Jee 2009 Paper 1

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

Date : 12-04-2009 Duration : 3 Hours Max. Marks : 240

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D. Marking schemeRead the instructions printed on the back page of this booklet.

PAPER - 1

Name of the Candidate

Roll Number

I have read all the instructions andshall abide by them.

-------------------------------- Signature of the Candidate

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QUESTIONS & SOLUTIONS OF IIT-JEE 2009

RESONANCE Page # 2

PART- I

CHEMISTRYSECTION - I

Straight Objective Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D)

for its answer, out of which ONLY ONE is correct.

1. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3

sol is :

(A) Na2SO4 (B) CaCl2 (C) Al2(SO4)3 (D) NH4Cl

Sol. (C)

Most effective coagulating agent for Sb2S3 is Al2(SO4)3 because of high charge.

2. The term that corrects for the attractive forces present in a real gas in the vander Waals equation is :

(A) nb (B) 2

2

V

an(C) � 2

2

V

an(D) �nb

Sol. (B)

Correction factor for attractive force in to the real gas is given by 2

2

V

an.

3. The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of

N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure

is :

(A) 4 × 10�4 (B) 4.0 × 10�5 (C) 5.0 × 10�4 (D) 4.0 × 10�6

Sol. (A)

PN2 = KH ×

2Nx

2Nx = 510

1 × 0.8 × 5 = 4 × 10�5 per mole

In 10 mole solubility is 4 × 10�4 .

RESONANCE Page # 3

4. The reaction of P4 with X leads selectively to P4O6. The X is :

(A) Dry O2 (B) A mixture of O2 and N2

(C) Moist O2 (D) O2 in the presence of aqueous NaOH

Sol. (B)

P4 + 3O2 2NofpresenceIn P4O6. Here nitrogen acts as diluent.

Note :

In dry O2 following reactions may take place.

P4 + 3O2 P4O6.

P4O6 + 2O2 P4O10.

In moist O2 the P4O6 gets hydrolysed forming H3PO3.

P4O6 + 6H2O 4H3PO3.

In presence of NaOH.

P4 + 3OH� + 3H2O PH3 + 3H2PO2�

5. Among celluose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force

of attraction is weakest is :

(A) Nylon (B) Poly(vinyl chloride) (C) Cellulose (D) Natural Rubber

Sol. (D)

The natural rubber has intermolecular forces which are weak dispersion force (van-der-waal forces of attraction)

and is an example of an elastomer (polymer).

6. The correct acidity order of the following is :

(A) (III) > (IV) > (II) > (I) (B) (IV) > (III) > (I) > (II)

(C) (III) > (II) > (I) > (IV) (D) (II) > (III) > (IV) > (I)

Sol. (A) H�

More stable conjugate base (carboxylate anion, �ve charge always on oxygen atom).

RESONANCE Page # 4

H�

More stable conjugate base (carboxylate anion, �ve charge always on oxygen atom). + I and hyperconjugative

effect of � CH3 group decreases stability of benzoate anion.

H�

Less stable conjugate base (phenoxide ion is less resonance stabilized than benzoate anion). �Cl group

exhibits � I effect.

H�

Less stable conjugate base (phenoxide ion is less resonance stabilized than benzoate anion).

7. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, the atomic

mass of Fe is :

(A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05

Sol. (B) 54Fe 5%

56Fe 90%

57Fe 5%

Av. atomic mass = x1A1 + x2A2 + x3A3

= 54 × 0.05 + 56 × 0.9 + 57 × 0.05 = 55.95

RESONANCE Page # 5

8. The IUPAC name of the following compound is :

(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile

(C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile

Sol. (B)

2-Bromo-5-hydroxybenzonitrile (�CN group gets higher priority over �OH and �Br)

SECTION - II

Multiple Correct Answer Type


for its answer, out of which ONE OR MORE is/are correct.

9. The correct statement(s) about the compound H3C(HO)HC�CH=CH�CH(OH)CH3 (X) is(are) :

(A) The total number of stereoisomers possible for X is 6.

(B) The total number of diastereomers possible for X is 3.

(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4.

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2.

Sol. (A, D) (X) has configurations

(1) R Z R

(2) S Z S

(3) R Z S

(4) R E R

(5) S E S

(6) R E S

RESONANCE Page # 6

10. The compound(s) formed upon combustion of sodium metal in excess air is(are) :

(A) Na2O2 (B) Na2O (C) NaO2 (D) NaOH

Sol. (A, B)

6Na + 2O2 (from air) combustion 2Na2O + Na2O2

11. The correct statement(s) regarding defects in solids is(are) :

(A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion.

(B) Frenkel defect is a dislocation defect.

(C) Trapping of an electron in the lattice leads to the formation of F-center.

(D) Schottky defects have no effect on the physical properties of solids.

Sol. (B, C) Frenkel defect is a dislocation defect.

Trapping of an electron in the lattice leads to the formation of F-center.

12. The compound(s) that exhibit(s) geometrical isomerism is(are) :

(A) [Pt(en)Cl2] (B) [Pt(en)2]Cl2 (C) [Pt(en)2Cl2]Cl2 (D) [Pt(NH3)2Cl2]

Sol. (C, D)

(A) [Pt(en)Cl2] : exist only in one form.

(B) [Pt(en)2]Cl2 : exist only in one form

(en is symmetrical lignad).

