : XII & XIII
Time : 3 hour Max. Marks : 210
INSTRUCTIONS
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).

(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 Column I hassomeconductor accrosswhich battery isconnected asshown. Variation of resisitivity is
alsoindicated.WhichofthequantitiesincolumnIIremainconstantthroughout thevolumeofconductor.
ColumnI ColumnII
(A) (P)Magnitudeofelectricfield
(B) (Q) Magnitudeofcurrent density
(C) (R)Electricpowerdissipated perunit volume
(D) (S) Drift speed of free electron
[Ans.A-QS, B-P, C-P, D-bonus]

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
struck byan small elastic ball whose velocityis horizontal just before the
impact.Theballbouncesofftheinclinedplaneand thenlandsonitagainat
the point of first impact. Findthe ratio of themasses of the ball and the
inclined plane.(Angle =30°)
[Sol. This happens ofafter collisionbothball and inclinedplanehave same horizontal velocities sayV2 say
V1 beinitialvelocityofballVy beverticalvelocityofballaftercollision,m1 –mass ofball,m2 –mass of

= 4.8 k̂
2
1
8
3








 = 4.8 k̂
8
4
3







 
= 6 k̂
4
3  = 13.6 or 13.61 or 13.62 ]
Q.6 Initially switch S was closed for long time. Switch is opened at t = 0. Find the current (in ampere)
through battery at time t = RC ln2. Take V = 12 V, R = 4 ohm, C = 1.0 F. [Ans.1.3 or 1.34A]
[Sol. WhenS isclosedforlongtime
fromKVL
I(2R) + IR = V
I =
R
3
V
charge on 2C capacitor
C
2
q 0
2
= IR  0
2
q =
3
VC
2
charge on C
C
q 0
1
= I(2R)  0
1
q =
3
VC
2
when S is opened then at t = t
I2 =
dt
dq2
=
R
2
V
e–t ]