INSTRUCTIONSA. General1. This booklet is your Question Paper containing 57 questions.

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B. Filling the ORS :

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C. Question paper format :

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

Name of the Candidate

Roll Number

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

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

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

PAPER - 2

QUESTIONS & SOLUTIONS OF IIT-JEE 2009

RESONANCE Page # 2

PART ICHEMISTRY

SECTION - IStraight 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) CH

3 at C-2 (D) H at C-2

Sol. (D)

shiftHydride2,1

(More stable carbocation due to

+m effect of � OH group and + and

hyperconjugative effect of �CH3 group)

RESONANCE Page # 3

3. For a first order reaction A P, the temperature (T) dependent rate constant (k) was found to follow the

equation log k = � (2000) T1

+ 6.0. The pre-exponential factor A and the activation energy Ea, respectively,,

are :

(A) 1.0 × 106 s�1 and 9.2 kJ mol�1 (B) 6.0 s�1 and 16.6 kJ mol�1

(C) 1.0 × 106 s�1 and 16.6 kJ mol�1 (D) 1.0 × 106 s�1 and 38.3 kJ mol�1

Sol. (D) From Arrhenius equation

K = Ae�Ea/RT

nk = nA � RTEa

2.303 log K = 2.303 log A � RTEa

log K = R303.2Ea

× T1

+ log AA ....... (1)

log K = � (2000) T1

+ 6 ........(2)

On comparing equation (1) and (2)

R303.2Ea

= �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

= )2n(n = 0 as there is no unpaired electrons.

RESONANCE Page # 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 + B2H

6 [BH

2(X)

2]+ [BH

4]�

the amine(s) X is(are) :

(A) NH3

(B) CH3NH

2(C) (CH

3)2NH (D) (CH

3)3N

Sol. (A, B, C)

Small amines such as NH3 ,CH

3NH

2 and (CH

3)

2NH give unsymmetrical cleavage of diborane according to

following reaction.

B2H

6 + 2 NH

3 [H

2B(NH

3)

2]+ [BH

4]�

Large ammines, such as (CH3)

3 N gives symmetrical cleavage of diborane according to following reaction.

B2H

6 + 2N (CH

3)

3 2H

3B N(CH

3)

3

6. The nitrogen oxide(s) that contain(s) N�N bond(s) is(are) :

(A) N2O (B) N

2O

3(C) N

2O

4(D) N

2O

5

Sol. (A, B, C)

(A)

N2O

(B)

(C)

N2O

4

(D)

N2O

5

RESONANCE Page # 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

Fe3+ (aq) + 3e� Fe Eº = �0.04 V

Au3+ (aq) + 3e� Au Eº = +1.40 V

Hg2+ (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 NO

3�.

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 # 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 # 7

Sol. (A)

(B)

(C)

(D)

RESONANCE Page # 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 HNO

3) 2NO + Cu(NO

3)

2 + 4H

2O

(B) Cu + 4HNO3 (concentrated) 2NO

2 + Cu(NO

3)

2 + 2H

2O

(C) 4Zn + 10HNO3 (dilute) 4Zn(NO

3)

2 + N

2O + 5H

2O

(D) Zn + 4HNO3 (concentrated) Zn(NO

3)

2 + 2NO

2 + 2H

2O

SECTION - IVInteger Answer Type

___________________________________________________________________________

This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from0 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 # 9

12. The total number of and particles emitted in the nuclear reaction U23892 Pb142

28 is

Ans. 8

Sol. U23892

Pb14228 + )He(6 4

2 + 2(�1

e0)

= 6, = 2Total = 8

13. The number of water molecule (s) directly bonded to the metal centre in CuSO4. 5H

2O is

Ans. 4

Sol.

14. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is

Ans. 6

Sol. 2MnO2 + 4KOH + O

2

fusion 2K

2MnO

4 + 2H

2O

Let the oxidation state of Mn in MnO4

2� 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 at298.0 K. The temperature of the calorimeter was found to increases from 298.0 K to 298.45 K due to thecombustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K�1, the numerical value for theenthalpy of combustion of the gas in kJ mol�1 is

Ans. 9

Sol. n = 28

5.3

T = T2 � T

1 = 298.45 � 298

= 0.45

CV = 2.5 kJ k�1 = 2500 JK�1

CP = C

V + R = 2500 + 8.314

= 2508.314 JK�1

QP = C

PT = 1128.74 J

H = 28/5.374.1128

n

Qp 9030 J mol�1

= 9.030 KJ mol�1

= 9 KJ mol�1.