(C) [Pt(en)2Cl2]Cl2 :

RESONANCE Page # 7

(D) [Pt(NH3)2Cl2] : Pt is in +2 oxidation state having 5d8 configuration. Hence the hybridisation of complex is

dsp2 and geometry is square planar.

SECTION - III

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice questions based on

a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE

is correct.

Paragraph for Question Nos. 13 to 15A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by

dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes

intramolecular aldol reaction to give predominantly S.

P

,SOH.3OH,H.2

MeMgBr.1

42

2

Q OH,Zn.2

O.1

2

3 R

.2

OH.1 S

13. The structure of the carbonyl compound P is :

(A) (B) (C) (D)

Ans. (B)

14. The structures of the products Q and R, respectively, are :

(A) , (B) ,

(C) , (D) ,

Ans. (A)

RESONANCE Page # 8

15. The structure of the product S is :

(A) (B)

(C) (D)

Ans. (B)

Sol. (13, 14 & 15)

OH,H)2(

MeMgBr)1(

2

,SOH 42

oncondensatialdolramolcularint

OH

OH,Zn/O 23

Paragraph for Question Nos. 16 to 18p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with

a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment

of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an

intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of

the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the

formation of Z.

RESONANCE Page # 9

16. The compound X is :

(A) NaNO3 (B) NaCl (C) Na2SO4 (D) Na2S

Ans. (D)

17. The compound Y is :

(A) MgCl2 (B) FeCl2 (C) FeCl3 (D) ZnCl2

Ans. (C)

Sol. (16 to 17)

+ S2� (X) + + 6Fe3+ (Y)

6Fe2+ + NH4+ + 2H+ + (methylene blue).

The compound X is Na2S and Y is FeCl3.

18. The compound Z is :

(A) Mg2[Fe(CN)6] (B) Fe[Fe(CN)6] (C) ) Fe4[Fe(CN)6]3 (D) ) K2Zn3[Fe(CN)6]2

Sol. (B)

Fe3+ + [Fe(CN)6]3� Fe[Fe(CN)6] (brown colouration) (Z).

SECTION - IV

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 :

RESONANCE Page # 10

19. Match each of the compounds in Column I with its characteristic reaction(s) in Column II.

Column I Column II

(A) CH3CH2CH2CN (p) Reduction with Pd-C/H2

(B) CH3CH2OCOCH3 (q) Reduction with SnCl2/HCl

(C) CH3�CH=CH�CH2OH (r) Development of foul smell on treatment with

chloroform and alcoholic KOH.

(D) CH3CH2CH2CH2NH2 (s) Reduction with diisobutylaluminium hydride (DIBAL-H)

(t) Alkaline hydrolysis

Ans. (A) � p, q, s, t ; (B) � p, s, t ; (C) � p ; (D) � r

Sol.

(A) CH3CH

2CH

2CN

2H/CPd CH

3CH

2CH

2CH

2NH

2(p)

HCl/SnCl2 CH

3CH

2CH

2CHO (q)

HDIBAl

CH3CH

2CH

2CHO (s)

OH

OH

2

CH3CH

2CH

2COO (t)

(B) CH3CH

2OCOCH

3

2H/CPd 2CH3�CH

2 � OH (p)

HDIBAl

CH3�CHO (s)

OH

OH

2

CH3�COO (t)

(C) CH3�CH=CH�CH

2OH

2H/CPd CH

3CH

2CH

2CH

2OH (p)

(D) CH3CH

2CH

2CH

2NH

2

KOH

CHCl3 CH3CH

2CH

2CH

2NC (r)

RESONANCE Page # 11

20. Match each of the diatomic molecules in Column I with its property/properties in Column II.

Column I Column II

(A) B2 (p) Paramagnetic

(B) N2 (q) Undergoes oxidation

(C) O2� (r) Undergoes reduction

(D) O2 (s) Bond order 2

(t) Mixing of 's' and 'p' orbitals

Ans. (A) - p, r, t ; (B) - s, t ; (C) - p, q, r ; (D) - p, r, s

Sol. (A) B2 1s2 1s 2 2s 2 2s 2 2px1 = 2py1

Bond order = 1 Paramagnetic

(B) N2 1s2 1s 2 2s 2 2s 2 2px2 = 2py2 2pz2

Bond order = 3 Diamagnetic

(C) O2� 1s2 1s 2 2s 2 2s 2 2pz2 2px2 = 2px 2 2px 2 = 2py 1

Bond order = 1.5 Paramagnetic

(D) O2 1s2 1s 2 2s 2 2s 2 2pz2 2px2 = 2px 2 2px 1 = 2py 1

Bond order = 2 Paramagnetic

RESONANCE Page # 12

PART-II

MATHEMATICSSECTION - I

Single Correct Choice Type


for its answer, out of which ONLY ONE is correct.

