iitjeesol_2011

IIT-JEE 2011 Pattern and AnalysisIIT-JEE, the most challenging of the Engineering Entrance exams in the country and probably the world has proved that it has not lost its charm at all and has retained its glory as ever. IIT-JEE 2011 was conducted on 10th April 2011 in 131 cities all over India in 1051 centres for 4,85,262 registered candidates. Paper 1 was from 9.00 12.00 hrs and paper 2 from 14.00 to 17.00 hrs.

For the first time, JEE has announced that scanned images of the Optical Response Sheets with the candidates details and their responses (answers) will be displayed on the JEE website after 25th May, 2011. Answer key to the JEE 2011 questions will be displayed on the JEE website on 15th May, 2011 after completing the evaluation of OMR sheets.

There were 10 versions for IIT-JEE 2011 0 to 9. The order of subjects in both papers were the same Chemistry, Mathematics and Physics but the total marks, the number of questions and types of questions in both papers were distinct from each other.

Paper 1Paper1 had 69 questions in total for 240 marks - in the order of Chemistry, Physics and Mathematics. Each subject had 23 questions and the pattern was exactly the same for all the subjects.

Number Type of question of questions Section I - Single Correct Answer Type Section II - Multiple Correct Answers Type Section III - Paragraph Type Section IV - Integer Answer Type 7 4 5 7

Marks for a correct answer 3 4 3 4

Negative marking 1 NIL 1 NIL

Total Marks for the Section 21 16 15 28

What was new in IIT-JEE 2010 in comparison with earlier years Paper 1:

NO partial marks are awarded for correct answers in Multiple Correct Type questions. Last year there were partial marks.

Part 1 - Chemistry This is a moderate level paper leaning more towards easy level of difficulty. The usual weightage of Physical > Organic > Inorganic Chemistry is maintained. Most of the questions in this paper evaluate the conceptual clarity of the student.

CHEMISTRY - PAPER I - Difficulty Analysis(on a total of 80 marks) 20% Easy 28% Medium

Difficult 53%

The topic wise distribution of the questions is shown below:

CHEMISTRY - PAPER I - Topic wise Analysis(on a total of 80 marks) 24% 46% Inorganic Chemistry Organic Chemistry Physical Chemistry

30%

Part 2 Physics Overall the difficulty level of the paper was moderate. Still there were a few numerical questions involving elaborate mathematical calculations making them time consuming and hence difficult. No representation of Optics in this paper. Major focus in this paper was on Mechanics (for a total marks of 31)

PHYSICS - PAPER I - Difficulty Analysis(on a total of 80 marks)

36% 38%

Easy

Medium

Difficult26%

The topic wise analysis of paper 1 for Physics is shown below.

PHYSICS - PAPER I - Topic wise Analysis(on a total of 80 marks)Electricity & Magnetism 8% 16% 29% Heat & thermodynamics Mechanics

Modern Physics 9% 38% Properties of Matter

Part 3 Mathematics Mathematics part of the 2011 JEE paper 1 was considered easier than the paper 1 of 2010. Few questions in Integer answer type require in-depth understanding of the concepts with perseverance in solving problems. Trigonometry had a good representation in the paper; the passage on matrices required quick thinking to interpret and solve the questions. One question about 3 3 non-singular skew symmetric matrices has a basic flaw since odd ordered skew symmetric matrices are always singular. As shown below paper 1 was a moderately difficult paper.

MATHEMATICS - PAPER I - Difficulty Analysis(on a reduced total of 76 marks) 5% Easy 41% Medium

Difficult 54%

Algebra and Matrices & Determinants, Vectors had the major shares with Calculus and Trigonometry following close. Topic wise analysis is given below.MATHEMATICS - PAPER I - Topic wise Analysis(on a reduced total of 76 marks) Algebra Coordinate Geometry Differential Calculus Integral Calculus Trigonometry 18% 13% 5% 9% Vectors, 3-D, Matrices & Determinants

22%

33%

Paper 2Paper 2 had 60 questions in total - in the order of Chemistry, Physics and Mathematics. Each subject had 20 questions and as in paper 1 the pattern was exactly the same across the subjects. Questions had differential markings, each subject had 80 marks and the maximum mark for the paper was 240.

Number Type of question of questions Section I - Single Correct Choice Type Section II Multiple Correct answers Type Section III - Integer Answer Type Section IV Matrix Match Type 8 4 6 2

Marks for a correct answer 3 4 4 8

Negative marking 1 NIL NIL NIL

Total Marks for the Section 24 16 18 16

What was new in IIT-JEE 2011 in comparison with earlier years Paper 2:

For the first time Multiple Answer Type, Integer Answer Type and Matrix Match Type appeared in one paper.

Part 1 - Chemistry Paper 2 of IIT-JEE 2011 was difficult compared to paper 1. Weightage wise the Physical > Organic > Inorganic formula is maintained. Generally the nature of the questions in Physical chemistry is calculation oriented and hence the difficulty. Also complexity of Organic questions made this paper a challenging one. The difficulty analysis and the topic wise analysis are shown below:

CHEMISTRY - PAPER II - Difficulty Analysis(on a total of 80 marks) 13% 34% Medium Easy

Difficult 54%

CHEMISTRY - PAPER II - Topic wise Analysis(on a total of 80 marks)23% 41%

Inorganic Chemistry Organic Chemistry Physical Chemistry

36%

Part 2 Physics Questions from certain areas which were not represented in paper 1 (gravitation, optics etc.) appeared in paper 2 thus balancing the syllabus coverage. The difficulty level of this paper was slightly less than paper 1 especially the Integer Answer Type.

PHYSICS - PAPER II - Difficulty Analysis(on a total of 80 marks) 5% Easy 48% Medium

Difficult 48%

And the less colourful distribution of marks across the topics in Physics shown below confirms the overall lions share of Mechanics in Physics part of the IIT-JEE 2011 paper 2.PHYSICS - PAPER II - Topic wise Analysis(on a total of 80 marks) 9% 5% 31%Electricity & Magnetism Heat & thermodynamics Mechanics

Modern Physics

45%

10%Optics

Part 3 Mathematics Paper 2 Mathematics part was tougher than the first paper for the complexity in combining topic areas, for e.g. the matrix match had questions from 4 different areas posing a challenge for the student in the exam hall. Likewise, this paper required more calculations in solving than the first paper.

Overall Mathematics part of IIT-JEE 2011 was less difficuly and less time consuming than that of 2010.