RESONANCE Page # 10

17. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecularformula C

5H

10 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 ofits sodium salt is

Ans. 8

Sol. Given Ka = 10�4

pKa = 4

C = 0.01 M

pH = 7 + 21

pKa +

21

log C

= 7 + 21

(4) + 21

(�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 probablespeed of gas Y at 60 K. The molecular weight of the gas Y is

Ans. MY = 4.

Sol. Vrms

= Vmp

XMRT3

= YM

RT2

40400R3

= YM60R2

MY = 4.

RESONANCE Page # 11

PART-II

MATHEMATICSSECTION - 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 linesegment PQ, then the locus of M intersects the latus rectum of the given ellipse at the points

(A)

72

,253

(B)

419

,253

(C)

71

,32 (D)

734

,32

Sol. (C)

16x2

+ 4y2

= 1

a = 4, b = 2equation of normal 4x sec � 2y cosec = 12

M

sin,

2cos7

= (h, k) (say)

h = 2

cos7 cos =

7h2

and k = sin

49h4 2

+ k2 = 1

locus 49x4 2

+ y2 = 1 ....(i)

for given ellipse e2 = 1 � 164

= 43

e = 23

x = ± 4 × 23

= ± 32 ....(ii)

solving (i) and (ii)

494

× 12 + y2 = 1

y2 = 1 � 4948

= 491

y = ± 71

required points

71

,32

RESONANCE Page # 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)

slope of line (2) is =

qq1

slope of altitude BN (as shown in figure) is = q1q

equation of BN is y � 0 = q1q

(x + p)

y = )q1(q

(x + p) ....... (4)

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)

= m = n = 3

1

2x + y + 2 = 9

equations of line are 3/1

2�x =

3/1

1y =

3/1

2�z

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 = 111 = 3

RESONANCE Page # 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

(A) 6

c)1n4(n 22 (B)

3c)1n4(n 22

(C) 3

c)1n4(n 22 (D)

6c)1n4(n 22

Sol. (C)

Sn = cn2

Sn�1

= c(n � 1)2 = cn2 + c � 2 cn

Tn = 2cn � c

Tn

2 = (2cn � c)2 = 4c2 n2 + c2 � 4c2n

Sum = Tn

2 = 6

)1n2)(1n(n.c4 2 + nc2 � 2c2n (n + 1)

= 3

)1n(nc6nc3)1n2)(1n(nc2 222 =

3]6n632n6n4[nc 22

= 3

)1n4(nc 22

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

(A) vertex is

0,

3

a2(B) directrix is x = 0 (C) latus rectum is

3a2

(D) focus is (a, 0)

Sol. (A,D)

Let centroid of PTN is R (h. K)

h = 3

ata2)at(�at 222 & k =

3at2

3h = 2a + a .

2

a2k3

3h = 2a + a4

k9 2

9k2 = 4 a (3h � 2a)

locus of centroid is

y2 = 3a4

3a2

�x

vertex

0,

3a2

; directric x � 3a2

= � 3a

x = 3a

RESONANCE Page # 14

Latus rectum = 3a4

focus

0,

3a2

3a

i.e. (a, 0) Ans. A, D

25. For the function f(x) = x cos x1

, x 1,

(A) for at least one x in the interval [1, ), f(x + 2) � f(x) < 2

(B) x

lim 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)

f(x) = x cos x1

, x 1

f(x) = x1

sin x1

+ cos x1

f(x) = � 3x

1 cos

x1

Now x

lim f(x) = 0 + 1 = 1 option �B� is correct

x [1, ) x1

(0, 1]

f(x) < 0 option �D� is correct

As f(1) = sin 1 + cos 1 > 1

f(x) is strictly decreasing and x

lim f(x) = 1

so graph of f(x) is as below

Now in [x, x + 2], x [1, ), f(x) is continuous and differentiable

so by LMVT, f(x) = 2

)x(f)2x(f

as f(x) > 1 for all x [1, )

2

)x(f)2x(f > 1 f(x + 2) � f(x) > 2

for all x [1, )

RESONANCE Page # 15

26. For 0 < < 2

, the solution(s) of

6

1m4

meccos

4)1m(

eccos = 24 is(are)

(A) 4

(B) 6

(C) 12

(D) 125

Sol. (C,D)

0 < < 2

6

1m4

meccos

4)1m(

eccos = 24

6

1m4

msin

4)1m(

sin

1 = 24

6

1m

4m

sin4

)1�m(sin

4sin

4)1m(

4m

sin

= 24

6

1m

2

14

mcot

4)1m(

cot

= 24 ; 44

mcot

4)1m(

cot6

1m

cot () � cot

4 + cot

4 � cot

4

2 + ... + cot

4

5 � cot

4

6 = 4

cot � cot

2

3 = 4

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

= 12

= 125

2,0

option C and D is correct

RESONANCE Page # 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

21

x2

�

21

y2

= 1 ......... (1)