21. Let z = cos + i sin . Then the value of

15

1m

1m2 )z(m at = 2º is

(A) º2sin

1(B)

º2sin31

(C) º2sin2

1(D)

º2sin41

Sol. (D)

Given that z = cos + i sin = ei

15

1m

Im (z2m�1) =

15

1m

Im (ei)2m�1 =

15

1m

Im ei(2m�1) = sin+ sin3+ sin5 + .......... + sin29

=

22

sin

2215

sin229

sin

=

sin)15(sin)15sin(

= 4sin2º

1

22. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and

3 only, is

(A) 55 (B) 66 (C) 77 (D) 88

Sol. (C)

There are two possible cases

Case 1 : Five 1�s, one 2�s, one 3�s

Number of numbers = !5!7 = 42

Case 2 : Four 1�s, three 2�s

Number of numbers = !3!4

!7 = 35

Total number of numbers = 42 + 35 = 77

RESONANCE Page # 13

23. Let P(3, 2, 6) be a point in space and Q be a point on the line )k�5j�i�3()k�2j�i�(r

. Then the value

of for which the vector PQ is parallel to the plane x � 4y + 3z = 1 is

(A) 41

(B) � 41

(C) 81

(D) � 81

Sol. (A)

Given OQ = (1�3) i� + ( �1) j� + (5 +2)k� , OP = k�6j�2i�3 (where O is origin)

PQ = (1�3� 3) i� + (�1�2) j� + (5 +2�6) k�

= (�2 � 3) i� + (�3) j� + (5 �4) k�

PQ is parallel to the plane x � 4y + 3z = 1

�2�3 �4 + 12 + 15 � 12 = 0

8 = 2 = 4

1

24. Let f be a non-negative function defined on the interval [0, 1]. If dt))t(f(1x

0

2 =

x

0

,dt)t(f 0 x 1 and

f(0) = 0, then

(A) f

21

< 21

and f

31

> 31

(B) f

21

> 21

and f

31

> 31

(C) f

21

< 21

and f

31

< 31

(D) f

21

> 21

and f

31

< 31

Sol. (C)

Given x

0

2))t('f(�1 dt =

x

0

1x0,dt)t(f

Apply Leibnitz theorem, we get 2))x('f(�1 = f (x)

1 � (f(x))2 = f2(x) (f(x))2 = 1�f2(x)

'f (x) = )x(f�1 2 dxdy

= 2y�1 , where y = f(x)

2y�1

dy = dx

Integrating both sides

sin �1 (y) = x + c

f (0) = 0 c = 0

RESONANCE Page # 14

y = ± sin x

y = sin x = f(x) given f(x) 0 for x [0, 1]

it is known that sinx < x, x R+

sin

2

1 < 2

1 f

2

1 < 2

1and sin

3

1 < 3

1 f

3

1 < 3

1

25. Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whose vertices

are the roots of the equation 350zzzz 33 is

(A) 48 (B) 32 (C) 40 (D) 80

Sol. (A)

z z (z2 + 2z ) = 350

2(x2 + y2) (x2 � y2 ) = 350

(x2 + y2) (x2 �y2) = 175

Since x, y I , the only possible case which gives integral solution, is

x2 + y2 = 25 ......... (1)

x2 � y2 = 7 ..........(2)

From (1) and (2) x2 = 16 ; y2 = 9

x = 4 ; y = 3 Area = 48

26. Tangents drawn from the point P(1, 8) to the circle x2 + y2 � 6x � 4y � 11 = 0 touch the circle at the points A

and B. The equation of the circumcircle of the triangle PAB is

(A) x2 + y2 + 4x � 6y + 19 = 0 (B) x2 + y2 � 4x � 10y + 19 = 0

(C) x2 + y2 � 2x + 6y � 29 = 0 (D) x2 + y2 � 6x � 4y + 19 = 0

Sol. (B)

For required circle, P(1, 8) and O(3, 2) will be the end points of its diameter.

(x � 1) (x � 3) + (y � 8) (y � 2) = 0

x2 + y2 � 4x � 10y + 19 = 0

RESONANCE Page # 15

27. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse

x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and

the origin O is

(A) 1031

(B) 1029

(C) 1021

(D) 1027

Sol. (D)

Equation of auxiliary circle is x2 + y2 = 9 (1)

Equation of AM is 3x

+ 1y

= 1 (2)

on solving (1) and (2), we get M

59

,5

12

Now, area of AOM = 21

. OA × MN = 10

27 square unit

28. If c,b,a

and d

are unit vectors such that )dc(.)ba(

= 1 and 21

c.a

, then

(A) c,b,a

are non-coplanar (B) d,c,b

are non-coplanar

(C) d,b

are non-parallel (D) d,a

are parallel and c,b

are parallel

Sol. (C)

Let angle between a

and b

be 1 , c

and d

be 2 and ba

and dc

be .

)dc(.)ba(

= 1

sin 1 . sin 2 . cos = 1

1 = 90º, 2 = 90º , = 0º

)dc(||)ba(,dc,ba

so ba

= )dc(k

and ba

= k )dc(

c.)ba(

= c.)dc(k

and d.)ba(

= k d.)dc(

0]cba[

and 0]dba[

c,b,a

and d,b,a

are coplanar vectors so options A and B are incorrect

Let b

|| d

db

as )dc(.)ba(

= 1 )bc(.)ba(

= ± 1

]bcba[

= ± 1

]babc[

= ± 1

RESONANCE Page # 16

1)]ba(b[.c

]b)a.b(a[.c

= ± 1

1a.c

( 0b.a

)

which is a contradiction so option 'C' is correct

Let option 'D' is correct

d

= ± a

and c

= ± b

as 1)dc(.)ba(

)ab(.)ba(

= ± 1

which is a contradiction so option 'D' is incorrect

Alternatively option 'C' and 'D' may be observed from the given figure

SECTION - II

Multiple Correct Choice Type


for its answer, out of which ONE OR MORE is/are correct.