MATHEMATICS - PAPER II - Difficulty Analysis(on a total of 80 marks) 14% 26% Easy

Medium

Difficult 60%

The second paper was restricted to lesser number of areas, for e.g. there were no questions from Trigonometry. But this time Differential Calculus and Coordinate Geometry were well represented. The topic wise distribution is shown below:MATHEMATICS - PAPER I - Topic wise Analysis(on a total of 80 marks) Algebra

13% 14%

25%

Coordinate Geometry Differential Calculus Integral Calculus 21%

27%

Vectors, 3-D, Matrices & Determinants

IIT-JEE 2011 EXAMINATION(HELD ON : 10 - 04 - 2011)

PAPER - 1

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27 / 28

IIT-JEE 2011 EXAMINATION(HELD ON : 10-04-2011)

PAPER - 1Im

Uqn0 Cpu0 "7 7 Yg"jcxg" vq" hkpf"okpkowo" xcnwg" qh" 4 | - 5 - k 4

3 c -7 + c -6 + c -5 + c -5 + c -5 + c : + c32 -7 -6 -5 -5 -5 : 32 9 ( c 0c 0c 0c 0c 0c 0c ) 9

c -7 + c -6 + c -5 + c -5 + c -5 + c : + c32 -7 -6 -5 : 32 3 ( c + c + 5c + c + c ) okp ? "9 9 yjgtg" c7" ? "c6" ? "c5" ? "c:" ? "c32" k0g0 "c "? "3 *c7" - "c6" - "5c5" - "c:" - "c32" - "3+okp" ? ": " " " "yjgp "c "? "3 8;0 Ngv" " 0

f ( x) has exactly 1 point of minima and f (0) is 1f ( x) has two distinct real roots.

Ans. 2d f ( x) and g ( x) is a given non-constant dx

54.

Let y( x ) + y ( x) g ( x) = g ( x) g ( x), y (0) = 0, x R , where f ( x) denotes differentiable function on R with g (0) = g (2) = 0 . Then the value of y (2) is

Sol.:

dy + g ' ( x) y = g ( x) g ' ( x) dx

y e g ( x ) = e g ( x ) g ( x) g ' ( x )dx

Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396

f = 0 f ( x) is continuous at x = 2 2

IITJEE 2011 SOLUTIONS= ( g ( x) 1)e g ( x ) + c y = g ( x) 1 + c e g ( x ) Put x = 0 0 = 1 + c c = 1 Put x = 2 y (2) = 1 + e 0 = 1 + 1 = 0Ans. 0

43

55.

1 0 1 1 0 1 Let M be a 3 x 3 matrix satisfying M 1 = 2 , M 1 = 1 , and M 1 = 0 . Then the sum of the diagonal entries of 12 1 1 3 0 0 M is Let M be a1 b1 c1 a2 b2 c2 a3 b3 c3

Sol.:

0 1 M 1 = 2 3 0 a 2 = 1 b2 = 2 c2 = 3 (i)

1 1 Again, M 1 = 1 1 0 a1 a 2 = 1 b1 b2 = 1 c1 c 2 = 1 a1 = 1 + a 2 = 1 + (1) = 0 b1 = 1 + b2 = 1 + 2 = 3 c1 = 1 + c 2 = 1 + 3 = 2 Also c1 + c 2 + c 3 = 12 c 3 = 12 c1 c 2 = 12 2 3 = 7 Required sum is = a1 + b2 + c 3 = 0+2+7 =9Ans. 9 56.r r r r r r r r r r , b = i be three given vectors. If r k + + 2 j and c = i j + 3k Let a = i is a vector such that r b = c b and r . a = 0 , r r then the value of r . b is r r r , b = i k + a = i j, c = i +2 j + 3k r r r r r b c b = 0 r r r (r c ) b = 0Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017 Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396

Sol.:

IITJEE 2011 SOLUTIONS Sor r r So r c || b r r r r c = b r r r ) + ( i +2 + r = c + b = (i j + 3k j)

44

+ (2 + ) = (1 )i r j + 3k rr r .a = 0 (1 ) + 0 1 3 = 0 1 + 3 = 0, = 4

r + ( 2 + 4) r = (1 4)i j + 3k r + 6 r = 3i j + 3k r r ).(i + 6 + r .b = (3i j + 3k j)= 3+ 6 = 9 rr r .b = 9

SoAns. 9 *57.

The

straight

line

2x 3y = 1

x2 + y2 6

divides

the

circular

region

into

two

parts.

If

3 5 3 1 1 1 1 S = 2, , , , , , , , then the number of point(s) in S lying inside the smaller part is 4 2 4 4 4 8 4 Sol.:

Only points 1 3 1 1 1 2, , , and , lies inside the circle 4 4 4 8 4 Put value of O (0, 0) in equation 2 x 3 y 1 = 0 So = 2 0 3 0 1 = ve Since points are in smaller region. So points should be opposite to origin w.r.t. line 2 x 3 y 1 = 0 3 Case I 2, 4 = 2 2 3 1 1 Case II , 4 4 = 2 = 1 1 3 1 4 4 3 9 1 = 3 > 0 + ve 4 4(0,0) (1/2, 0) (0, 1/3)

x2 + y2 = 6 2 x 3y = 1

2 3 + 1 = + ve 4 4

1 1 Case III , 8 4 = 2 = = 1 1 3 1 8 4

2 3 1 8 4 268 = ve 8


IITJEE 2011 SOLUTIONS

45

1 3 1 Out of these 3 points only 2, , , lie inside the smaller part as origin (0, 0) and these points lie on opposite side of 4 4 4 line 2 x 3 y 1 = 0Ans. 2 *58.

Let = ei / 3 and, a, b, c, x, y, z be non-zero complex numbers such that a+b+c = x

a + b + c2 = ya + b2 + c = z . Then the value of

| x |2 + | y |2 + | z |2 | a |2 + | b |2 + | c |2

is

*Sol.: Value of this expression does not come out to be constant as it depends on choice of a, b, c.

(For example if a = b = c, x = 3, y = z = 1 + 3i then expression is equal to

17 . 3

Similarly a = 1, b = 1, c = 1, x = 1, y = 1 + 3i, z = 1 + 3i then expression is equal to 3 etc.) But , If we take = ei 2 3

then it is constant and equal to 3.

SECTION IV (Total Marks : 16) (Matrix-Match Type)

This section 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 ORS59.

Match the statements given in Column I with the intervals/union of intervals given in Column-II.