Eccentricity of hyperbola = 2

so eccentricity of ellipse = 2

1

Let equation of ellipse be

2

2

a

x + 2

2

b

y = 1 (a > b)

2

1 = 2

2

a

b�1

2

2

a

b =

21

a2 = 2b2

x2 + 2y2 = 2b2 ..........(2)

Let ellipse and hyperbola intersect at A

tan

2

1,sec

2

1

Now 4 x � 4y dxdy

= 0

dxdy

= yx

Aatdxdy

=

tansec

= cosec

and 2x + 4ydxdy

= 0

Aatdxdy

= � y2x

= � 21

cosec

Ellipse and hyperbola are orthogonal, so

RESONANCE Page # 17

�21

cosec2= � 1

cosec2= 2

Q = 4

A

2

1,1 or

2

1�,1

1 + 2

2

2

1

= 2b2 b2 = 1

Equation of ellipse is x2 + 2y2 = 2

Coordinate of foci )0,ae( =

0,

2

1.2 = )0,1(

option (A) and (B) are correct

Note :

If major axis is along y-axis, then 2

1 = 2

2

b

a1

b2 = 2a2

2x2 + y2 = 2a2

y = � yx2

� sin

2 . cosec = � 1

cosec2 = 1

= ± 4

2x2 + y2 = 2a2

2 + 21

= 2a2

a2 = 45

2x2 + y2 = 25

, corresponding foci are (0, ±1).

RESONANCE Page # 18

28. If n =

xsin)1(

nxsinx dx, n = 0, 1, 2, ..., then

(A) n =

n+2(B)

10

1m1m2 10 (C)

10

1mm2 0 (D)

n =

n+1

Sol. (A,B,C)

(A) n =

xsin)1(

nxsinx dx

n = dx

xsin)1(

nxsinx

x

(by property

b

a

b

a

dx)xba(fdx)x(f )

2n =

dxxsin

nxsin

2n = 2

0

dxxsin

nxsin

n = dx

xsinnxsin

0

n+2

� n =

0

dxxsin

nxsinx)2nsin( =

0

dxxsin

xsinx)1ncos(2 = 2

0)1n(x)1nsin(

= 0

n+2

= n

(B) 3 =

5 = ..... =

21

10

1m1m2 = 10

3 = 10

0xsinx3sin

dx = 10

0

2 dx)xsin43(

= 0x2sin2x2x310 = 10

(C) 2 =

4 = ........ =

20

10

1mm2 = 10

0xsinx2sin

dx = 20 0xsin = 0

RESONANCE Page # 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

2,0

(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

2,0

f is strictly increasing

f(0) = � 1

f

2 = 2

e one solution

RESONANCE Page # 20

(B)122

2k4

14k

= 0

k (k � 4) � 4(0) + 1 (8 � 2k) = 0

k2 � 6k + 8 = 0

k = 2, 4

(C) for solutions, 4k 6 k 23

.

Integer values of k are 2, 3, 4, 5

(D)dxdy

= y + 1

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

(B) Points of discontinuity of the function f(x) =

x6 cos

x3, (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 j�i� , j�2i� and k�j�i�

(D) Angle between vectors a

and b

where b,a

and c

are unit (s)2

vectors satisfying 0c3ba

(t)

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

RESONANCE Page # 21

Sol. (A) 2 sin2sin22= 2

sin22= 2 cos2 4sin2cos2= 2cos2

cos2 = 0 or sin2 = 21

cos = 0 or sin= 2

1 =

4

or 2

(B) f(x) =

x6 cos

x3

possible points of disontinunity of

x6 are

x6 = n , n I

x = 6

n x =

6

,3

, 2

,

6

x

lim f (x) = 0 cos 0 = 0

6

x

lim f(x) = 1 cos 0 = 1

discontinous at x = 6

. Similarly discontinous at x = 3

, 2

,

(C) V = 11

021

011

= cubic units

(D) a

+ b

+ 3 c

= 0

a

+ b

= � 3 c

2

ba

= 2

c3

a2 + b2 + 2 a

.b

= 3c2

2 + 2 cos = 3

cos= 21

RESONANCE Page # 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 :

________________________________________________________________________________________________

31. Let f : R R be a continuous function which satisfies f(x) = x

0

dt)t(f . Then the value of f(n 5) is

Ans. 0

Sol. From given integral equation, f(0) = 0.Also differentiating the given integral equation w.r.t. xf(x) = f(x)