29. Let L = 4

222

0x x4x

xaalim

, a > 0. If L is finite, then

(A) a = 2 (B) a = 1 (C) L = 641

(D) L = 321

Sol. (A, C)

L = 4

222

0x x4x

xaalim

, a > 0 = 0x

lim

4

2

4

4

2

2

x

4x

....a

x.

2

121

21

a

x.

21

1.aa

= 0xlim

4

2

3

42

x4x

......a

x.

81

a2x

Since L is finite 2a = 4 a = 2

L = 0xlim

3a.8

1 = 64

1

RESONANCE Page # 17

30. Area of the region bounded by the curve y = ex and lines x = 0 and y = e is

(A) e � 1 (B)

e

1

dy)y1e(n (C) e � 1

0

x dxe (D) e

1

dyyn

Sol. (B, C, D)

Shaded area = e �

1

0

x dxe = 1

Also

e

1

dyy)1( en

put e + 1 � y = t � dy = dt

= 1

e

tn (�dt) = dttne

1 = 1dyyn

e

1

31. If 2

xsin4

+ 3

xcos4

= 51

, then

(A) tan2x = 32

(B) 8

xsin8

+ 27

xcos8

= 125

1

(C) tan2 x = 31

(D) 8

xsin8

+ 27

xcos8

= 125

2

Sol. (A, B)

2xsin4

+ 3

xcos4

= 51

2

xsin4

+ 3

)xsin1( 22 =

51

2

xsin4

+ 3

xsin2xsin1 24 =

51

5 sin4x � 4 sin2x + 2 = 56

25 sin4x � 20 sin2x + 4 = 0

(5 sin2x � 2)2 = 0

sin2x = 52

, cos2x = 53

tan2x = 3

2

and8

xsin8 +

27

xcos8 = 125

1

RESONANCE Page # 18

32. In a triangle ABC with fixed base BC, the vertex A moves such that cos B + cos C = 4 sin2 2A

. If a, b and c

denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then

(A) b + c = 4a (B) b + c = 2a

(C) locus of points A is an ellipse (D) locus of point A is a pair of straight lines

Sol. (B, C)

cos B + cos C = 4 sin2 2A

2 cos 2

CB cos

2CB

= 4 sin2 2A

2 sin 2A

2A

sin22

CBcos = 0

cos

2CB

� 2 cos

2CB

= 0 as sin 2A

0

� cos 2B

cos 2C

+ 3 sin 2B

sin 2C

= 0

tan 2B

tan 2C

= 31

)cs(s)as)(bs(

.)bs(s

)cs)(as(

=

31

s

as =

31

2s = 3a b + c = 2a

Locus of A is an ellipse

SECTION - III

This section contains 2 groups of questions. Each group has 3 multiple choice questions based on

a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE

is correct.

Paragraph for Question Nos. 33 to 35

Let be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries

are 1 and four of them are 0.

RESONANCE Page # 19

33. The number of matrices in is

(A) 12 (B) 6 (C) 9 (D) 3

Sol. (A)

Case I : All three diagonal elements are 1

No. of matrices = 3C1 = 3

Case II : Two diagonal elements are zero & one element is one

No. of matrices = 3C1 . 3C1 = 9

Total matrices = 3 + 9 = 12

34. The number of matrices A in for which the system of linear equations AA

z

y

x

=

0

0

1

has a unique solution,

is

(A) less than 4 (B) at least 4 but less than 7

(C) at least 7 but less than 10 (D) at least 10

Sol. (B)

A

z

y

x

=

0

0

1

For unique solution det (A) 0

Case I det(A) = = 1 � a2 � b2 � c2 + 2abc 0

Here a, b, c is selected from 1, 0, 0. (No case is possible)

Case II

(i) det(A) = = 2abc � c2 0

Here a, b, c are selected from 1, 1, 0. (2 cases are possible)

(ii) det(A) = = 2abc � b2 0


(iii) det(A) = = 2abc � a2 0


Hence there are exactly 6 matrices for unique solution . Hence option B is correct

RESONANCE Page # 20

35. The number of matrices A in for which the system of linear equations A

z

y

x

=

0

0

1

is inconsistent, is

(A) 0 (B) more than 2 (C) 2 (D) 1

Sol. (B)

A

z

y

x

=

0

0

1

Case I :

1cb

c1a

ba1

z

y

x

=

0

0

1

a, b, c are selected from 1, 0, 0

x + ay + bz = 1

ax + y + cz = 0

bx + cy + z = 0

(i) If a = 1 , b = c = 0

then x + y = 1 Inconsitent system of equation

x + y = 0

(ii) If a = 0 = c, b = 1

then x + z = 1

y = 0 Inconsitent system of equation

x + z = 0

(iii) If c = 1, a = b = 0

then x = 1, z = 0, y = 0

Case II :

(i)

0cb

c0a

ba1

z

y

x

=

0

0

1

a, b, c are selected from 1, 1, 0

x + ay + bz = 1

ax + cz = 0

bx + cy = 0 Clearly, In all three cases, solutions are possible so system is consistent.