Column I *(A)

Column II (p)

2i z The set Re 2 1 z

: z is a complex number, | z | = 1, z 1 is

( , 1) (1, )

(B)

8(3) x 2 is The domain of the function f ( x) = sin 1 1 32( x 1) 1 tan 1 If f () = tan 1 tan , then the set f () : 0 < is 2 1 1 tan If f ( x) = x3 / 2 (3x 10), x 0 , then f ( x) is increasing in

(q)

( , 0) (0, )

*(C)

(r)

[2, )

(D)

(s) (t)

( , 1] [1, ) ( , 0] [2, )

Sol.:

(A) (s); (B) (t); (C) (r); (D) (r). (A)

2i 2iz Re Re = 1 1 z 2 z zBrilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017

Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396

IITJEE 2011 SOLUTIONS2i 2i = Re = Re = cosec zz 2i sin Real part (, 1] [1, )(B)

46

1

8 .3 x / 9 1 32x 9

1

1

8 .3 x 9 32 x

1

Case I:

8 .3 x 9 32x

+1 0 0 0 0 0(i)

8.3 x + 9 3 2 x (3 + 3 x )(3 3 x ) (3 2 x 8.3 x 9) (3 + 3 x )(3 x 3) 3x 3 3x 3

3 2 x 9.3 x + 3 x 9

3 x (3 x 9) + (3 x 9)x ( , 1) [ 2, )

8 .3 x 9 32x

1 08 .3 x 9 + 3 2 x 9 32x

Case II:

0

3 2 x + 83 x 9 (3 + 3 x )(3 3 x ) 3 3x

0 0

3 2 x + 9 .3 x 3 x 9

(3 x 1)(3 x + 9) 3 3x

0(ii)

x ( , 0] (1, )

So common solution is x (, 0] [2, )(C)

f () = 1(1 + tan 2 ) + tan (0) + 1(1 + tan 2 )

= 2 sec 2 2 So solution is [2, )(D)

f ( x) = ( x) 3 / 2 (3 x 10), x 0

f ( x) = 3.x 5 / 2 10 x 3 / 2 f ' ( x) = 3 5 3/ 2 3 15 3 / 2 30 1 / 2 x 10 x 1 / 2 = x x 2 2 2 2

Since f ( x) is an increasing function f ' ( x) 0

15 x ( x 2) 0 ( x 2) 0 x 2 2

x [2, )


IITJEE 2011 SOLUTIONS60.

47

Match the statements given in Column I with the values in Column-II.Column I r r r , b = and c form a triangle, then the If a = j + 3k j + 3k = 2 3k r r internal angle of the triangle between a and b is Column II (p)

(A)

62 3

(B)

If

( f ( x) 3x) dx = aa 5/ 6 7/6

b

2

b 2 , then the value of f is 6

(q)

(C)

2 The value of ln 3

sec(x) dx is

(r)

3 2

*(D)

(t)

1 The maximum value of Arg for | z | = 1, z 1 is given by 1 z

(s)

Sol.:

(A) (q); (B) (p); (C) (s); (D) (t). r r r (A) As a + b = cr r So angle between a and b is r a

r c

).( ) = ( ) ( ) cos ( j + 3k j 3k j + 3k j 3k cos = b

r b

1 2 = 2 3b2 a2 f ( x) = x 2

(B)

a

f ( x)dx =

f = 6 6(C)

Let x = t + 1 1/ 6

2 ln 3

1 / 6

1/ 6

sec( + t )dt1/ 6 0 =

2 2 = ln 3(D)

sec t dt = ln 3 | ln(sec t + tan t ) |0

2

Let z = e i

1 z = 1 cos i sin = 2 sin

sin i cos 2 2 2sin

1 = 1 z

1 = 2 sin sin i cos 2 2 2

+ i cos 2 2 2 sin 2

=

1 i + cot 2 2 2

1 1 lies on x = 1 z 2

1 as cot . arg tends towards 2 2 1 z Brilliant Tutorials Pvt. Ltd. Head Office: 12, Masilamani Street, T. Nagar, Chennai-600 017

Delhi Office: 50-C, Kalu Sarai (Behind Azad Apartments), New Delhi-110 016 Ph.: 2653 7392/93/94/95 Fax: 2653 7396


48

BREAK UP 1 (LEVEL OF DIFFICULTY)CHEMISTRY PAPER 1DIFFICULT 24% MODERATE 36%

EASY 40%

CHEMISTRY PAPER 2DIFFICULT 28% EASY 24%

MODERATE 48%

CHEMISTRY COMBINEDDIFFICULT 26%MODERATE 42%

EASY 32%



49

BREAK UP 1 (LEVEL OF DIFFICULTY)PHYSICS PAPER 1DIFFICULT 15% MODERATE 34% EASY 51%

PHYSICS PAPER 2DIFFICULT 10% EASY 51%

MODERATE 39%

PHYSICS COMBINEDDIFFICULT 13%MODERATE 37%

EASY 50%



50

BREAK UP 1 (LEVEL OF DIFFICULTY)MATHEMATICS PAPER 1DIFFICULT 5% MODERATE 62%

EASY 33%

DIFFICULT 5%

MATHEMATICS PAPER 2

MODERATE 52%

EASY 43%

MATHEMATICS COMBINEDDIFFICULT 5% EASY 38% MODERATE 57%



51

BREAK UP 2 (XI-XII)CHEMISTRY PAPER 1

XII 59%

XI 41%

CHEMISTRY PAPER 2XI 29%

XII 71%

CHEMISTRY COMBINED

XII 65%

XI 35%



52

BREAK UP 2 (XI-XII)PHYSICS PAPER 1XII 48% XI 52%

PHYSICS PAPER 2XII 42%

XI 58%

PHYSICS COMBINED

XII 45%

XI 55%



53

BREAK UP 2 ( XI-XII)MATHEMATICS PAPER 1

XII 55%

XI 45%

MATHEMATICS PAPER 2XI 38%

XII 62%

MATHEMATICS COMBINED

XII 58%

XI 42%



54

BREAK UP 3 (TOPICWISE/PARTWISE)CHEMISTRY PAPER 1

INORGANIC 37%

PHYSICAL 33% ORGANIC 30%

CHEMISTRY PAPER 2INORGANIC 27%

PHYSICAL 37%

ORGANIC 36%

CHEMISTRY COMBINED

INORGANIC32%

PHYSICAL 35%

ORGANIC 33%



55

BREAK UP 3 (TOPICWISE/PARTWISE)PHYSICS PAPER 1OPTICS 0% MODERN PHYSICS 16% ELECTRO MAGNETISM 31%

MECHANICS 35%

HEAT & THERMO DYNAMICS 9%

MODERN PHYSICS 5%

PHYSICS PAPER 2OPTICS 9%

MECHANICS 31%

ELECTRO MAGNETISM 28%


SHM & WAVE MOTION 17%

PHYSICS COMBINEDOPTICS 5% MODERN PHYSICS 11%

MECHANICS 32%

ELECTRO MAGNETISM 29%






56

BREAK UP 3 (TOPICWISE/PARTWISE)MATHEMATICS PAPER 1TRIGONO METRY 14% CO CALCULUS ORDINATE 17% 14%

VECTORS & 3D 9%

ALGEBRA 46%

VECTORS & 3D 8%

MATHEMATICS PAPER 2

TRIGONO METRY 3% ALGEBRA 28% CO ORDINATE 20% CALCULUS 41%TRIGONO METRY 9% VECTORS & 3D 9%

MATHEMATICS COMBINEDCALCULUS 29%

ALGEBRA 36%

CO ORDINATE 17%


MODEL SOLUTIONS TO IIT JEE 2011Paper I1 A 8 A, B, D 12 D 17 9 18 5 19 5 2 D 3 A 9 A, D 13 B 4 B 5 D 10 A, C, D 14 B 20 4 15 A 21 6O C N C O 4. is tetrahedral, [Ni(CN)4] [NiCl4] 2+ planar & [Ni(H2O)6] is octahedral.2 2

6 C

7 C 11 A, B, C

16 C 22 5 23 7

Section I1.27 Al + 4 He 30 P + 1n( Y ) 13 2 15 0

CH2

Br

30 Si + 1H( X ) 30 Si + 0 e( Z ) 14 1 14 1

2.

