The document provides instructions for a chemistry exam consisting of 57 questions divided into 4 sections. Some key details: 1. The exam is 3 hours long and has a maximum of 240 marks. 2. Students must fill out identifying information on the question paper and answer sheet. 3. The answer sheet is machine-readable and students must follow instructions to fill it out correctly. 4. Section I contains 4 multiple choice questions with a single correct answer each. Section II has 5 multiple choice questions where students select one or more correct answers. Section III involves matching statements between two columns. Section IV contains 8 questions where students provide a single digit integer answer.

1. QUESTIONS & SOLUTIONS OF IIT-JEE 2009
Date : 12-04-2009 Duration : 3 Hours Max. Marks : 240
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
PAPER - 2
INSTRUCTIONS
A. General
1. This booklet is your Question Paper containing 57 questions.
2. The Question Paper CODE is printed on the right hand top corner of this page and on
the back page of this booklet.
3. Each page of this question paper contains half page for rough work (except front and
back page). No additional sheets will be provided for rough work.
4. Blank paper, clipboard, log tables, slide rules, calculators, cellular phones, pagers, and
electronic gadgets in any form are not allowed to be carried inside the examination hall.
5. Fill in the boxes provided below on this page and also write your Name and Roll No. in
the space provide on the back page of this booklet.
6. The answer sheet, a machine-readable Objective Response Sheet (ORS), is provided
separately.
7. DO NOT TAMPER WITH/ MUTILATE THE ORS OR THE BOOKLET.
8. Do not open the seals of question-paper booklet before being instructed to do so by the
invigilators.
B. Filling the ORS :
9. Write your Roll No. in ink, in the box provided in the upper part of the ORS and darken
the appropriate bubble UNDER each digit of your Roll No. with a good quality HB pencil.
C. Question paper format :
D. Marking scheme
Read the instructions printed on the back page of this booklet.
Name of the Candidate
Roll Number
I have read all the instructions and --------------------------------
shall abide by them. Signature of the Invigilator
--------------------------------
Signature of the Candidate

2. PART I
CHEMISTRY
SECTION - I
Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONLY ONE is correct.
1. The correct stability order of the following resonance structures is :
(A) (I) > (II) > (IV) > (III) (B) (I) > (III) > (II) > (IV)
(C) (II) > (I) > (III) > (IV) (D) (III) > (I) > (IV) > (II)
Sol. (B)
octet complete octet incomplete octet complete octet incomplete
–ve charge on nitrogen –ve charge on nitrogen –ve charge on carbon –ve charge on carbon
2. In the following carbocation; H/CH3 that is most likely to migrate to the positiviely charged carbon is
(A) CH3 at C-4 (B) H at C-4 (C) CH3 at C-2 (D) H at C-2
Sol. (D)
1, 2 Hydride shift
 

(More stable carbocation due to
+m effect of – OH group and +  and
hyperconjugative effect of –CH3 group)
RESONANCE Page # 2

3. 3. For a first order reaction A  P, the temperature (T) dependent rate constant (k) was found to follow the
1
equation log k = – (2000) + 6.0. The pre-exponential factor A and the activation energy Ea, respectively,
,
T
are :
(A) 1.0 × 106 s–1 and 9.2 kJ mol–1 (B) 6.0 s–1 and 16.6 kJ mol–1
6
(C) 1.0 × 10 s and 16.6 kJ mol
–1 –1
(D) 1.0 × 106 s–1 and 38.3 kJ mol–1
Sol. (D) From Arrhenius equation
K = Ae–Ea/RT
Ea
nk = nA –
RT
Ea
2.303 log K = 2.303 log A –
RT
Ea 1
log K = 2.303 R × + log A ....... (1)
T
1
log K = – (2000) +6 ........(2)
T
On comparing equation (1) and (2)
Ea
2.303 R = –2000.
Ea = 2.303 × 8.314 × 2000 = 38.29 kJ.
and log A = 6 A = 106
4. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is :
(A) 0 (B) 2.84 (C) 4.90 (D) 5.92
Sol. (A)
The chromium is in zero oxidation state having configuration [Ar]18 3d5 4s1. The CO is a strong field ligand so
compels for the pairing of electrons. Thus the complex has d2 sp3 hybridisation and is diamagnetic.
[Cr(CO)6]
BM = n(n  2) = 0 as there is no unpaired electrons.
RESONANCE Page # 3

4. SECTION - II
Multiple Correct Answer Type
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONE OR MORE is/are correct.
5. In the reaction
2X + B2H6  [BH2(X)2]+ [BH4]–
the amine(s) X is(are) :
(A) NH3 (B) CH3NH2 (C) (CH3)2NH (D) (CH3)3N
Sol. (A, B, C)
Small amines such as NH3 ,CH3NH2 and (CH3)2NH give unsymmetrical cleavage of diborane according to
following reaction.
B2H6 + 2 NH3  [H2B(NH3)2]+ [BH4]–
Large ammines, such as (CH3)3 N gives symmetrical cleavage of diborane according to following reaction.
B2H6 + 2N (CH3)3  2H3B  N(CH3)3
6. The nitrogen oxide(s) that contain(s) N—N bond(s) is(are) :
(A) N2O (B) N2O3 (C) N2O4 (D) N2O5
Sol. (A, B, C)
(A)
N2 O
(B)
(C)
N2 O 4
(D)
N2 O 5
RESONANCE Page # 4