RESONANCE Page # 21

(ii)

0cb

c1a

ba0

z

y

x

=

0

0

1

ay + bz = 1

ax + y + cz = 0

bx + cy = 0

Clearly , b = 0, a= c= 1 gives

y = 1

x + y + z = 0 Inconsistent system

y = 0

More than 2 matrices are possible. Hence option B is correct


A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required.

36. The probability that X = 3 equals

(A) 21625

(B) 3625

(C) 365

(D) 216125

Sol. (A)

P(X = 3) = 65

. 65

. 61

= 216

25

37. The probability that X 3 equals

(A) 216125

(B) 3625

(C) 365

(D) 21625

Sol. (B)

P(X 3) = 65

. 65

. 1 = 36

25

38. The conditional probability that X 6 given X > 3 equals

(A) 216125

(B) 21625

(C) 365

(D) 3625

Sol. (D)

P((X 6) / (X > 3)) = )3X(P

)6X(P.))6X/()3X((P

=

......61

.65

61

.65

.....61

.65

61

.65

.1

43

65

= 36

25

RESONANCE Page # 22

SECTION - IV

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.

39. Match the statements/expressions in Column - I with the open intervals in Column - II

Column - I Column - II

(A) Interval contained in the domain of definition of (p)

2,

2�

non-zero solutions of the differential

equation (x � 3)2 y + y = 0

(B) Interval containing the value of the integral (q)

2,0

5

1

dx)5�x()4�x()3�x()2�x()1�x(

(C) Interval in which at least one of the points of local (r)

45

,8

maximum of cos2x + sinx lies

(D) Interval in which tan�1 (sinx + cosx) is increasing (s)

8,0

(t) (� , )

RESONANCE Page # 23

Sol. (A)

(x � 3)2 . y + y = 0

dxdy

= � 2)3x(

y

ydy

=

2)3x(

dx

n y = )3x(

1

+ n C

y = 3x1

Ce , C 0

x R � {3} (hence correct options p, q, s)

Aliter

Given differential equation is homogeneous linear differential equation and has x = 3 as a singular

point hence x = 3 cannot be in domain of solution.

(B) Let I =

5

1

dx)5x()4x()3x()2x()1x(

Let x � 3 = t dx = dt

I =

2

2

dt)2t()1t(t)1t)(2t( ,

Integrand is an odd function

I = 0 (hence correct options are p, t)

Aliter

Let =

5

1

dx)5x()4x()3x()2x()1x( .......(1)

Using b

a

dx)x(f =

b

a

dx)xba(f

=

5

1

dx)x1()x2()x3()x4()x5( ........(2)

On adding (1) and (2), we get 2 = 0 = 0

RESONANCE Page # 24

(C) f(x) = cos2x + sin x

f(x) = � 2 cos x sin x + cos x = cos x (1 � 2 sin x) = 0 (hence correct options are p, t)

Sign scheme for first derivative

Points of local maxima are 6

, 6

5

Clearly option (p, q, r, t) are correct.

Aliter

y = cos2x + sin x

y = 45

�

2

21

xsin

For y to be maximum.

2

21

xsin

= 0

sin x = 21

x = n + (�1)n 6

, n

Hence option (p, q, r, t) are correct

(D) y = tan�1 (sin x + cos x)

dxdy

= 2)xcosx(sin1

xsinxcos

Clearly by graph cos x > sin x is true for option(s)

Ans. (A) (p, q, s), (B) (p, t), (C) (p, q, r, t), (D) (s)

RESONANCE Page # 25

40. Match the conics in Column - I with the statements/expressions in Column - II.

Column - I Column - II

(A) Circle (p) The locus of the point (h, k) for which the

line hx + ky = 1 touches the

circle x2 + y2 = 4

(B) Parabola (q) Points z in the complex plane satisfying

|z + 2| � |z � 2| = 3

(C) Ellipse (r) Points of the conic have parametric

representation x = 3

2

2

t1

t�1,

y = 2t1

t2

(D) Hyperbola (s) The eccentricity of the conic lies in the

interval 1 x <

(t) Points z in the complex plane satisfying

Re(z + 1)2 = |z|2 + 1

Sol. (p)22 kh

1

= 2 h2 + k2 =

41 locus is a circle.

(q) 32�z�2z and 2 �(�2) = 4 > 3 locus is a hyperbola.

(r) x = 3

2

2

t1

t1, y = 2t1

t2

Let tan =t

x = 3 cos 2 y = sin 2

Hence 3x2

+ y2 = 1 locus is an ellipse

(s) Eccentricity x = 1 Parabola , 1< x < Hyperbola

(t) z = x + iy

Re(z + 1)2 = |z|2 + 1

(x + 1)2 � y2 = x2 + y2 + 1

2x = 2y2

x = y2 Hence locus is a parabola

Ans. (A) (p), (B) (s, t), (C) (r), (D) (q, s)

RESONANCE Page # 26

PART-III

PHYSICS

SECTION - I

Single Correct Choice Type

Thus section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its

answer, out which ONLY ONE is correct.

41. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at

(�a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4

to x = 5a/4. Two point charges �7C and 3C are placed at (a/4, �a/4, 0) and (�3a/4, 3a/4, 0), respectively.

Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ±a/2. The electric flux through this

cubical surface is

(A) 0

C2

(B)

0

C2

(C) 0

C10

(D) 0

C12

Sol. (A) From Gauss's law

0

inQ

=

0

)2/C6(C7�)4/C8(

= �

0

C2

.

42. The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the

particle at t = 4/3 s is

0 4 8 12t(s)

1

�1

x(cm

)

(A) 2

323 cm/s2 (B)

32

2 cm/s2 (C)

32

2 cm/s2 (D) � 2

323 cm/s2

RESONANCE Page # 27

Sol. (D)

T = 8 second., A = 1 cm, x = A sint. = 1sin82

t.

a = � 2x = �

2

82

sin

82

t cm/s2

At, t = 34

second. a = �

2

82

sin

3

= � 323 2

cm/s2.

43. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is

found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the

charges given to the shells, Q1 : Q2 : Q3, is

(A) 1 : 2 : 3 (B) 1 : 3 : 5 (C) 1 : 4 : 9 (D) 1 : 8 : 18

Sol. (B)

The charge distribution on the surfaces of the shells are given. As per the given condition.

21

R4

Q

= 2

21

)R2(4

QQ

= 2

321

)R3(4

QQQ

5Q

3Q

1Q 321 .

44. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is

4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, When the ball is

12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2]

(A) 9 m/s (B) 12 m/s (C) 16 m/s (D) 21.33 m/s

Sol. (C)

x' = relnx

, v' = relnv

= 34

1

)8.12�20(102

= 16 m/s.

45. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of

ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink

used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0,

0), left inner circle (�a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, �a). The y-

coordinate of the centre of mass of the ink in this drawing is

RESONANCE Page # 28

(A) 10a

(B) 8a

(C) 12a

(D) 3a

Sol. (A)

ycm = 54321

5544332211mmmmm

ymymymymym

ycm = m6mmmm

)a(�m)a(m)0(m)a(m)0(m6

=

10a

.

46. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction

between them is 3 . The inclination of this inclined plane from the horizontal plane is gradually increased

from 0º. Then

(A) at = 30º, the block will start sliding down the plane

(B) the block will remain at rest on the plane up to certain and then it will topple

(C) at = 60º, the block will start sliding down the plane and continue to do so at higher angles

(D) at = 60º, the block will start sliding down the plane and on further increasing , it will topple at certain

Sol. (B)

Angle of repose 0 = tan�1 = tan�1 3 = 60º

tan = 2/15

5 =

32

. < 45º.

Block will topple before it starts slide down.

47. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular

orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the

particles move with constant speeds. After making how many elastic collisions, other than that at A, these

two particles will again reach the point A?

RESONANCE Page # 29

A 2vv

(A) 4 (B) 3 (C) 2 (D) 1

Sol. (C)

Since masses of particles are equal, collisons are elastic, so particles will exchange velocities after

each collision. The first collision will be at a point P and second at point Q again and before third

collision the particles will reach at A.

48. The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a

perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field

increases with time. 1 and 2 are the currents in the segments ab and cd. Then,

(A) 1 > 2(B) 1 < 2(C) 1 is in the direction ba and 2 is in the direction cd

(D) 1 is in the direction ab and 2 is in the direction dc

Sol. (D)

Current 1 = 2 ,

Since magnetic field increases with time

So induced net flux should be outward i.e. current will flow from a to b

RESONANCE Page # 30

SECTION - II

Multiple 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 ONE OR MORE is/are correct.

49. A student performed the experiment of determination of focal length of a concave mirror by u-v method using

an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the

location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56),

(48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incor-

rectly recorded, is (are)

(A) (42, 56) (B) (48, 48) (C) (66, 33) (D) (78, 39)

Sol. (C, D)

f1

u1

v1

or |f|1

|u|1

|v|1

|v| = |f||u|

|f||u|

For |u| = 42, |f| = 24 ; |v| = 2442

)24()42(

= 56 cm so (42, 56) is correct observation

For |u| = 48 or |u| = 2f or |v| = 2f so (48, 48) is correct observation

For |u| = 66 cm ; |f| = 24 cm

|v| = 2466

)24()66(

36 cm which is not in the permissible limit

so (66, 33), is incorrect recorded

For |u| = 78, |f| = 24 cm

|v| = 2478

)24()78(

32 cm which is also not in the permissible limit.

so (78, 39), is incorrect recorded.

50. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame,

one can surely say that

(A) linear momentum of the system does not change in time

(B) kinetic energy of the system does not changes in time

(C) angular momentum of the system does not change in time

(D) potential energy of the system does not change in time

RESONANCE Page # 31

Sol. (A)

(A) Since there in no resultant external force, linear momentum of the system remains constant.

(B) Kinetic energy of the system may change.

(C) Angular momentum of the system may change as in case of couple, net force is zero but torque

is not zero. Hence angular momentum of the system is not constant.

(D) Potential energy may also change.