Since the mobility of K is nearly same as that of + Ag which it replaces, the conductance will remain as more or less constant and will increase only after the end point.

+

is square

O3.

5.

C NH C O O C NK C O8.Br CH2 Cl KOH

Ba(N3)2 Ba + 3N2 Very pure N2 is produced by the thermal decomposition of Barium or sodium azide. M=2 1.15 1000 1120 = 2.05 M

6.

7.

o-hydroxy benzoic acid: pKa = 2.99 phydroxy benzoic acid: 4.58 ptoluic acid = 4.34 pnitrophenol = 7.15

Section IIAdsorption is always exothermic. Chemisorption is more exothermic than physisoprtion and it requires activation energy.

9.

A, D

Section IV10. Cassiterite contain 0.5 10% of metal as SnO2 the rest being impurities of pyrites of Fe, Cu & wulframite. 11. In a, b & c all atoms are in the same plane.17. Maximum no. of electron with n = 3 is 18, of which are having 1 spin 218. 3Na2CO3 + 3Br2 5NaBr + NaBrO3 + 3CO2 19.

Section IIIH3C H3C C H3C H 3C H3C CH3C

12.

C (P)C O OH

CH

dil.H2SO4 HgSO 4

Br CH3 CH 2 C CH2 CH 2 CH3alc.KOH

CH3

(i) NaBH4 (ii) dil.acid

H 3C H3C C H3C

CH3 CH

C

CH 2

CH2

CH3 (2 isomers)

C H OH C H

CH 3 (Q)CH3 CH2H+ (-H2 O)

C

CH

CH2

CH3 (2 isomers)

H3C C 13. H3C H3C

CH3

CH3 CH2

C

CH2

CH 2

CH3 (1 isomer)

H 3C H3C C H3C

C H (2) CH3

CH3

1, 2-shift of -CH3

20. h = CH3-H+

6.626 1034 3 108 300 10 9 1.6 10 19 = 4.14 eV

H3C C H 3CCH3 CH3 C

C HC

CH3 ozonolysis CH3

O 2 H 3C C CH3

E = h h0 For Li, Na, K and Mg h0 values are less than 4.14 eV 21. Let the number of glycine units be x Total mass of the hydrolysed products = 796 + 9 18 = 958 75 x 100 % by mass of glycine = = 47 958 x=6 22. S4O2 6 has sulphur atoms with oxidation states

14. 2AgNO3 + Cu Cu(NO3)2 +

1 Ag 2

The metal dipped is Cu. 2+ Blue colour due to formation of Cu 15. The solution in which the metal is dipped in AgNO3. 16. The deep blue colour due to formation of 2+ [Cu(NH3)4] and the white precipitate of AgCl + dissolve due to formation of [Ag(NH3)2]

0 & +5. 23. Vol. of 0.1 mole at 0.32 atm = 2.24 =7L 0.32

PART II 24 D 31 B, D 25 D 26 D 32 A, B, C, D 35 C 40 3 41 5 42 6 36 B 27 A 28 A 33 A, B, C, D 37 D 43 3 38 C 44 9 39 B 45 4 46 1 29 A 30 B 34 A

Section I24. x component of area vector is a2 2 = E 0a 2 22

1 1 1 = Z2R 2 2 2 4 2 5 36 2 = 1 22 3 160

25. T sin = m Lsin =

324 0 .5 0 .5

T L

= 1215 A30. X 52 + 1 = 10 48 + 2

= 36mg

26. Position (1): Let charge on 2 F be Q. Then energy =Q2 J 4 Q2 J 20proton

Section II31.electron B

Position (2): Ceq = 10 F Total charge = Q. Energy = Loss % = 80% 27. n =1 ( 22.4 = 1 mole) 4 1 3 W = nCVT = R.T 4 2 1 1 T1V1 = T2V2 T2 = 4T1 T = 3T1 f v 28. f = 1 + 0 vs c 1 c 8 (320 + 10 ) = 320 10

T=

2m , different for them. qB

32.

dQ is same for A and E and both are maximum. dt Thermal resistances are as below. KA RB RC RA RD RE

29.

1 1 1 = R 2 2 1 3 2

1 4 1 4 , RB = , RC = , RD = 8 3 2 5 1 . So C is also correct. RE = 24 RA =

1 . 4 D also correct RB parallel RD is equal RC Note: It is assumed that there is no radiation loss.

( Eq.R (RC, RB, RD) is

q2 a= r N=3

1/ 3

4+ 2 16 0

33. On interconnecting V same Q Q A = B (B) correct. R A RB V=

R is standard formula 0V V , EB = RA RB

41. Fd = mg sin mg cos Fu = mg sin + mg cos Fu = 3Fd 1 = 2 1 N = 10 = 10 = 5 2 42.L x x x x x x x x B

(C) correctV same EA =

(D) correct.34. Under case (B) eventhough the disc is free to rotate, it will not rotate. Cases (A) and (B) are identical in all respects. 0 L

B=

Section III e2 3 35. [N] = L = Force area 0 36. Find by the given formula c = 2 37. As ball goes up, x is positive and increasing, v is positive and decreasing. Symmetrical for ball coming down.

0 r 2g L d E 2 = E, i = ; M = ir dt R N=6coil = Br =2

43. =

38. At x = 0, E = K.E =

p2 2m

MgL AY L = L + L' Mg = 1 + L AY L'L = (L 40 L'30 ) Mg = Mg AY 40 1 + 30 AY Solving, we get M 3 kg

E varies as p E1 = 4E2

2

2 2 4 44. N 10 = 2 mr 2 + md2 + 2 mr 2 ; 5 5 Here d = r= 4 2 2 2 10 = 2 2 10 m 2

39. Starting from positive positions (i.e. in air), amplitude in air will be more than amplitude in water. Momentum is negative for downward journey. Also water produces damping. graph is spiralling in.

