IIT JEE 2009 Paper-2 Questions and Solutions

8/14/2019 IIT JEE 2009 Paper-2 Questions and Solutions

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INSTRUCTIONSA. General1. 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 onthe back page of this booklet.

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7. DO NOT TAMPER WITH/ MUTILATE THE ORS OR THE BOOKLET.

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B. Filling the ORS :

9. Write your Roll No. in ink, in the box provided in the upper part of the ORS and darkenthe appropriate bubble UNDER each digit of your Roll No. with a good quality HB pencil.

C. Question paper format :

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

Name 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


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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/CH 3 that is most likely to migrate to the positiviely charged carbon is

(A) CH 3 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 CH 3 group)


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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 E a , respectively, ,

are :(A) 1.0 10 6 s 1 and 9.2 kJ mol 1 (B) 6.0 s 1 and 16.6 kJ mol 1

(C) 1.0 10 6 s 1 and 16.6 kJ mol 1 (D) 1.0 10 6 s 1 and 38.3 kJ mol 1

Sol. (D) From Arrhenius equationK = 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 = 10 6

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 3d 5 4s 1. The CO is a strong field ligand socompels for the pairing of electrons. Thus the complex has d 2 sp 3 hybridisation and is diamagnetic.

[Cr(CO)6]

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


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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 + B 2H6 [BH2(X)2]+ [BH4]

the amine(s) X is(are) :

(A) NH3 (B) CH 3NH2 (C) (CH 3)2NH (D) (CH 3)3N

Sol. (A, B, C)

Small amines such as NH 3 ,CH 3NH2 and (CH 3)2NH give unsymmetrical cleavage of diborane according tofollowing reaction.

B2H6 + 2 NH 3 [H2B(NH3)2]+ [BH4]

Large ammines, such as (CH 3)3 N gives symmetrical cleavage of diborane according to following reaction.

B2H6 + 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 2O3 (C) N 2O4 (D) N2O5

Sol. (A, B, C)

(A)

N2O

(B)

(C)

N2O4

(D)

N2O5


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7. For the reduction of NO 3 ion in an aqueous solution E is +0.96 V. Values of E for some metal ions are given

below

V2+ (aq) + 2e V E = 1.19 VFe 3+ (aq) + 3e Fe E = 0.04 VAu3+ (aq) + 3e Au E = +1.40 VHg 2+ (aq) + 2e Hg E = +0.86 V

The pair(s) of metals that is(are) oxidized by NO 3

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 NO 3 (E = 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.


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SECTION - III

Matrix - Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have tobe matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II arelabelled 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 haveto 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 ofbubbles will look like the following :

10. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given incolumn 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


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Sol. (A)

(B)

(C)

(D)


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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 HNO 3 (p) NO

(B) Cu + conc HNO 3 (q) NO 2

(C) Zn + dil HNO 3 (r) N2O

(D) Zn + conc HNO 3 (s) Cu(NO 3)2

(t) Zn(NO 3)2

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

Sol. (A) 3Cu + 8HNO 3 (dilute HNO 3) 2NO + Cu(NO 3)2 + 4H 2O

(B) Cu + 4HNO 3 (concentrated) 2NO 2 + Cu(NO 3)2 + 2H 2O

(C) 4Zn + 10HNO 3 (dilute) 4Zn(NO 3)2 + N 2O + 5H 2O

(D) Zn + 4HNO 3 (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 :


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

28 is

Ans. 8

Sol. U23892 Pb142 28 + )He(6

42 + 2( 1e

0)

= 6, = 2Total = 8

13. The number of water molecule (s) directly bonded to the metal centre in CuSO 4. 5H 2O is

Ans. 4

Sol.

14. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO 2 is

Ans. 6

Sol. 2MnO 2 + 4KOH + O 2 fusion 2K2MnO 4 + 2H 2O

Let the oxidation state of Mn in MnO 42 is x.

So x + 4 ( 2) = 2 or x = 6

15. The Coordination number of Al in the crystalline state of AlCl 3 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 = 285.3

T = T 2 T1 = 298.45

298

= 0.45CV = 2.5 kJ k

1 = 2500 JK 1

CP = C V + R = 2500 + 8.314= 2508.314 JK 1

QP = C P T = 1128.74 J

H =28 / 5.374.1128

n

Qp 9030 J mol 1

= 9.030 KJ mol 1

= 9 KJ mol 1.


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17. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecularformula C 5H10 is

Ans. 7

Sol.

18. The dissociation constant of a substituted benzoic acid at 25 C is 1.0 10 4 . The pH of 0.01 M solution ofits sodium salt is

Ans. 8

Sol. Given K a = 10 4

pKa = 4C = 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.