5. 7. For the reduction of NO3– ion in an aqueous solution Eº is +0.96 V. Values of Eº for some metal ions are given
below
V2+ (aq) + 2e–  V Eº = –1.19 V
3+
Fe (aq) + 3e  Fe
–
Eº = –0.04 V
3+
Au (aq) + 3e  Au
–
Eº = +1.40 V
2+
Hg (aq) + 2e  Hg
–
Eº = +0.86 V
The pair(s) of metals that is(are) oxidized by NO3 in aqueous solution is(are) :
–
(A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V
Sol. (A, B, D)
The species having less reduction potential with respect to NO3– (Eº = 0.96 V) will be oxidised by NO3–.
These species are V, Fe, Hg.
8. The correct statement(s) about the following sugars X and Y is(are) :
(A) X is a reducing sugar and Y is a non-reducing sugar.
(B) X is a non-reducing sugar and Y is a reducing sugar.
(C) The glucosidic linkages in X and Y are  and , respectively.
(D) The glucosidic linkages in X and Y are  and , respectively.
Sol. (B, C)
X has acetal linkage and Y has hemiacetal linkage. Carbohydrate with hemiacetal linkage are reducing
sugars and carbohydrate with acetal linkage are non reducing sugars.
X is  – anomer and Y is - anomer of D (+) glucose.
9. Among the following, the state function(s) is(are) :
(A) Internal energy (B) Irreversible expansion work
(C) Reversible expansion work (D) Molar enthalpy
Sol. (A, D )
State function are internal energy and molar enthalpy.
Work is path function whether it is reversible or Irreversible.
RESONANCE Page # 5

6. SECTION - III
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to
be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are
labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE
statement(s) in Coloumn-II. The appropriate bubbles corresponding to the answers to these questions have
to be drakened as illustrated in the following example.
If the correct matches are A-p, s and t ; B-q and r; C-p and q; and D-s and t; then the correct darkening of
bubbles will look like the following :
10. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in
column II
Column I Column II
(A) (p) Nucleophilic substitution
(B) (q) Elimination
(C) (r) Nucleophilic addition
(D) (s) Esterification with acetic anhydride
(t) Dehydrogenation
Ans. A  p, q, t ; B  p, s, t ; C  r, s ; D  p
RESONANCE Page # 6

7. Sol. (A)
(B)
(C)
(D)
RESONANCE Page # 7

8. 11. Match each of the reactions given in column I with the corresponding products (s) given in column II
Column I Column II
(A) Cu + dil HNO3 (p) NO
(B) Cu + conc HNO3 (q) NO2
(C) Zn + dil HNO3 (r) N2O
(D) Zn + conc HNO3 (s) Cu(NO3)2
(t) Zn(NO3)2
Ans. A  p, s ; B  q, s ; C  r, t ; D  q, t
Sol. (A) 3Cu + 8HNO3 (dilute HNO3)  2NO + Cu(NO3)2 + 4H2O
(B) Cu + 4HNO3 (concentrated)  2NO2 + Cu(NO3)2 + 2H2O
(C) 4Zn + 10HNO3 (dilute)  4Zn(NO3)2 + N2O + 5H2O
(D) Zn + 4HNO3 (concentrated)  Zn(NO3)2 + 2NO2 + 2H2O
SECTION - IV
Integer Answer Type
___________________________________________________________________________
This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from
0 to 9. The appropriate bubbles below the respectively question numbers in the ORS have to be darkened.
For example, if the correct answers to question number X, Y, Z and W (say) are 6, 0, 9 and 2, respectively,
then the correct darkening of bubbles will like the following :
RESONANCE Page # 8

9. 238
12. The total number of  and particles emitted in the nuclear reaction 92U  28 2Pb is
14
Ans. 8
Sol. 238
U  214 Pb + 6 ( 4 He ) + 2(–1e0)
92 82 2
 = 6,  = 2
Total = 8
13. The number of water molecule (s) directly bonded to the metal centre in CuSO4. 5H2O is
Ans. 4
Sol.
14. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is
Ans. 6
fusion
Sol. 2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O
Let the oxidation state of Mn in MnO42– is x.
So x + 4 (–2) = –2 or x=6
15. The Coordination number of Al in the crystalline state of AlCl3 is
Ans. 6
16. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at
298.0 K. The temperature of the calorimeter was found to increases from 298.0 K to 298.45 K due to the
combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K–1, the numerical value for the
enthalpy of combustion of the gas in kJ mol–1 is
Ans. 9
3 .5
Sol. n=
28
T = T2 – T1 = 298.45 – 298
= 0.45
CV = 2.5 kJ k–1 = 2500 JK–1
CP = CV + R = 2500 + 8.314
= 2508.314 JK–1
QP = CPT = 1128.74 J
Qp 1128 .74
H =  9030 J mol–1
n 3.5 / 28

= 9.030 KJ mol–1
= 9 KJ mol–1.
RESONANCE Page # 9

10. 17. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular
formula C5H10 is
Ans. 7
Sol.
18. The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10–4 . The pH of 0.01 M solution of
its sodium salt is
Ans. 8
Sol. Given Ka = 10–4
pKa = 4
C = 0.01 M
1 1
pH = 7 + pKa + log C
2 2
1 1
=7+ (4) + (–2)
2 2
= 8 Ans.
19. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable
speed of gas Y at 60 K. The molecular weight of the gas Y is
Ans. MY = 4.
Sol. Vrms = Vmp
3RT 2RT
MX
= MY
3R  400 2R  60
= MY
40
MY = 4.
RESONANCE Page # 10

11. PART-II
MATHEMATICS
SECTION - I
Single Correct Choice Type
_____________________________________________________________________________
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONLY ONE is correct.
_____________________________________________________________________________
20. The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q. If M is the mid point of the line
segment PQ, then the locus of M intersects the latus rectum of the given ellipse at the points
 3 5 2  3 5 19  1 4 3
(A)   ,  (B)   , (C)   2 3 ,   (D)   2 3 , 
 
2 7 2 4  7 7 
 
   
     
Sol. (C)
x2 y2
+ =1
16 4
a = 4, b = 2
equation of normal 4x sec  – 2y cosec  = 12
 7 cos 
M  , sin   = (h, k) (say)