If f(x) 0 )x(f)x(f

= 1 f(x) = ec ex

f(0) = 0 ec = 0, a contradiction f(x) = 0 x R f(n 5) = 0

32. The centres of two circles C1 and C

2 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 C

2 and C be a circle touching circles C

1 and C

2

externally. If a common tangent to C1 and C passing through P is also a common tangent to C

2 and C, then

the radius of the circle C isAns. 8

Sol. (r + 1)2 = 2 + 9r2 + 8 = 2

r2 + 2r + 1 = r2 + 8 + 92r = 16r = 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, isAns. 2

Sol. (i) x2 � 8kx + 16(k2 � k + 1) = 0

D = 64 (k2 � (k2 � k + 1)) = 64 (k � 1) > 0

k > 1

(ii) � a2

b > 4

2k8

> 4 k > 1

(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 # 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 = 22 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

Asina

= º30sin

22 =

Csin4

C = 45º, C = 135º

When C = 45º A = 180º � (45º + 30º) = 105º

When C = 135º A = 180º � (135º + 30º) = 15º

Area of ABC = 21

AB × AC sin A = 21

× 4 × 22 sin (105º) = 24 × 22

13

Area of ABC = 21

AB × AC sinA = 21

× 4 × 22 sin (15º) = 2 1�3

Difference of areas of triangles = | 2 13 � 2 1�3 | = 4

Aliter

AD = 2 , DC = 2

Difference of Areas of triangle ABC and ABC = Area of triangle ACC

= 21

AD × CC = 21

× 2 × 4 = 4

RESONANCE Page # 24

36. If the function f(x) = x3 + 2x

e and g(x) = f�1(x), then the value of g(1) is

Ans. 2Sol. g(f(x)) = x

g(f(x)) f(x) = 1 ........(i)if f(x) = 1 x = 0, f(0) = 1substitute x = 0 in (i), we get

g(1) = )0(f1

g(1) = 2 (f(x) = 3x2 + 21

ex/2 f(0) = 21

)

37. Let p(x) be a polynomial of degree 4 having extremum at x = 1, 2 and 0xlim

2x

)x(p1 = 2. Then the value of

p(2) isAns. 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)

0xlim

2x

)x(p1 = 2

0xlim

2

234

x

edxx)1c(bxax = 2

c + 1 = 2, d = 0, e = 0

c = 1

Now equation (i) and (ii) are

4a + 3b = � 2 and 32 a + 12 b = � 4

a = 41

and b = � 1

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 = 0So 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 # 25

PART - IIIPHYSICSSECTION - I

Single Correct Choice Type

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

eV25.2eV550

1240E nm5501

eV8.2eV450

1240E nm4502

eV5.3eV350

1240E nm3503

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 # 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 :

(A) Mk2

21

(B) Mk

21

(C) Mk6

21

(D) M

k2421

Sol. (C)

Torque about P = (kx) 2L

+ (kx) 2L

= kxL = k 2L2

For small angle , x = 2L

=

� 12

ML2

KL 22

MK6

=

= MK6

and f =

2 =

MK6

21

RESONANCE Page # 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) gka

(B) gk2a

(C) gka2

(D) gk4a

Sol. (B)

ma cos = mg cos (90 � )

tanga

ga

= dxdy

dxd

(kx2) = ga

x = gk2a

= D

42. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the

point P is

M

k2k1

P

(A) 2

1

kAk

(B) 1

2

kAk

(C) 21

1

kkAk

(D) 21

2

kkAk

RESONANCE Page # 28

Sol. (D)

Extentions in springs are x1 and x2 then

k1x1 = k2x2

and x1 + x2 = A

x1 + 2

11

kxk

= AA x1 = 21

2

kkAk

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 # 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) AC VV

= CB VV2

(B) BC VV

= AB VV

(C) AC VV

= CB VV2

(D) AC VV

= BV4

Sol. (B, C)

)i�(R)i�(VVA

; i�VVB

; i�Ri�VVC

i�R2VV AC

2 )]i�(R)i�(V)i�(V[2VV CB

= �2R( i� )

Hence AC VV

= )VV(2 CB

so |VV| AC

= |)VV(2| CB

BC VV

= R( i� )

AB VV

= R( i� )

ABBC VVVV

Hence )i�(R2VV AC

ABBC VVVV

; BV4

= 4V( i� ) = 4R ( i� )

Hence )V(2VV BAC

RESONANCE Page # 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 (2r)

Ft = MV

i =

A)r2()t/(

BH i (2r) t = MV

V = MK

MABH

RESONANCE Page # 31

h = 22

22

M

K

g2

V

hA > hB

2B

2B

2

2A

2A

2

M

K

M

K

BMB > AMA

Using this we get

Answer (B) and (D)

Section�III

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.