5 1 2 10 m; m = kg 2 2 N=9

Section IV40. U = q2 2 4 + 2 + 2a r 40a dU =0 da

45.

a = 0.3 2N

[

]

f2 f1

For equilibrium,

=

f1R f2R2

mR f1 f2 2 f1 = a = 0 .3 = ,m=2 m m f1 = 1.4, f2 = 0.8 .2 = 0.8 = 0.410

,=

a R

M = mN0 = 10 10 kg 25 19 6 = 10 10 10 mg = 1 mg25 19

46. A0 = |N0| = 10 N0 =

s

1

1010 10 9 19 = 10 10 = 10

PART III 47 C 48 B 49 B 55 B,D 59 D 65 8 60 D 66 6 50 A 51 C 56 A,D 61 A 67 1=

52 D

53 C 57 B,C

54 QUESTION INCORRECT 58 B 63 7 64 9

62 B 68 2(b 1)3 1 + 3 31 1

69 5

Section I 47. Let v = Ai + B j + C kab. v = 0

1 1 A

1 1 1 1 = 0 B C 0 2 1 1 = 0 B C

( x 1)3 R2 = ( x 1) dx = 3 b b

2

R1 R2 =

1 gives = 0 4

3 (b 1) 3

2 1 A

2 ( C B) + 2 (B + A) = 0 A=C For the vector in (C), A=CProjection of 3i j + 3k on C 3 + 1 3 1 = = 3 3 48.

2(b 1)3 1 1 + = 3 3 4 1 b= satisfies the above equation. 2 49. y + 2 = m (x 3) and m = y+2= y3 3 (x 3)

(

)

3 x+2+3 3 =0ln 3

50. y

Let I = Set x = t 1 dx = dt 2 t x=

ln 2

x sin x 2 sin x 2 + sin (ln 6 x 2 )

dx

ln 2 , t = ln2 ln 3 , t = ln3ln 3

x 0 b 1

x=

I=b

R1 =

( x 1) 2 (x 1) dx = 0

b

3

3

=

1 2

ln 2 ln 3

t sin t 1 dt sin t + sin(ln 6 t ) 2 t

ln 2

sin t dt sin t + sin (ln 6 t )

0

=

1 2

ln 3

sin(ln 6 t ) dt sin (ln 6 t ) + sin t

But if The question is reframed in the following manner; Let M and N be two nonsingular skewsymmetric matrices The solution is 2 2 -1 -1 -1 T M N (M N) (MN ) 2 2 -1 -1 T T = M N (-MN) (N ) M 2 2 -1 -1 -1 T -1 = M N (-N M ) (N ) (-M) 2 2 -1 -1 -1 = M N N M (-N) M 2 2 -1 -1 -1 2 -1 -1 =-M N N M N M=-M NM N M 2 -1 = - M N (N M) M 2 -1 = - M N(M N) M 2 -1 -1 =-M NN M M 2 =-M Choice (c) x 2 y2 + =1 4 1 3 and focus = ( 3 ,0) 2 x2 a2

2I = =

1 2

ln 3

dt

ln 2

1 3 log 2 2 1 I = log 3 2 4 51. log x log 2 = log y log 3

log x = k log 2 x = 2 k log y = k log 3 y = 3

k

(2 )log 2 = (3 )log 3k +1 k +1k +1

2log 2

k +1

= 3log 3

( k +1) (log2) = (k+1) (log 3) k = 1 1 x0 = 2 = 12

2

2

55. For e=

52. P = : sin = cos 4 3 P = : cos = cos 4 3 P = : = 2n 4 3 = : = n + 8 Q = : = 2n 4 2 3 = : = n + 8 P=Q a10 2a8 2a9

for

y2 b2

= 1, e =

2 3

b a

2 2

=

1 3

Substitution Focus (2, 0)

( 3,0), a

2

= 3, b = 1 x2 y2 = 1 3

2

required hyperbola is

53.

=

10 10 2 8 8 2

(

9

(

9

)

)

56. j k = i + j + k (i + j + 2k) & k j = i + j +2k (i + 2j +k) j k, k j coplanar with the given vectors 1 1 0 1 0 0 and 1 1 2 = 1 1 3 0

1

2

1

0 1 1

=

8 ( 2 2) 8 (2 2) 2(9 9 ) 8 ,6 8 6 2 9 9 6(9 9 ) 2(9 9 )

(A) and (D) true. 57. f (x +y) = f (x) +f(y) f(x) = kx f(x) is continuous x R and f (x) = k

=

(

)

=

=3

Section II54. Skew symmetric matrix of order 3 is singular. Its inverse does not exist. Therefore there is NO SOLUTION TO THIS QUESTION.

Section IIIFor problems (58) and (59)3 5

1 n =0

a + 48 + 7c = 0 a+6+c=0 Solving a = 1 and c = - 7. 2 The equation is x + 6x 7 = 0 1 1 + 6 + = = 0 = 3

1 3 +i 2 2

a = 2 2 + 8b + 7c = 0 14 + 7b + 7c = 0 (taking first and third columns) Solving we get b = 12 and c = 14 3 1 3 The equation in 2 + 12 + 14 2 = 3 + 1 + 3 2 = 3 ( + ) + 1 = 2 Correct choice (a) 62. b = 6. Taking 1 and 3 columns we getst rd

5n [6 + (5n 1) (a2 3)] sm 64. m = 5n; = 2 n sn [6 + (n 1) (a2 3) 2 9 a2 + 5n (a2 3) =5 9 a2 + n (a2 3 ) 9 a2 + 5 (a2 3) (9 a2 ) + 10(a2 3 ) = , 9 a2 + a2 3 9 a2 + 2 (a2 3 ) is

independent of n 4a2 6 9a2 21 = 6 a2 + 3 (2a2 3) (a2 + 3) = (3a2 7)

2a2 2 24a2 + 54 = 0 a2 2 12a2 + 27 = 0a2 3, a2 = 9

65.

a 5 + a 4 + a 3 + a 3 + a 3 + 1 + a8 + a10 8

a5a 4 a3 13 a8a10 minimum value of sum = 8.

(

( )3

)

1 8

=1

68. Extremum point of latus rectum are (2,4) and (2,4) 1 Area of triangle so formed with ,2 2 1 = = 1 (y1 y2) (y2y3) (y3y1) 8a

66. 6 f ( t )dt = 3x f(x) x1

x

6f(x) = 3xf(x) + 3f(x) 3x 2 f(x) = xf(x) x dy y= x x2 dx dy dy y x y = x2 =x dx dx x 1 I. F = dx x

2

1 (4+4) (4 2) (2 4) = 6 8 .2 Eqn. of tangent at (2,4) is y = x +2 _______(1) Eqn. of tangent at (2, 4) is y = x +2 _____(2) 1 Eqn. of tangent at ,2 is y = 2x +1 _____(3) 2

e

1 1 = x dx = x + c x x y = x(x + c) = f(x) f(1) = 2 2 = 1 + c c = 1 f(x) = x(x + 1) f(2) = 6

y.