 2 
7 cos  2h
h= cos  = and k = sin 
2 7

4h 2
+ k2 = 1
49
4x 2
locus + y2 = 1 ....(i)
49
4 3
for given ellipse e2 = 1 – =
16 4
3
e=
2
3
x=±4× = ± 2 3 ....(ii)
2
solving (i) and (ii)
4
× 12 + y2 = 1
49
48 1
y2 = 1 – =
49 49
1
y=±
7
1
required points   2 3 ,  

7


RESONANCE Page # 11

12. 21. The locus of the orthocentre of the triangle formed by the lines
(1 + p) x – py + p (1 + p) = 0, (1 + q) x – qy + q(1 + q) = 0, and y = 0, where p  q, is
(A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line
Sol. (D)
(1 + p) x – py + p (1 + p) = 0 ......(1)
(1 + q) x – qy + q(1 + q) = 0 ......(2)
on solving (1) and (2), we get C(pq, (1 + p) (1 + q))
 equation of altitude CM passing through C and perpendicular to AB is x = pq .......(3)
 1 q 
 slope of line (2) is = 
 q  
 
q
slope of altitude BN (as shown in figure) is =
1 q

q
equation of BN is y – 0 = (x + p)
1 q

q
y= (x + p) ....... (4)
(1  q)

Let orthocentre of triangle be H(h, k) which is the point of intersection of (3) and (4)
 on solving (3) and (4), we get
x = pq and y = – pq  h = pq and k = –pq
 h+k=0
 locus of H(h, k) is x + y = 0
22. A line with positive direction cosines passes through the point P(2, –1, 2) and makes equal angles with the
coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ
equals
(A) 1 (B) 2 (C) 3 (D) 2
Sol. (C)
1
=m=n=
3
2x + y + 2 = 9
x–2 y 1 z–2
equations of line are = =
1/ 3 1/ 3 1/ 3

x – 2 = y +1 = z – 2 = r
Q  (r + 2, r – 1, r + 2)
 Q Lies on the plane 2x + y + z = 9
2(r + 2) + (r – 1) + (r + 2) = 9
 4r + 5 = 9  r=1
 Q (3, 0, 3)
 PQ = 1 1 1 = 3
RESONANCE Page # 12

13. 23. If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is
n( 4n 2  1) c 2 n( 4n 2  1) c 2 n( 4n 2  1) c 2 n( 4n 2  1) c 2
(A) (B) (C) (D)
6 3 3 6
Sol. (C)
Sn = cn2
Sn–1 = c(n – 1)2 = cn2 + c – 2 cn
Tn = 2cn – c
Tn2 = (2cn – c)2 = 4c2 n2 + c2 – 4c2n
4c 2 . n(n  1)(2n  1)
Sum =  Tn2 = + nc2 – 2c2n (n + 1)
6
2c 2n(n  1)(2n  1)  3nc 2  6c 2n(n  1) nc 2 [ 4n 2  6n  2  3  6n  6] nc 2 ( 4n 2  1)
= = =
3 3 3
SECTION - II
Multiple Correct Choice Type
______________________________________________________________________________________________
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)
for its answer, out of which ONE OR MORE is/are correct.
______________________________________________________________________________________________
24. The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N,
respectively. The locus of the centroid of the triangle PTN is a parabola whose
 2a  2a
(A) vertex is  , 0  (B) directrix is x = 0 (C) latus rectum is (D) focus is (a, 0)
 3  3
Sol. (A,D)
Let centroid of  PTN is R (h. K)
at 2  (– at 2 )  2a  at 2 2at
h= &k=
3 3

2
 3k 
3h = 2a + a .  
 2a 

9k 2
3h = 2a +
4a

 9k2 = 4 a (3h – 2a)
locus of centroid is
4a  2a 
y2 = x –
3  3 

 2a  2a a a
vertex  , 0  ; directric x – =– x=
3  3 3 3
 

RESONANCE Page # 13

14. 4a
Latus rectum =
3
 a 2a 
focus   ,0 
3 3 
i.e. (a, 0) Ans. A, D
1
25. For the function f(x) = x cos , x  1,
x
(A) for at least one x in the interval [1, ), f(x + 2) – f(x) < 2
(B) xlim f(x) = 1

(C) for all x in the interval [1, ), f(x + 2) – f(x) > 2
(D) f(x) is strictly decreasing in the interval [1, )
Sol. (B,C,D)
1
f(x) = x cos ,x1
x
1 1 1
f(x) = sin + cos
x x x

1  1
f(x) = – 3 cos  
x x

Now xlim f(x) = 0 + 1 = 1

 option ‘B’ is correct
1
x  [1, )  (0, 1]
x

 f(x) < 0  option ‘D’ is correct
As f(1) = sin 1 + cos 1 > 1
f(x) is strictly decreasing and xlim f(x) = 1

so graph of f(x) is as below
Now in [x, x + 2], x  [1, ), f(x) is continuous and differentiable
f ( x  2)  f ( x )
so by LMVT, f(x) =
2
as f(x) > 1 for all x  [1, )
f ( x  2)  f ( x )
>1 f(x + 2) – f(x) > 2
2
 
for all x  [1, )
RESONANCE Page # 14

15. 6
(m  1)   m 
26. For 0 <  < , the solution(s) of  cos ec     cos ec     = 4 2 is(are)
  
2 m 1  4   4 
5
(A) (B) (C) (D)
  
4 6 12 12
Sol. (C,D)
0 < <

2
6
(m  1)   m 
 cos ec     cos ec     = 4 2
 
m 1  4   4 
6
1
= 4 2
(m  1)   m 
m 1 sin    sin  
 
4 4 

  
m  (m  1) 
sin 

6
4  4
   
= 4 2
 
m 1 (m – 1)   m  
sin sin   sin  
  
4  4 4 
 
 
(m  1)  m 
6 cot     cot   6
 
4 4  (m  1)  m 

= 4 2 ;  cot     cot   4
 
1 4 4 
  
m 1 m 1
   
2
2  5  6 
cot () – cot     + cot     – cot     + ... + cot     =4
      
4 4 4  4  4 
 – cot   
    
 3
cot  – cot    = 4

 2 
cot  + tan  = 4 ; tan2 – 4 tan  + 1 = 0
(tan  – 2)2 – 3 = 0
(tan  – 2 + 3 ) (tan  – 2 – 3)=0
tan  = 2 – 3 or tan  = 2 + 3
5
= =