51. Cv and Cp denote the molar specific heat capacities of a gas at constant volume and constant pressure,

respectively. Then

(A) Cp � Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas

(B) Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas

(C) Cp / Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas

(D) Cp . Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas

Sol. (B, D)

For monoatomic gas, Cp = 25

R, Cv = 23

R. Cp � Cv = R

For diatomic gas Cp = 27

R, Cv = 25

R. Cp � Cv = R

Cp � Cv is same for both

Cp + Cv = 6R (for diatomic)

Cp + Cv = 4R (for mono)

so (Cp + Cv)dia > (Cp + Cv)mono

57

C

C

v

p = 1.4 (for diatomic)

35

C

C

v

p = 1.66 (for monoatomic)

(Cp) (Cv) = 4

35R2 = 8.75 R2 (for diatomic)

(Cp) (Cv) = 4

15 R2 (for monoatomic)

so (Cp . Cv)diatomic > (Cp . Cv)monoatomic

52. For the circuit shown in the figure

[Current Electricity]

(A) the current I through the battery is 7.5 mA

(B) the potential difference across RL is 18 V

(C) ratio of powers dissipated in R1 and R2 is 3

(D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9

RESONANCE Page # 32

Sol. (A, D)

Req =

2

5.75.16

= 56

+ 2 = 5

16 k

= eqR

V =

1624

× 5 mA = 23

× 5 = 7.5 mA

for potential difference across R1 V1 = 7.5 × 2 = 15 V

for potential difference across R2 V2 = 24 � 15 = 9 V

for power :21 RR P:P =

2

22

1

21

RV

:RV

= 6

9:

2)15( 22

= 3

25

L

22

R RV

PL =

5.192

= 54 mW

If R1 and R2 are interchanged

R´ = R1 || R2 =

5.12)5.1()2(

= 5.3

3

V´L = ´RR

´R

2 × 24 V = 3 V

Now power dissipated in RL is P´L = L

2´L

RV

= 5.1

32

= 6 mW

SECTION�III

Comprehension TypeThis section contains 2 groups of questions. Each group has 3 multiple choice questions based on a

paragraph. Each question has 4 choices (A),(B),(C) and (D) for its answer, out of which ONLY ONE is

correct.


Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, H21 , known as

deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is

energynHeHH 32

21

21 . In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into

deuteron nuclei and electrons. This collection of H21 nuclei and electrons is known as plasma. The nuclei

move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place.

Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the

plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from

the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of

the criteria, a reactor is termed successful if Lawson number is greater than 5×1014 s/cm3.

It may be helpful to use the following: Boltzman constant k = 8.6×10�5 eV/K ;0

2

4e

= 1.44 × 10�9 eVm.

RESONANCE Page # 33

53. In the core of nuclear fusion reactor, the gas becomes plasma because of

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between deuterons-electrons pairs

(D) the high temperature maintained inside the reactor core

Sol. (D)

the high temperature maintained inside the reactor core

54. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each

other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb

potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T

required for them to reach a separation of 4 × 10�15 m in the range.

(A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K

(C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K

Sol. (A)

From energy conservation

2 re

41

kT23 2

0

T = )106.8(

)1044.1(5

9

× 151043

1

= 1.39 × 109 K

55. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which

of these is most promising based on Lawson criterion ?

(A) deuteron density = 2.0 × 1012 cm�3, confinement time = 5.0 × 10�3 s

(B) deuteron density = 8.0 × 1014 cm�3, confinement time = 9.0 × 10�1 s

(C) deuteron density = 4.0 × 1023 cm�3, confinement time = 1.0 × 10�11 s

(D) deuteron density = 1.0 × 1024 cm�3, confinement time = 4.0 × 10�12 s

Sol. (B)

nt0 > 5 × 1014 s/cm3

for deuteron density = 8.0 × 1014 cm�3, confinement time = 9.0 × 10�1 s

nt0 = 7.2 × 1014 s/cm3


When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer

dimension, its energy can take only certain specific values. The allowed energies of the particle moving in

such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and

x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to

the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = m2

p2

.

Thus, the energy of the particle can be denoted by a quantum number �n� taking values 1,2,3,......., (n = 1,

called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line x = 0

to x = a. Take h = 6.6 × 10�34 J s and e = 1.6 × 10�19 C.

RESONANCE Page # 34

56. The allowed energy for the particle for a particular value of n is proportional to :

(A) a�2 (B) a�3/2 (C) a�1 (D) a2

Sol. (A)

n2

= a and p =

h =

a2nh

Energy E = m2

p2

= 2

22

ma8

hn 2a

1E

57. If the mass of the particle is m = 1.0 × 10�30 kg and a = 6.6 nm, the energy of the particle in its ground state

is closest to :

(A) 0.8 meV (B) 8 meV (C) 80 meV (D) 800 meV

Sol. (B)

E = 2

22

ma8

hn

For ground state n = 1 E1 = 192930

2342

106.1)106.6(108

)106.6()1(

= 8 meV

58. The speed of the particle, that can take discrete values, is proportional to :

(A) n�3/2 (B) n�1 (C) n1/2 (D) n

Sol. (D)

P = a2

nh = mv

v n

Section�IV

Matrix � Match TypeThis 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.

RESONANCE Page # 35

p q r sp q r sp q r sp q r s

ABCD

p q r s

t

t

ttt

59. Six point charges, each of the same magnitude q, are arranged in different manners as shown in

Column�II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field

and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it

is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let

B be the magnetic field at M and be the magnetic moment of the system in this condition. Assume each

rotating charge to be equivalent to a steady current.

Column�I Column�II

(A) E = 0 (p) + �

+�

+ �

Q

P

Charges are at the corners of a

regular hexagon. M is at the

centre of the hexagon. PQ is

perpendicular to the plane of

the hexagon.