Tangent at the extremities of latus rectum intersect directrix at (2, 0). Point of intersection of (1) and (2) is (1,3) and of (2) and (3) is (1, 1) 1 2 0 1 2 = 1 1 3 = 3 = 3 2 1 1 1 1 6 = =2 2 3

67. f() = sin tan

-1

=

tan tan1 1 + tan2 (tan sin

sin cos 2 sin cos 2 1 sin

69.3,2

cos 2

=

cos 2 cos 2 + sin cos 22

=

sin cos sin2 + sin2 2

5 2 5 minimum of 2 z 3 + i 23,

d(f ( )) =1 d(tan )

= tan

5 = 2 + 2 2 = 5 2

MODEL SOLUTIONS TO IIT JEE 2011Paper II1 C 9 C, D 13 8 2 D 3 B 10 A, B, C, D 14 6 19 15 4 PART I 4 5 A D 11 A, B, D 16 7 6 C 7 A 12 A, C, D 8 B

17 18 6 8 20

A r, s, t B p, s C r, s D q, r

A p, r, s B r, s Ct D p, q, r, t

Section I1. [Co(NH3)6]Cl3, Na3[Co(ox)3], [K2Pt(CN)4] [Zn(H2O)6](NO3)2 are diamagnetic.PO2 [H+ ]4 0.06 log 4 [Fe2+ ]20.1 (10 3 )4 (10 3 )2

6.

&

Aromatic primary amines form diazonium salt with NaNO2 and HCl at low temperature which couples with naphthol to form coloured azo dye.Tf = i Kf m

7.

2.

Ecell = E0 cell +

= 4 1.86

0. 1 10 329

= 1.67 + 0.015 log= 1.67 0.015 7 = 1.57 V

= 0.023 2 F.P = 2.3 10 C 8. The structure given is that of Dglucose.

Section II3.

RCH2OH H+ O O

9.OCH2R

Cu

+2

is reduced to Cu

+1

by CN & SCN

4.

CuS & HgS are insoluble in dilute mineral acids. 2+ 2+ Cu & Hg ions belong to group II of qualitative analysis. Haematite is Fe2O3. Oxidation state of Fe is +3 Magnetite is Fe3O4. Oxidation state of Fe is +2 and +3.

10. In (B), (C) and (D), X(CH2)4X is converted to a diamine which can form condensation polymer with adipic acid. In (A), X(CH2)4X is converted to a diol which gives polyester. 11. K = 0.693 t12

5.

K increases with increase of temperature and hence half life decreases. t1 100 = 8t 1 t = 2 log 0 .3 0 .4 2R = R0e-kt

CH318. ClCH2

CH2

C H

CH2

CH3

(2 isomers)CH3

H2+

CH 3 C H CH2

12. MnO4 Mn (acid medium) MnO4 MnO2 (Neutral and aqueous mediums)

CH3

C Cl

Section III13. A truncated octahedrous has 8 hexagonal and 6 square faces. (36 edges and 24 vertices) 14. There are six C H bonds that can involve in hyperconjugation. 15. PCl5 + SO2 POCl3 + SOCl2 PCl5 + H2O POCl3 + 2HCl PCl5 + H2SO4 SO2Cl2 + 2POCl3 + 2HCl 6PCl5 + P4O10 10POCl3 16. [Cl ] from CuCl = 10 + Ksp(AgCl) = [Ag ] [Cl ] 3

(2 different chiral carbon - 4 isomers)CH3 CH3 (1 isomer)

CH3

CH2

C

CH2

Cl CH2ClCH3 CH2 C H CH2 CH3 (1 isomer)

Section IV19. (A) is intramolecular aldol condensation (B) involves Grignard reagent addition. To carbonyl compound. (C) involves nucleophilic addition and dehydration. (D) involves dehydration and intramolecular FriedelCraft reaction. 20. (A) involves transition from solid to gas phase with the absorption of heat. (B) is exothermic and a gaseous product is formed from a solid. (C) involves association (D) white phosphorous is converted to the polymeric red allotropic form as heating. Different solids are considered as different phases.

[Ag ] =

+

1.6 10 10 10 3

= 1.6 10

7

17. Millimoles of Cl = 30 0.01 2 = 0.6 0 .6 Vol. of 0.1 M AgNO3 = = 6 mL 0 .1

PART II 21 B 22 B 29 B, C 33 2 23 C 30 A, C 34 7 39 A p, t B p, s C q, s D q, r Section I21. Escape KE =and V = GMm r

24 A

25 C 31 C, D 36 5

26 D

27 C 32

28 A

A, B, D 37 4 40 38 4

35 5

A p, r, t B p, r C q, s D r, tdB =B=

0N i dr. (b a ) 2r N.i b dB = 0 ln 2(b a ) a

GM r

Eliminate r 22. Phasor:A Resultant /3 A Required

25. At c, 100% reflected. (B or C) But C is correct. At = 0, T + R = 100, R0 26. 0.01 V = 0.01 V + 0.2 V But time of fall =

2h =1 g

4 A sin t + 3 23. L.C = 0.01 mm diameter = 2.70 mm 1 % error = 2.70 3dia m = + dia m 3 % error = + 2 = 3.1% 2.70 24. Take dr at r No. of turns =

V = 100, V = 20 V = 500 27. If non-conservative induced electric field this pattern is possible. 28. T = 2

m holds. K

At equilibrium position, kx0 = QE Equilibrium shifted.

Section I29.R ZAA XC

N.dr (b a)

R ZB

35. T =

XB C

2u sin 60 2 10 3 = = 3 s g 10 2

1 A < XC C ZB < ZA XB C =A A B B R > R and VC > VC

1 2 aT + 1.15 2 1 (u = 10 m s ) 2 a = 5 m s (u cos) T = 36. 1 surface 1 1+ 2 = 2 u v R 7 7 1 1 4 4 + = ( 24 ) V1 6st

30. Linear momentum in the horizontal direction is zero before and after collision CM is stationary. for conservation of angular momentum, of ring increases after collision slip friction towards left. 31.+q1 (or q1) d E=0 +q2 (q2)

W agent = U = q(V2 V1) = 1 (VB VA) = (VB VA) 37. 32. VdA + VdB = 2VdF = 0 dA + dB = 2dF Obviously dA < dF and dB > dF

v1 = 21 cm nd 2 surface nd ( R = for 2 surface) 7 4 4 + =0 21 3 v 2 v2 = 16 cm x = 18 16 = 2 cm1 1 2 2 mv = mg S + kS 2 2 m = 0.18 kg, = 0.1, 2 g = 10 m s , S = 0.06 m 2 v = 0.12 + 0.04 = 0.16

Section I4 4 7 7 1 1 3 3 4 4 33. = + v 24 6 v = 16 from top surface of liquid 2 cm from bottom surface of liquid 34. h = 1242 eV nm = 6.21 eV 200nm

0.16 = 0.4 m s 4 = N=4 10 2 2 2 38. Z = R + XC 2 2 2 XC = 1.25 R R 2 = 0.25 R XC = 0.5 R 1 1 1 C= = = XC 0.5 R 500 250 R

v=

1

= RC =

KE = 6.21 4.7 = 1.51 eV kq = 1.5 V for stopping electrons. r

1 s = 4 ms 250

Section IV39. Knowledge based.