12 12
  0, 
 
 2

option C and D is correct
RESONANCE Page # 15

16. 27. An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally. The eccentricity of the ellipse is reciprocal of
that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then
(A) Equation of ellipse is x2 + 2y2 = 2 (B) The foci of ellipse are (±1, 0)
(C) Equation of ellipse is x2 + 2y2 = 4 (D) The foci of ellipse are (± 2 , 0)
Sol. (A,B)
2x2 – 2y2 = 1
x2 y2
=1 ......... (1)
 1  1
–
2 2
   
Eccentricity of hyperbola = 2
1
so eccentricity of ellipse =
2
Let equation of ellipse be
x2 y2
+ = 1 (a > b)
a2 b2
1 b2
= 1–
2 a2
b2 1
=
a2 2

 a2 = 2b2
 x2 + 2y2 = 2b2 ..........(2)
 1 1
Let ellipse and hyperbola intersect at A  sec , tan  

 2 2
 

dy
Now 4 x – 4y =0
dx
dy x
= y
dx
dy sec 
dx = = cosec 
at A tan 
dy
and 2x + 4y =0
dx
dy x 1
dx = – 2y = – cosec 
at A 2
Ellipse and hyperbola are orthogonal, so
RESONANCE Page # 16

17. 1
cosec2= – 1
2
–
cosec2= 2
Q= 

4
 1  1 
A 1,  or 1, –

2 2
    

 
2
 1 
1+2   = 2b2  b2 = 1
 2
 
Equation of ellipse is x2 + 2y2 = 2
1 
Coordinate of foci (ae,0) =   2. , 0  = (1, 0)

2 
 

option (A) and (B) are correct
Note :
1 a2
If major axis is along y-axis, then = 1
2 b2
b2 = 2a2
2x2 + y2 = 2a2
2x
y = – y
2
. cosec  = – 1
sin 
–
cosec2 = 1
=±

4
2x2 + y2 = 2a2
1
2+ = 2a2
2
5
a2 =
4
5
2x2 + y2 = , corresponding foci are (0, ±1).
2
RESONANCE Page # 17

18. sin nx

28. If n = dx, n = 0, 1, 2, ..., then


(1   x ) sin x
10 10
(A) n = n+2 (B)  2m 1  10  (C)  2m 0 (D) n = n+1
m 1 m 1

Sol. (A,B,C)
sin nx

(A) n = dx


(1   x ) sin x
b b
 x sin nx

dx
n = 

(1   x ) sin x
(by property  f ( x) dx   f (a  b  x ) dx )
a a
sin nx

2n = dx
 sin x

sin nx

2n = 2 dx
sin x
0

sin nx

n = dx
sin x
0

sin(n  2) x  sin nx 2 cos(n  1) x sin x  sin(n  1)x 
  
n+2 – n = dx = dx = 2   =0
sin x sin x  (n  1)  0
0 0
 
 n+2 = n
(B) 3 = 5 = ..... = 21
10
sin 3 x

2


m1
 2m1 = 10 = 10
3
sin x
dx = 10  (3  4 sin
0
x ) dx
0
 
= 103 x  2x  2 sin 2x 0 = 10

(C) 2 = 4 = ........ = 20
10
sin 2 x

 2m = 10
sin x dx = 20 sin x 0 = 0

m 1 0
 
RESONANCE Page # 18

19. SECTION - III
Matrix - Match Type
_________________________________________________________________________________________________
This section contains 2 questions. Each question contains statements given in two columns, which
have to be matched. The statements in Column - I are labelled A, B, C and D, while the statements
in Column - II are labelled p, q, r, s and t. Any given statement in Column - I can have correct
matching with ONE OR MORE statement(s) in Column - II. The appropriate bubbles corresponding
to the answers to these questions have to be darkened as illustrated in the following example :
If the correct matches are A – p, s and t ; B – q and r ; C – p and q ; and D – s and t; then the correct
darkening of bubbles will look like the following.
_________________________________________________________________________________________________
29. Match the statements/expressions given in Column - I with the values given in Column - II
Column - I Column - II
(A) The number of solutions of the equation xesinx – cos x = 0 (p) 1
in the interval  0, 
 
 2
(B) Value(s) of k for which the planes kx + 4y + z = 0, (q) 2
4x + ky + 2z = 0 and 2x + 2y + z=0 intersect in a straight line
(C) Value(s) of k for which |x – 1| + |x – 2| + |x + 1| + |x + 2| = 4k (r) 3
has integer solution(s)
(D) If y = y + 1 and y(0) = 1, then value(s) of y (n 2) (s) 4
(t) 5
Ans. (A)  (p), (B)  (q, s), (C)  (q, r, s, t), (D)  (r)
Sol. (A) Let f(x) = xesinx – cos x
f(x) = esinx + xesinx cos x + sin x  0 for interval x   0, 
 
 2
 f is strictly increasing
f(0) = – 1
f   = e one solution
 
2 2

RESONANCE Page # 19

20. k 4 1
(B) 4 k 2 =0
2 2 1
k (k – 4) – 4(0) + 1 (8 – 2k) = 0
k2 – 6k + 8 = 0
k = 2, 4
3
(C) for solutions, 4k  6 k .
2
Integer values of k are 2, 3, 4, 5
dy
(D) =y+1
dx
n |(y + 1)| = x + c
n 2 = c  n |y + 1| = x + n 2
put x = n 2
n (y + 1) = n 2 + n 2 = n 4
y+1=4
y=3
30. Match the statements/expressions given in Column - I with the values given in Column - II
Column - I Column - II
(A) Root(s) of the equation 2 sin2 + sin22= 2 (p)