(B) V 0 (q) P

Q

M

� + � + � +

Charges are on a line perpendicular to

PQ at equal intervals. M is the

midpoint between the two innermost

charges.

(C) B = 0 (r) +� +

��

+P

Q

M

Charges are placed on two coplanar

insulating rings at equal intervals. M is

the common centre of the rings. PQ is

perpendicular to the plane of rings.

(D) 0 (s) � �+

� �+

P QM

Charges are placed at the corners of a

rectangle of sides a and 2a and at the

mid�points of the longer sides. M is at

the centre of the rectangular. PQ is

parallel to the longer sides.

RESONANCE Page # 36

(t) +

+ + M

P

Q

�

� �

Charges are placed on two coplanar,

identical insulating rings at equal

intervals. M is the mid�point between

the centres of the rings. PQ is

perpendicular to the line joining the

centres and coplanar to the rings.

Ans. : (A) (p), (r), (s); (B) (r), (s); (C) (p), (q), (t) ; (D) (r), (s)

Sol. (p) + �

+�

+ �

Q

P

By symmetry E = 0 , V = 0 , B = 0

and µ = NA but effective = 0, So µ = 0

(q) P

Q

M

� + � + � +

E 0 , V = 0

Since effective = 0 B = 0 and µ = 0

(r) +� +

��

+P

Q

M

E = 0 (By Symmetry)

V 0 (since distances are different)

B 0 (since Radius is different)

µ 0

(s) � �+

� �+

P QM

E = 0 (By symmetry)

V 0 (since distances are different)

B 0

µ 0

(t) +

+ + M

P

Q

�

� �

E 0, V = 0, µ = 0, B = 0 and given each rotating charge

to be equivalent to a steady current so B = 0 so (t C)

and if it was not given that each rotating charge to be

equivalent to steady current then B 0.

RESONANCE Page # 37

60. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is

shown. Column I gives some statements about X and and/or Y. Match these statements to the appropriate

system(s) from Column II.

Column I Column II

(A) The force exerted (p) Block Y of mass M left on a

by X on Y has a fixed inclined plane X, slides

magnitude Mg. on it with a constant velocity.

(B) The gravitational (q) Two ring magnets Y and Z,

potential energy of each of mass M, are kept in

X is continuously frictionless vertical plastic

increasing, stand so that they repel each

other. Y rests on the base X

and Z hangs in air in

equilibrium. P is the topmost

point of the stand on the

common axis of the two rings.

The whole system is in a lift

that is going up with a constant

velocity.

(C) Mechanical energy (r) A pulley Y of mass m0 is fixed

of the system X + Y to a table through a clamp X.

is continuously A block of mass M hangs from

decreasing. a string that goes over the

pulley and is fixed at point P

of the table. The whole system

is kept in a lift that is going

down with a constant velocity.

(D) The torque of the (s) A sphere Y of mass M is put

weight of Y about in a nonviscous liquid X kept

point P is zero. in a container at rest. The

sphere is released and it moves

down in the liquid.

(t) A sphere Y of mass M is falling

with its terminal velocity in a

viscous liquid X kept in a

container.

Ans. : (A) (p), (t); (B) (q), (s), (t); (C) (p), (r), (t); (D) (q)

RESONANCE Page # 38

Sol. (p)

Net force on Y due to X = 22 )sinMg()cosMg( = Mg

(B) As the inclined is fixed. So, gravitational P.E. of X is constant(C) As K.E. is constant and P.E. of Y is decreasing. So mechanical energy of (X + Y) is decreasing.

(q)(A) force on Y due to X will be greater than Mg which is equal to (Mg + repulsion force)(B) As the system is moving up, P.E. of X is increasing.(C) Mechanical energy of (X + Y) is increasing(D) Torque of the weight of Y about point P = 0

(r)

(A) force on Y due to X = 220 )Mg(]g)mM[(

(B) As the system moves down, gravitational P.E. of X decreases(C) As the system moves down, total mechanical energy of (X + Y) also decreases(D) P 0

(s)

(A) force on Y due to X = Buoyancy force which is less than mg(B) As the sphere moves down, that volume of water comes up, so gravitational P.E. of X increases.(C) As there is no non�conservative force, so total mechanical energy of X + Y remains conserved.

(D) p 0

(t)

(A) As the sphere is moving with constant velocityB + fv = Mgso force on Y due to X is B + fv = Mg

(B) As the sphere moves down, that volume of water comes up, so gravitational P.E. of X will increase(C) Increase in mechanical energy = wfr = �ve

(D) p = 0

C. Question paper format :

10. The question paper consists of 3 parts (Part I : Chemistry, Part II : Mathematics,

Part-III : Physics). Each part has 4 sections.

11. Section�I contains 8 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 4 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 groups of questions. Each group has 3 questions based on a paragraph.

Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only one is

correct.

14. Section�IV 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.

D. Marking Scheme :

15. For each question in Section�I , you will be awarded 3 marks if you darken the bubble

corresponding to the correct answer and zero marks if no bubble is darkened. In case of bubbling

of incorrect answer, minus one (�1) mark will be awarded.

16. 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 bubble is darkened.

In all other cases, minus one (�1) mark will be awarded.

17. For each question in Section�III, 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.

18. For each question in Section�IV, 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

section.

Name : ___________________________ Roll No. :

Education

Jee 2009 Paper 1