q=

1.5 10 2 9 107

( ) = Ne9

N = 1 10

40. Knowledge based.

PART III 41 B 49 A, D 53 9 42 D 43 A 50 C, D 54 2 59 A p, r, s B r, t Cr Dr Section I41. Let denote the common root 2 + b 1 = 0 2 ++b=0 b 1 = + b (b 1) = b + 1 b +1 = b 1 2 2 2 (b + 1) + b(b 1) (b 1) = 0 2 4b + b(b 1) = 0 2 4+b =0 2 b =3b=I3

44 A

45 46 B C 51 A, B, D

47 48 D C 52 A, B, C, D

55 3

56 1

57 2 60

58 9

Aq B p Cs Ds= f(sin(sin(x ) 2 2 = sin (sin(x ) 2 gogof(x) = gog(x ) 2 = g(sin(x) 2 = sin(sin (x )) 2 2 2 sin (sin(x ) = sin sin (x ) 2 2 sin(sin(x ) (sin sin(x ) 1) = 0 2 2 sin(sin(x ) = 0 or sin (sin(x ) = 1 2 2 sin(x ) = 0 sin (x ) 2 x = n x= n n = {0, 1, 2 ..}2

1 44. for non-singular matrics, 22

a b 1 c 0 1

42. Let the circle be 2 2 x + y + 2gx + 2fy + c = Passed thro (1, 0) 2g + c = 1(1) Passes thro (0, 2) 4f + c = 4(2)

g2 + f 2 c = g f c=0 2 f =c 2 4f + f = 4 2 (f + 2) = 0 f = 2 c = 4 4f = 4 5 2g = 5, g = 2 2 2 Circles is x + y + 5x 4y + 4 = 0 (4, 0) satisfies the equation2

1 - c - a - ac 0 (1 - c) (1 - c) 0 1 1 2 2 a = and c = 2 a = c = and b = pr Hence there are two such matrices.45. Let the point be Normal at is ax by + = a2 + b2 sec tan Normal passes thro (9, 0) 9a = a2 + b2 (1) sec a sec = 6 6 sec = a (1) reduces

43. fogogo f(x) 2 = fogog(x ) 2 = fog(sin(x)

9a

a 2 2 =a +b 6

48.A

3a 2

2

=a +b2

2

2 2 2

(t , 2t)

2

= a + a (e 1) 3 2 =1+e 1 2 2 =e e=P 0 (0, 0

3 2

46.y

OP 1 = PA 3 Let P be (x, y) X=1 2

x= 1 0 1 2

x

t2 2t ,Y= 4 4 Lows of P is 2 4X = 4Y 2 Y =X

Section IIf(x) = f(1 x) 1 curve is symmetrical about x = 2 R1 =2

49. P E F + P F E = Also P E F =

(

) (

)

xf ( x ) dx1

2

(

)

11 25

=2

(1 + 2 x) f (1 + 2 x) dx1

But P E F = P E F = P E F 2 23 P(EF) = 1 = 25 25 2 23 P(EF) = 1 = 25 25 (refer figure) 23 11 = + P(E F ) 25 25 12 P(E F ) = = P(E ) P(F ) 25 From the options 3 4 P(E) = and P(F) = 5 5 or 4 3 P(E) = and P(F) = 5 5 50. Let y = bx 1 bx (1 bx) y = b x x(1 by) = b y by x= 1 by-1

(

) (

2 25

) (

)

(1 x) f (1 x) dx1

P(EF) = P(E F ) + P(F E ) + P(E F)

=

1

(1 x) f ( x) dx1

2

1

2

f ( x ) dx

2

E EF FE

F

xf ( x ) dx

= R2 R1 2R1 = R2 47. Which is of the form 1 ex 0 = e =lim ( f ( x )1) g( x ) (1+ x log(1+ b 2 )1) 1 x

EP

x log(1+ b 2 ) x e2

= elog(1+b2

)2

1 + b = 2b sin 1+ b sin = 2b2

2

Since b > 0 = 2

f (x) = f(x) (1 bx )(1) + (b x )b f(x) = (1 bx )2

f(b) =

1 b2 1 = 2 22 b 1 1 b

(

)

(

)

1 0 M = 1 = 0 2 + 3 + c3 = 12 1 12 c3 = 7. A1 + b 2 + c 3 = 0 + 2 + 7 = 954. Origin lies on the side 2x 3y - < 0 The smaller intersection Is on 2x 3y 1 > 062x 3y = 1

f(0) =

1 + b2 1 = 1 f ' (b )

51. Normal at t is 3 y + xt = 2t + t 3 6 + 9t = 2t + t 3 t 7t 6 = 0 t = 1, t = 2, t = 3 Normal are Y x = 3, And y + 3x = 33 And y 2x = 12 52. f =0 2 + f = cos =0 2 2 (A) is true f(x) is continuous at x = 0 f(0 ) = 0 + f(0 ) = 1 (B) is true F(x) is continuous at x = 1 f(1 ) = 1 + f(1 ) = 1 f(x) is differentiable at x = 1. (c) is true. 22 11 = =1 14 7 3 x= lies in , 2 2 f(x) is differentiable there, since f(x) is a linear function in that interval (D) is true.

6 6

6

All points S lies inside the circle, except 5 3 1 1 , and , satisfies the inequality 2 4 8 4 2x 3y 1 < 0. The remaining 2 points lie on the smaller intersection. 55. Let A = ai + bj + ckP 1=i+j+k

P1 = 02

2

P2 = i + wj + w k P3 = i + w j + wk2

P2 = 0 P3 = 02

2

Them x = A, P1 , y = A, P2 z = A, P3 x + y +z a +b +c2 2 2 2 2 22 2 2 2 A P1 + P2 + P3 = 2 A

=3+0+0=3 56. Let y(x) = n as g (x) = u The given equation becomes du + udu = udu dy u u + u = v ue = e +c du g(x) g(x) i.e y(x) e = e + c y(0) = 0 as g(0) = 0 C = 0 g(2) g(2) y(2) e = e + 0 y (2) = 1 57. x 4x + 12x + x 1 = 0 f(0) < 0, f(1) > 0 one root in (0, 1) 1 3 2 f (x) = 4x 12x + 24x + 1 11 2 f (x) = 12(x 2x +2) > 0 1 f (x) is increasing4 3 2

Section III a1 a2 a3 53. M = b1 b2 b3 c c c 3 1 2 0 1 M 1 = 2 a2 = - 1; b2 = 2; c2 = 3 0 3 1 1 M 1 = 1 a1 + 1 = 1 a1 = 0 0 1 C1 3 = - 1 c1 = 2

it has exactly one root f(x) may have almost 2 distinct roots Since real roots are even in number f(x) has 2 distinct real roots 58.

x ( , 0] [ z, )

(c)