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

ABCD

p q r s

t

t

ttt

48. Column I shows four situations of standard Young�s double slit arrangement with the screen placed far away

from the slits S1 and S

2 . In each of these cases S

1P

0 = S

2P

0 , S

1P

1 � S

2P

1 =/4 and S

1P

2 � S

2P

2 = /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 # 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) > (P

1)

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

(t) (P2) > (P

1)

Sol. (A) (P1) =

1 +

2 + 2 21 cos

42

= 0 +

0 + 2

0 .

2

1 = (2 + 2 )

0

(P2) =

1 +

2 + 2

0 .

2

1 = 2 21 . cos

32

= 0 +

0 + 2

0 .

21

= 0

(P1) > (P

2)

(B) (P0) = ( � 1)t .

2 =

22

.4

(P1) = 0

2.

4t)1(

RESONANCE Page # 33

(P2) = 6

2.

3t)1(

(P0) =

1 +

2 + 2 21 . cos (P

0)

= 0 +

0 + 2

0 . cos

2

=20

(P0) = 4

0

(P2) =

0 +

0 + 2

0 cos

6

= (2 + 3 )0

(C). (P0) = ( � 1)t .

2 =

2.

2

(P1) = 2

24

2.t)1(

(P2) = 6

2.

3t)1(

(P0) = 3

26

2.

32

(D) (P0) =

232

43

(P1) = 0

(P2) > 0, (P

0) > 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 # 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 TypeThis 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 # 35

50. A steady current goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x

and QR = 5x. If the magnitude of the magnetic field at P due to this loop is

x48k 0

, find the value of k.

Sol. B =

5x12

4

i0 (sin 37º + sin 53º)

B = x48i0

53

54

5

B = 7

x48i0

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.

Sol. T = gmmmm2

21

21

=

36.072.036.072.02

× 10

T = 4.8 N

RESONANCE Page # 36

a = gmmmm

21

21

=

3g

s = 2at21

= 21

3g

(1)2 = 6

10

Work done by T = (T) (S)

= (4.8) × 6

10 = 8J

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

a are constants and r is the distance from its centre. If the electric field at r = 2R

is 81

times that at r = R, find

the value of a.

Sol.

Total charge Q = )R(3a

k4)drr4()Kr(Vd 3a

Rr

0r

2a

Q´ =

3a2/Rr

0r

2a

2

R

3a

k4)drr4()Kr(Vd

According to question

041

2)2/R(

´Q =

20 R

Q4

181

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.

i1 = )kA/x(

0400

,

i2 = kA/x)10(

100400

RESONANCE Page # 37

v

f

v

f

2

1

LL

Ldtdm

Ldtdm

ii

54080

x)10/(300x/400

= 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/n

A, where n

A and n

B are the number of moles of air in bubbles A and B,

respectively. [Neglect the effect of gravity.]

Sol.

PA = P

0 +

ArT4

PA = 8 +

02.004.04

PA = 16 N/m2

PB = P

0 +

BrT4

= 8 + 04.0

04.04

PB = 12 N/m2

for bubble A, PV = nRT

(16) 34 (0.02)3 = n

A RT ....(1)

for bubble B

(12)

3)04.0(

34

= nBRT ....(2)

dividing eqn (i) and (2)

61

nn

B

A ; 6nn

A

B 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.

v =

T =

2.0/10

5.03 = 10 m/sec.

v = f

10 = (100) = 0.1 m = 10 cm

distance between two successive nodes = 2

= 5 cm

Ans. 5

RESONANCE Page # 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)V

2

9 = 2V1 � V

2..... (1)

e = 19

VV 21

......(2)

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)

P = mm300mm)H500(P0

....(2)

RESONANCE Page # 39

from (1) and (2)

300)H500(P0

+ 2000 = P0

300)H500(105

+ 2000 = 105

5 × 107 � H × 105 + 6 = 300

H = 206 mm

fall in height = 6 mm

Ans. 6

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 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 inColumn I and five statements (p, q, r, s and t) in Column II. Any given statement in Column Ican have correct matching with one or more statement(s) given in Column II. For example, iffor a given question, statement B matches with the statements given in q and r, then for thatparticular question, against statement B, darken the bubbles corresponding to q and r in theORS.

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 theinstructions 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 bubblecorresponding to the correct answer and zero mark if no bubble is darkened. In all case ofbubbling 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 bubblecorresponding to the correct answer, and zero mark if no bubble is darkened. In all other cases,minus one (�1) mark will be awarded.

Name : ___________________________ Roll No. :