6
 6x   3x 
(B) Points of discontinuity of the function f(x) =   cos   , (q)

    4
where [y] denotes the largest integer less than or equal to y
(C) Volume of the parallelopiped with its edges represented by the (r)

3
vectors ˆ  ˆ , ˆ  2ˆ and ˆ  ˆ  k
i j i j i j ˆ
Angle between vectors a and b where a, b and c are unit
 
(D) (s)
   
2
vectors satisfying a  b  3 c  0
   
(t) 
Ans. (A)  (q, s), (B)  (p, r, s, t), (C)  (t), (D)  (r)
RESONANCE Page # 20

21. Sol. (A) 2 sin2sin22= 2
 sin22= 2 cos2  4sin2cos2= 2cos2
1 1
cos2 = 0 or sin2  = cos = 0 or sin=  = or
 
2 2 4 2
  
(B)  f(x) =  6 x  cos  3 x 
   
   
6x
possible points of disontinunity of  6 x  are = n , n I
   
 
n
x= x= , , ,
  
6 6 3 2
 
lim  f (x) = 0 cos 0 = 0
x

6
lim f(x) = 1 cos 0 = 1
x

6
 discontinous at x = . Similarly discontinous at x = , , 
  
6 3 2
1 1 0
1 2 0
(C) V= =  cubic units
1 1 
(D) a + b + 3 c= 0
   
a + b=– 3 c
  

 2 2
 ab = 3c
a2 + b2 + 2 a . b = 3c2
 

 2 + 2 cos = 3
1
cos=
2

RESONANCE Page # 21

22. SECTION - IV
Integer Answer Type
________________________________________________________________________________________________
This section contains 8 questions. The answer to each of the questions is a single-digit integer,
ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS
have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say)
are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :
________________________________________________________________________________________________
x
31. Let f : R  R be a continuous function which satisfies f(x) =  f (t) dt . Then the value of f(n 5) is
0
Ans. 0
Sol. From given integral equation, f(0) = 0.
Also differentiating the given integral equation w.r.t. x
f(x) = f(x)
f ( x )
If f(x)  0 =1 f(x) = ec ex
f ( x)
 
 f(0) = 0  ec = 0, a contradiction
 f(x) = 0  x  R
 f(n 5) = 0
32. The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be
the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2
externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then
the radius of the circle C is
Ans. 8
Sol. (r + 1)2 = 2 + 9
r2 + 8 = 2
 r2 + 2r + 1 = r2 + 8 + 9
2r = 16
r=8
33. The smallest value of k, for which both the roots of the equation x2 – 8kx + 16(k2 – k + 1) = 0 are real, distinct
and have values at least 4, is
Ans. 2
Sol. (i) x2 – 8kx + 16(k2 – k + 1) = 0
D = 64 (k2 – (k2 – k + 1)) = 64 (k – 1) > 0
k>1
b 8k
(ii) >4 >4 k>1
2a 2
–  
(iii) f(4)  0
16 – 32k + 16 (k2 – k + 1)  0
k2 – 3k + 2  0
(k – 2) (k – 1)  0
k  1 or k  2
Hence k = 2
RESONANCE Page # 22

23. 34. The maximum value of the function f(x) = 2x3 – 15x2 + 36x – 48 on the set A = {x |x2 + 20  9x} is
Ans. 7
Sol. A = {x |x2 + 20  9x} = {x |x  [4, 5]}
Now, f(x) = 6(x2 – 5x + 6)
f(x) = 0  x = 2, 3
f(2) = –20, f(3) = –21, f(4) = –16, f(5) = 7
from graph, maximum of f(x) on set A is f(5) = 7
35. Let ABC and ABC be two non-congruent triangles with sides AB = 4, AC = AC = 2 2 and angle B= 30º. The
absolute value of the difference between the areas of these triangles is
Ans. 4
Sol. In  ABC, by sine rule
a 2 2 4
= =  C = 45º, C = 135º
sin A sin 30 º sin C
When C = 45º  A = 180º – (45º + 30º) = 105º
When C = 135º  A = 180º – (135º + 30º) = 15º
1 1 3 1
Area of  ABC = AB × AC sin A = × 4 × 2 2 sin (105º) = 4 2 ×
2 2 2 2
1 1
Area of  ABC = AB × AC sinA = × 4 × 2 2 sin (15º) = 2  3 – 1
2 2
Difference of areas of triangles = | 2  3  1 – 2  3 – 1 | = 4
Aliter
AD = 2 , DC = 2
Difference of Areas of triangle ABC and ABC = Area of triangle ACC
1 1
= AD × CC =
2 2
×2×4=4
RESONANCE Page # 23

24. x
36. If the function f(x) = x3 + e 2 and g(x) = f–1(x), then the value of g(1) is
Ans. 2
Sol. g(f(x)) = x
 g(f(x)) f(x) = 1 ........(i)
if f(x) = 1  x = 0, f(0) = 1
substitute x = 0 in (i), we get
1
g(1) = 
f (0)
1 x/2 1
g(1) = 2 (f(x) = 3x2 + e  f(0) = )
2 2

 p( x ) 
37. lim
Let p(x) be a polynomial of degree 4 having extremum at x = 1, 2 and x 0 1  2  = 2. Then the value of
 x 
p(2) is
Ans. 0
Sol. p(x) = ax4 + bx3 + cx2 + dx + e
p (x) = 4ax3 + 3bx2 + 2cx + d
p (1) = 4a + 3b + 2c + d = 0 .....(i)
p (2) = 32 a + 12 b + 4c + d = 0 .....(ii)
 p( x ) 
lim 1   =2
x 0 x2 
4 3 2
lim ax  bx  (c  1)x  dx  e = 2
x 0
x2
c + 1 = 2, d = 0, e=0
c=1
Now equation (i) and (ii) are
4a + 3b = – 2 and 32 a + 12 b = – 4
1
a= and b = – 1
4

38. Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations :
3x – y – z = 0, –3x + z = 0, –3x + 2y + z = 0. Then the number of such points for which x2 + y2 + z2  100
is
Ans. 7
Sol. 3x – y – z = 0 .........(1)
–3x + 2y + z = 0 .........(2)
–3x + z = 0 .........(3)
(1) + (2) y=0
So 3x = z
Now x2 + y2 + z2  100
 x2 + (3x)2 + 0  100
10x2  100
x2  10
x = –3, –2, –1, 0, 1, 2, 3
So No.of such 7 points are possible
RESONANCE Page # 24

25. PART - III
PHYSICS
SECTION - I
Single Correct Choice Type
Thus section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out which ONLY ONE is correct.
39. Photoelectric effect experiments are performed using three different metal plates p, q and r having work
functions p = 2.0 eV, q = 2.5 eV and r = 3.0 eV respectively. A light beam containing wavelengths of 550
nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct -V graph for the
experiment is [Take hc = 1240 eV nm]
(A) (B)
(C) (D)
Sol. (A)
1240
E1 550 nm  eV  2.25 eV
550
1240
E 2  450 nm  eV  2.8 eV
450
1240
E  3 350 nm  eV  3.5 eV
350
For metal r, only 3 is able to generate photoelectron.
For metal q, only  and  are able to generate photoelectron.
For metal p, all wavelength are able to generate photoelectron.
Hence photoelectric current will be maximum for p and least for r.
RESONANCE Page # 25

26. 40. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of
equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to
oscillate in the horizontal plane. The rod is gently pushed through a small angle  in one direction and
released. The frequency of oscillation is :
1 2k 1 k 1 6k 1 24k
(A) (B) (C) (D)
2 M 2 M 2 M 2 M
Sol. (C)
L L L2
Torque about P = (kx) + (kx) = kxL = k
2 2 2

L
For small angle , x =
2

=
KL2 ML2
2 12
 –  
 6K
=
M

6K 1 6K
= and f = =

M 2 2 M

RESONANCE Page # 26

27. 41. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The
bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest.
The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new
equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is
a a 2a a
(A) gk (B) 2gk (C) gk (D) 4gk
Sol. (B)
ma cos  = mg cos (90 – )
a
 tan 
 g
a dy
 g = dx
d a
(kx2) = g
dx

a
 x = 2gk = D
42. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the
point P is
k1 k2
M
P
k1A k2A k1A k 2A
(A) k (B) k (C) k  k (D) k  k
2 1 1 2 1 2
RESONANCE Page # 27

28. Sol. (D)
Extentions in springs are x1 and x2 then
k1x1 = k2x2
and x1 + x2 = A
k1x1 k 2A
x1 + k =A x1 = k  k
2 1 2
 
SECTION - II
Multiple Correct Choice Type
This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its
answer, out of which ONE OR MORE is/are correct.
43. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit.
Find out the correct statement(s).
(A) The angular momentum of the charge –q is constant
(B) The linear momentum of the charge –q is constant
(C) The angular velocity of the charge – q is constant
(D) The linear speed of the charge –q is constant
Sol. (A)
Torque about Q of charge –q is zero, so angular momentum charge –q is constant, but distance between
charges is changing, so force is changing, so speed and velocity are changing.
44. The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle
and CDA is half of an ellipse. Then,
(A) the process during the path A  B is isothermal
(B) heat flows out of the gas during the path B  C  D
(C) work done during the path A  B  C is zero
(D) positive work is done by the gas in the cycle ABCDA
RESONANCE Page # 28

29. Sol. (B, D)
(A) process is not isothermal
(B) volume decreases and temperature decreases
U = negative
W = negative
so Q = negative
(C) Work done in process A  B  C is positive
(D) Cycle is clockwise, so work done by the gas is positive.
45. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact,
B is the centre of the sphere and C is its topmost point. Then,
(A) VC  VA = 2 VB  VC (B) VC  VB = VB  VA
 
 
    
  
(C) VC  VA = 2 VB  VC (D) VC  VA = 4 VB
   
Sol. (B, C)
VA  V( ˆ)  R(  ˆ) ; VB  V ˆ ; VC  V ˆ  Rˆ
  
i i i i i
 
VC  VA  2Rˆ
i
2 VB  VC  2 [ V( ˆ)  V( ˆ)  R( ˆ)] = –2R( ˆ )
 
 i i i i
Hence VC  VA =  2( VB  VC )
   
so | VC  VA | = | 2( VB  VC ) |
   
 
VC  VB = R( ˆ )
i
VB  VA = R( ˆ )
i
 
   
VC  VB  VB  VA
Hence
 
VC  VA  2R( ˆ)
i
; 4VB = 4V( ˆ ) = 4R ( ˆ )
    
VC  VB  VB  VA i i
Hence
  
VC  VA  2( VB )
RESONANCE Page # 29

30. 46. A student performed the experiment to measure the speed of sound in air using resonance air-column
method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the
shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,
(A) the intensity of the sound heard at the first resonance was more than that at the second resonance
(B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube
(C) the amplitude of vibration of the ends of the prongs is typically around 1 cm
(D) the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of
the sound in air.
Sol. (A, D)
(A) The intensity of second decreases with increases of order. The intensity of
sound is maximum for first resonance.
(B) The prongs vibrate in vertical plane.
(C) The prongs does not vibrate in amplitude of that order.
(D) Consider end correction, the length of air column is slightly less than /4
So ans are (A) and (D)
47. Two metallic rings A and B, identical in shape and size but having different resistivities A and B, are kept on
top of two identical solenoids as shown in the figure. When current  is switched on in both the solenoids in
identical manner, the rings A and B jump to heights hA and hB, respectively, with hA > hB. The possible
relation(s) between their resistivities and their masses mA and mB is(are)
A B
(A) A > B and mA = mB (B) A < B and mA = mB
(C) A > B and mA > mB (D) A < B and mA < mB
Sol. (B, D)
The horizotanl component of magnetic field due to
solenoid will exert force on ring in vertical direction
F = BHi (2r)
Ft = MV
(  / t )
i=
 (2r ) 
A 
 