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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 x 2 + 4y 2 = 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

,2

53(B)

419

,2

53(C)

71

,32 (D)

7

34,32

Sol. (C)

16x2

+ 4y2

= 1

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

M

sin,2

cos7= (h, k) (say)

h =2

cos7cos =

7h2

and k = sin

49h4 2

+ k 2 = 1

locus49x4 2

+ y 2 = 1 ....(i)

for given ellipse e 2 = 1 164

=43

e =23

x = 4 23

= 32 ....(ii)

solving (i) and (ii)

494

12 + y 2 = 1

y2 = 1 4948

= 491

y = 71

required points

71

,32


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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 = pqh + k = 0locus 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 thecoordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQequals

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

Sol. (C)

= m = n =3

1

2x + y + 2 = 9

equations of line are3 / 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) = 94r + 5 = 9 r = 1Q (3, 0, 3)

PQ = 111 = 3


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23. If the sum of first n terms of an A.P. is cn 2, 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)S n = cn

2

S n 1 = c(n 1) 2 = cn 2 + c 2 cn

Tn = 2cn c

Tn2 = (2cn c) 2 = 4c 2 n2 + c 2 4c 2n

Sum = Tn2 =

6)1n2)(1n(n.c4 2

+ nc 2 2c 2n (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 y 2 = 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

3

a2(D) focus is (a, 0)

Sol. (A,D)

Let centroid of PTN is R (h. K)

h =3

ata2)at( at 222& k = 3

at2

3h = 2a + a .2

a2k3

3h = 2a +a4

k9 2

9k 2 = 4 a (3h 2a)locus of centroid is

y2 = 3a4

3a2 x

vertex

0,3a2

; directric x 3a2

= 3a

x = 3a


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Latus rectum = 3a4

focus

0,3a2

3a

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

25. For the function f(x) = x cosx1

, x 1,

(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)

f(x) = x cosx1

, x 1

f (x) =x1

sinx1

+ cosx1

f (x) = 3x1

cos

x1

Now xlim f (x) = 0 + 1 = 1 option B is correct

x [1, )x1

(0, 1]

f (x) < 0 option D is correctAs f (1) = sin 1 + cos 1 > 1

f (x) is strictly decreasing and xlim f (x) = 1

so graph of f (x) is as belowNow 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, )


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26. For 0 < 0k > 1

(ii)

a2

b

> 4 2

k8

> 4 k > 1(iii) f(4) 0

16 32k + 16 (k 2 k + 1) 0k2 3k + 2 0(k 2) (k 1) 0k 1 or k 2Hence k = 2


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36. If the function f(x) = x 3 + 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) = 3x 2 +21

e x/2 f (0) =21

)

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

2x)x(p

1 = 2. Then the value of

p(2) isAns. 0

Sol. p(x) = ax 4 + bx 3 + cx 2 + dx + e

p (x) = 4ax3

+ 3bx2

+ 2cx + dp (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

xedxx)1c(bxax

= 2

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

c = 1Now equation (i) and (ii) are4a + 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 x 2 + y2 + z 2 100isAns. 7

Sol. 3x

y

z = 0 .........(1) 3x + 2y + z = 0 .........(2)

3x + z = 0 .........(3)

(1) + (2) y = 0So 3x = z

Now x 2 + y 2 + z 2 100 x2 + (3x) 2 + 0 100

10x 2 100

x2 10

x = 3, 2, 1, 0, 1, 2, 3So No.of such 7 points are possible


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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 its

answer, out which ONLY ONE is correct.

39. Photoelectric effect experiments are performed using three different metal plates p, q and r having workfunctions p = 2.0 eV, q = 2.5 eV and r = 3.0 eV respectively. A light beam containing wavelengths of 550nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct -V graph for theexperiment 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.


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40. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs ofequal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free tooscillate in the horizontal plane. The rod is gently pushed through a small angle in one direction andreleased. 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 = k2

L2

For small angle , x = 2L

=

12

ML2

KL 22

MK6 =

=MK6

and f = 2 = MK6

21


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Sol. (D)

Extentions in springs are x 1 and x 2 thenk1x1 = k 2x2

and x 1 + x 2 = A

x1 + 211

kxk

= AA x1 = 212

kkAk

SECTION - IIMultiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, 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 betweencharges 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-circleand 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


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Sol. (B, D)

(A) process is not isothermal(B) volume decreases and temperature decreases

U = negativeW = 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 = 2 R( 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 ) = 4 R ( i

)

Hence )V(2VV BAC


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46. A student performed the experiment to measure the speed of sound in air using resonance air-columnmethod. Two resonances in the air-column were obtained by lowering the water level. The resonance with theshorter 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 ofthe sound in air.