BH i (2r) t = MV
BHA K
V= M

M
RESONANCE Page # 30

31. V2 K2
h = 2g  2 2
 M
hA > hB
K2 K2
2 M2 2 2
B MB

A A
 BMB > AMA
 Using this we get
Answer (B) and (D)
Section–III
Matrix – Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to
be matched. The statements in Column–I are labelled A,B,C and D, while the statements in Column-II are
labelled p,q,r,s and t. Any given statement in Column-I can have correct matching with ONE OR MORE
statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have
to be darkened as illustrated in the following examples :
If the correct matches A–p,s and t; B–q and r; C–p and q ; and D–s and t; then the correct darkening of
bubbles will look like the following.
p q r s t
A p q r s t
B p q r s t
C p q r s t
D p q r s t
48. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away
from the slits S1 and S2 . In each of these cases S1P0 = S2P0 , S1P1 – S2P1 =/4 and S1P2 – S2P2 = /3, where
 is the wavelength of the light used. In the cases B,C and D, a transparent sheet of refractive index  and
thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase
difference between the light waves reaching a point P on the screen from the two slits is denoted by (P) and
the intensity by (P). Match each situation given in Column-I with the statement(s) in Column-II valid for that
situation.
RESONANCE Page # 31

32. Column–I Column–II
(A) (p) (P0) = 0
(B) ( – 1)t = /4 (q) (P1) = 0
(C) ( – 1)t = /2 (r) (P1) = 0
(D) ( – 1)t = 3/4 (s) (P0) > (P1)
Ans. (A) p, s; (B) q; (C) t ; (D) r, s, t
(t) (P2) > (P1)
2
Sol. (A)  (P1) = 1 + 2 + 2 1 2 cos
4
1
= 0 + 0 + 20 . = (2 + 2 ) 0
2
1 2
(P2) = 1 + 2 + 20 . = 2 1  2 . cos
2 3
1
= 0 + 0 + 20 .   2  = 0

 
 (P1) > (P2)
2  2 
(B) (P0) = ( – 1)t . = .
4  2


  2
(P1) = (  1)t  . 0

 4 
RESONANCE Page # 32

33.   2
(P2) = (  1)t  3 .   6
 
 
(P0) = 1 + 2 + 2 1  2 . cos (P0)
= 0 + 0 + 20 . cos

2
=20
(P0) = 40
(P2) = 0 + 0 + 2 0 cos   6 
 
 
= (2 + 3 )0
2  2
(C). (P0) = ( – 1)t . = .
2 


2   2
(P1) =  (  1)t.   4    2
 
 
  2
(P2) = (  1)t  3 .   6
 
 
  2 2
(P0) =  2  3  .   6    3
  
 
3  2 3 
(D) (P0) =
4 2
 

(P1) = 0
(P2) > 0, (P0) > 0
49. Column II gives certain systems undergoing a process. Column I suggests changes in some of the
parameters related to the system. Match the statements in Column-I to the appropriate process(es) from
Column II.
Column–I Column–II
(A) The energy of the system is increased. (p) System: A capacitor, initially uncharged
Process: It is connected to a battery.
(B) Mechanical energy is provided to the system, (q) System: A gas in an adiabatic container fitted with
which is converted into energy of random motion an adiabatic piston
of its parts Process: The gas is compressed by pushing the
piston
(C) Internal energy of the system is converted (r) System: A gas in a rigid container
into its mechanical energy Process: The gas gets cooled due to colder
atmosphere surrounding it
RESONANCE Page # 33

34. (D) Mass of the system is decreased (s) System: A heavy nucleus, initially at rest
Process: The nucleus fissions into two fragments of
nearly equal masses and some neutrons are emitted
(t) System: A resistive wire loop
Process: The loop is placed in a time varying
magnetic field perpendicular to its plane
Ans. (A) p, q, t; (B) q, t (C) s, (D) s
Sol .
(A) (p) : Capacitor is charged, hence its energy is increased
(q) : The temperature is increased, hence its energy is increased or as the external positive work is
done, hence energy increases
(r) : The temperature decreases, its energy is decreased
(s) : All natural process, energy of the system decreases
(t) : The current is poroduced. Hence energy of the system increases
(B) (p), (r), (s) no mechanical energy is provided to the system
(q) the mechanical energy is provided which increases the temperature and hence random motion of
molecules
(t) Mechanical work is done to change the magnetic field, which increases the mechancal energy of
electron and these electrons strike with stationary positive charge and energy is converted in random
motion.
(C) (s) Internal binding energy is converted into mechanical energy
(D) (s) Mass changes only in nuclear process.
SECTION–IV
Integer Answer Type
This section contains 8 questions. The answer to each of the question is a single-digit integer, ranging from
0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For
example, if the correct answers to question numbers X,Y,Z and W (say) are 6,0,9 and 2, respectively, then
the correct darkening of bubbles will look like the following :
RESONANCE Page # 34

35. 50. A steady current  goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x
  0 
and QR = 5x. If the magnitude of the magnetic field at P due to this loop is k 48x  , find the value of k.
 
0i
Sol. B= (sin 37º + sin 53º)
 12 x 
4 
 5 

 0i   4 3 
B= 5    
48 x   5 5 
 
  0i 
B=7 
 48 x 

K=7 Ans. 7
51. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of
masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of
mass 0.36 kg during the first second after the system is released from rest.
2m1m2 2  0.72  0.36
Sol. T = m m g =
1 2 0.72  0.36
× 10
T = 4.8 N
RESONANCE Page # 35

36. m1  m2 g
a = m m g =
1 2 3
1 2 1 g 10
s= at =   (1)2 =
2 2 3 6
Work done by T = (T) (S)
10
= (4.8) × = 8J
6
Ans. 8
52. A solid sphere of radius R has a charge Q distributed in its volume with a charge density  = kra, where k and
R 1
a are constants and r is the distance from its centre. If the electric field at r = is times that at r = R, find
2 8
the value of a.
Sol.
r R
a 4k
d V  ) ( 4r 2 dr )  (R a  3 )
Total charge Q =   (Kr
r 0
a3
r R / 2 a 3
4k  R 
Q´ = d V  (Kr a ) ( 4r 2dr ) 
a3 2
 
r 0
 
According to question
1 Q´ 1 1 Q 
40 (R / 2)2
= 8  4 2 
0 R 
 

Putting the value of Q and Q´ get
a=2 Ans. 2
53. A metal rod AB of length 10 x has its one end A in ice at 0ºC and the other end B in water at 100ºC. If a point
P on the rod is maintained at 400ºC, then it is found that equal amounts of water and ice evaporate and melt
per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g.
If the point P is at a distance of x from the ice end A, find the value of . [ Neglect any heat loss to the
surrounding ]
Sol.
400  0
i1 = ( x / kA ) ,
400  100
i2 = (10   )x / kA
RESONANCE Page # 36

37.  dm 
 Lf
i1  dt  L
 f

i2  dm  Lv

 Lv
 dt 

400 / x 80
300 /(10   )x 540
  =9 Ans. 9
54. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The
radii of bubbles A and B are 2cm and 4cm, respectively. Surface tension of the soap-water used to make
bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A and B,
respectively. [Neglect the effect of gravity.]
Sol.
4T 4  0.04
PA = P0 + r PA = 8 +
A 0.02

PA = 16 N/m2
4T 4  0.04
PB = P0 + =8+
rB 0.04
PB = 12 N/m2
for bubble A, PV = nRT
4
(16)  (0.02)3 = nA RT ....(1)
3
for bubble B
4 3
(12)  (0.04 )  = nBRT ....(2)

3 
dividing eqn (i) and (2)
nA 1 nB
6
nB 6 ; nA Ans. 6

55. A 20cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The
string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm)
between the successive nodes on the string.
Sol.
T 0 .5
v= = 3 = 10 m/sec.
 10 / 0 .2
v = f
10 = (100)   = 0.1 m = 10 cm
distance between two successive nodes = = 5 cm

2
Ans. 5
RESONANCE Page # 37

38. 56. Three objects A,B and C are kept in a straight line on a frictionless horizontal surface. These have masses
m, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision
with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line.
Find the final speed (in m/s) of the object C.
Sol.
from momentum conservation :
9m = (2m) V1 – (m)V2
 9 = 2V1 – V2 ..... (1)
V1  V2
e=  1 ......(2)
9
from eqn(1) and eqn(2) V1 = 6 m/sec.
for second collision between second block and third block :
(2m) 6 + m(0) = (2m + m) VC
 VC = 4 m/sec.
Ans. 4
57. A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed
and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the
bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady
with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the
orifice.
[Take atmospheric pressure = 1.0 × 105 N/m2 , density of water = 1000 kg/m3 and g = 10 m/s2. Neglect any
effect of surface tension]
Sol.
P0 = atmospheric pressure
P + 200 × 10–3 × 1000 × 10 = P0 ....(1)
P0 (500  H) mm
 P= 300 mm ....(2)
RESONANCE Page # 38

39. from (1) and (2)
P0 ( 500  H)
+ 2000 = P0
300
105 (500  H)
+ 2000 = 105
300
 5 × 107 – H × 105 + 6 = 300
 H = 206 mm
fall in height = 6 mm
Ans. 6
RESONANCE Page # 39

40. Name : ___________________________ Roll No. :
C. Question paper format :
10. The question paper consists of 3 parts (Part I : Chem istry, Part II : Mathematics,
Part-III : Physics). Each part has 4 sections.
11. Section I contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and
(D) for its answer, out of which only one is correct.
12. Section II contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C)
and (D) for its answer, out of which one or more is/are correct.
13. Section III contains 2 questions Each question has four statements (A, B, C and D) given in
Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I
can have correct matching with one or more statement(s) given in Column II. For example, if
for a given question, statement B matches with the statements given in q and r, then for that
particular question, against statement B, darken the bubbles corresponding to q and r in the
ORS.
14. Section IV contains 8 questions. The answer to each of the questions is a single-digit integer,
ranging from 0 to 9. The answer will have to be appropriately bubbled in the ORS as per the
instructions given at the beginning of the section.
Fill your Answer as is given in the following example.
Example : If the correct answers to question numbers X, Y, Z and W (say) are 6,0,9 and 2,
respectively, then the correct darkening of bubbles will look like the following :
D. Marking Scheme :
16. For each question in Section I , you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all case of
bubbling of incorrect answer, minus one (–1) mark will be awarded.
17. For each question in Section II , you will be awarded 4 marks if you darken the bubble(s)
corresponding to the correct choice(s) for the answer and zero mark if no bubbled is darkened.
In all other cases, minus one (–1) mark will be awarded.
18. For each question in Section III, you will be awarded 2 marks for each row in which you
have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this
section carries a maximum of 8 marks. There is no negative marking for incorrect answer(s)
for this this section.
19. For each question in Section IV, you will be awarded 4 marks if you darken the bubble
corresponding to the correct answer, and zero mark if no bubble is darkened. In all other cases,
minus one (–1) mark will be awarded.