Sol. (A, D)

(A) The intensity of second decreases with increases of order. The intensity ofsound 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 ontop of two identical solenoids as shown in the figure. When current is switched on in both the solenoids inidentical manner, the rings A and B jump to heights h A and h B, respectively, with h A > h B. The possiblerelation(s) between their resistivities and their masses m A and m B is(are)

A B

(A) A > B and m A = m B (B) A < B and m A = m B(C) A > B and m A > m B (D) A < B and m A < m B

Sol. (B, D)

The horizotanl component of magnetic field due tosolenoid will exert force on ring in vertical direction

F = B Hi (2 r)F t = MV

i =

A)r2()t / (

BH i (2 r) t = MV

V =M

K

M

ABH


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h = 2222

M

Kg2

V

hA > hB

2B

2B

2

2A

2A

2

M

K

M

K

BMB > AMAUsing this we getAnswer (B) and (D)

Section IIIMatrix Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have tobe matched. The statements in Column I are labelled A,B,C and D, while the statements in Column-II arelabelled p,q,r,s and t. Any given statement in Column-I can have correct matching with ONE OR MOREstatement(s) in Column-II . The appropriate bubbles corresponding to the answers to these questions haveto 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 s

p q r s

ABCD

p q r stt

ttt

48. Column I shows four situations of standard Young s double slit arrangement with the screen placed far awayfrom the slits S 1 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 andthickness t is pasted on slit S 2. The thicknesses of the sheets are different in different cases. The phasedifference between the light waves reaching a point P on the screen from the two slits is denoted by (P) andthe intensity by (P). Match each situation given in Column-I with the statement(s) in Column-II valid for thatsituation.


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Column I Column II

(A) (p) (P 0) = 0

(B) ( 1)t = /4 (q) (P 1) = 0

(C) ( 1)t = /2 (r) (P 1) = 0

(D) ( 1)t = 3 /4 (s) (P 0) > (P 1)

Ans. (A) p, s; (B) q; (C) t ; (D) r, s, t(t) (P 2) > (P 1)

Sol. (A) (P 1) = 1 + 2 + 2 21 cos 42

= 0 + 0 + 2 0 . 21

= (2 + 2 ) 0

(P2) =

1+

2+ 2

0.

2

1= 2 21 . cos

3

2

= 0 + 0 + 2 0 .

21

= 0

(P 1) > (P 2)

(B) (P 0) = ( 1)t .

2= 2

2.

4

(P 1) = 02

.

4

t)1(


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(D) Mass of the system is decreased (s) System: A heavy nucleus, initially at restProcess: The nucleus fissions into two fragments of

nearly equal masses and some neutrons are emitted

(t) System: A resistive wire loopProcess: The loop is placed in a time varyingmagnetic field perpendicular to its plane

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

(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 IVInteger Answer Type

This section contains 8 questions. The answer to each of the question is a single-digit integer, ranging from0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. Forexample, if the correct answers to question numbers X,Y,Z and W (say) are 6,0,9 and 2, respectively, thenthe correct darkening of bubbles will look like the following :


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v

f

v

f

2

1

LL

Ldt

dm

Ldt

dm

ii

540

80

x)10 /(300

x / 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/m 2. Theradii of bubbles A and B are 2cm and 4cm, respectively. Surface tension of the soap-water used to makebubbles is 0.04 N/m. Find the ratio n B /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.

P A = P 0 + ArT4

P A = 8 + 02.004.04

P A = 16 N/m 2

P B = P 0 +BrT4

= 8 + 04.004.04

P B = 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 eq n (i) and (2)

61

nn

B

A; 6n

n

A

BAns. 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. Thestring 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 = f10 = (100) = 0.1 m = 10 cm

distance between two successive nodes = 2 = 5 cm

Ans. 5


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56. Three objects A,B and C are kept in a straight line on a frictionless horizontal surface. These have massesm, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collisionwith 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) V 1 (m)V 2

9 = 2V 1 V2 ..... (1)

e = 19VV 21 ......(2)

from eqn(1) and eqn(2) V 1 = 6 m/sec.

for second collision between second block and third block :

(2m) 6 + m(0) = (2m + m) V CVC = 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 closedand water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at thebottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steadywith 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 10 5 N/m 2 , density of water = 1000 kg/m 3 and g = 10 m/s 2. Neglect anyeffect of surface tension]

Sol.P 0 = atmospheric pressureP + 200 10 3 1000 10 = P 0 ....(1)

P = mm300mm)H500(P0

....(2)


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from (1) and (2)

300)H500(P0 + 2000 = P 0

300)H500(10 5

+ 2000 = 105

5 10 7 H 10 5 + 6 = 300

H = 206 mm

fall in height = 6 mmAns. 6


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IIT JEE 2009 Paper-2 Questions and Solutions

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