1 cos3 1 cos3

cos sin 0 sin

sin cos 0 cos

cos sin ` 2 cos sin cos

(r c) b = 0r = c +mb 0 = r . a = c . a + m b .a = 4 + m, m = 4r = c + 4b

cos sin cos

=

cos sin

=

2 cos cos 3 x23

= 2 sec 2 [2, )

r . b = c .b + 4 b

2

=1+8=9 (, 1]U[1, )

(d)

(3x 10)5 33

Section IV 2iz 59. (a). (a) Re 1 z2

= 3 x 2 10 x 2 f(x) =

2i Re 1 z z 2i = Re zz 1 = Re ( m z ) 1 = Im z (b) Put y = 3 8 < y 1 3 1 1 y2 8 xy 1 xy 2 3x -1

15 2 x 15 x 2 x = 15 x 1 0 2

If x 2. 60. (a) cos = = 3 2 = 3 3 1 + 3 1 = 4 2

required angle is -

(b)

(f ( x) 3x )dx = aa

x

2

x2

f(x) 3x = - 2x f(x) = x f = 6 65 log[sec x + tan x ]7 6 = log 3 6

8 2 y 1 y2 if 1 y 0 3 2 3y + 8y 3 0 i.e y [1, 1] 2 3y +ay y 3 0 1 (3y 1) (y +3) 0 y 3, 3 1 1 1 y 0, = 3 3 3 x0 8 2 y y 2 1 if 1 y 0 3 2 3y 8y 3 0 2 3y 9y + y 3 0 (3y +1) (y 3) 0 y3 x 1 3 3 x 2 (r, t)

(c)

(d)

1 Arg = Arg(z 1) z 1

Maximum value = .

QUESTIONS AND SOLUTIONS OF IIT-JEE 2011Date : 11-04-2011 Duration : 3 Hours Max. Marks : 240

PAPER - 1Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

INSTRUCTIONSA. General : 1. The question paper CODE is printed on the right hand top corner of this sheet and on the back page of this booklet. 2. 3. 4. 5. 6. 7. 8. 9. No additional sheets will be provided for rough work. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and electronic gadgets are NOT allowed. Write your name and registration number in the space provided on the back page of this booklet. The answer sheet, a machine-gradable Optical Response Sheet (ORS), is provided separately. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET.DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR

Do not break the seals of the question-paper booklet before being instructed to do so by the invigilators. This Question Paper contains having 69 questions. On breaking the seals, please check that all the questions are legible.

B. Filling the Right Part of the ORS: 10. 11. 12. The ORS also has a CODE printed on its Left and Right parts. Make sure the CODE on the ORS is the same as that on this booklet. If the codes do not match, ask for a change of the booklet. Write your Name, Registration No. and the name of centre and sign with pen in the boxes provided. Do not write them anywhere else. Darken the appropriate bubble UNDER each digit of your Registration No. with a good quality HB pencil.

C. Question paper format and Marking Scheme: 13. 14. The question paper consists of 3 parts (Chemistry, Physics and Mathematics). Each part consists of four sections. In Section I (Total Marks: 21), for each question you will be awarded 3 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one (-1) mark will be awarded. In Section Il (Total Marks: 16), for each question you will be awarded 4 marks if you darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. There are no negative marks in this section. In Section III (Total Marks: 15), for each question you will be awarded 3 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one (-1) mark will be awarded. In Section IV (Total Marks: 28), for each question you will be awarded 4 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in this section.

15.

16.

17.

Write your Name registration number and sign in the space provided on the back page of this booklet.

Useful Data : R = 8.314 JK1 mol1 or 8.206 102 L atm K1 mol1 1 F = 96500 C mol1 h = 6.626 1034 Js 1 eV = 1.602 1019 J c = 3.0 108 m s1 NA = 6.022 1023

RESONANCE

J10411Page # 2

SECTION - I (Total Marks : 21)(Single Correct Answer Type)

CHEMISTRY

This section contains 7 multiple choice questions, Each question has four choices, (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. Extra pure N2 can be obtained by heating (A) NH3 with CuO (C) (NH4)2Cr2O7 Ans. Sol. 2. (D)

(B) NH4NO3 (D) Ba(N3)2

Ba (N3)2 Ba + 3N2 Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl, CN and H2O, respectively, are (A) octahedral, tetrahedral and square planar (B) tetrahedral, square planar and octahedral (C) square planar, tetrahedral and octahedral (D) octahedral, square planar and octahedral Ans. (B)

Sol.

Ni+2 + 4Cl [NiCl4]2 SP 3 [NiCl4]2 = 3d8 confiuration with Nickel in + 2 Oxidation state. Cl being weak field ligand does not compel for pairing of electrons. So,

Hence, complex has tetrahedral geometry

Ni+2 + 4Cl [Ni(CN)4]2 [Ni(CN)4]2 = 3d8 configuration with nickel in + 2 oxidation state. CN being strong field ligand with comples for pairing of electrons. So,

Hence, complex has square planar gemetry.

RESONANCE

J10411Page # 3

CHEMISTRY

Ni+2 + 6H2O [NI(H2O)6]+2 [Ni(H2O)6] = 3d8 configuration with nickel in + 2 oxidation state. As with 3d8 configuration two d-orbitals are not available for d2sp3 hybridisation. So, hybridisation of Ni (II) is sp3d2. As Ni (II) with six co-ordination will have octahedral geometry.

Note : With water as ligand Ni (II) forms octahedral complexes. 3. Bombardment of aluminum by -particle leads to its artificial disintegration in two ways, (I) and (ii) as shown. Products X, Y and Z respectively are,

(A) proton, neutron, positron (C) proton, positron, neutron Ans. (A)

(B) neutron, positron, proton (D) positron, proton, neutron

Sol.

+

4.

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is : (A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M Ans. (C)

RESONANCE

J10411Page # 4

CHEMISTRYSol. Mole =120 =2 60

mass of solution = 1120 g V=1120 112 = L 1.15 1000 115

M=

2 115 = 2.05 mol/litre 112

5.

AgNO3 (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance () versus the volume of AgNO3 is :

(A) (P) Ans. (D)

(B) Q

(C) (R)

(D*) (S)

Sol.

6.

Among the following compounds, the most acidic is : (A) p-nitrophenol (B) p-hydroxybenzoic acid (C) o-hydroxybenzoic acid (D) p-toluic acid Ans. (C)

RESONANCE

J10411Page # 5

CHEMISTRY

Sol.

Due to intramolecular hydrogen bonding in conjugate base of o-Hydroxybenzoic acid, it is strongest acid. 7. The major product of the following reaction is :

(A)

(B)

(C)

(D)

Ans.

(A)

Sol.

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CHEMISTRYSECTION II (Total Marks : 16)(Multiple Correct Answers Type) This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE may be correct. 8. Extraction of metal from the ore cassiterite involves (A) carbon reduction of an oxide ore (B) self-reduction of a sulphide ore (C) removal of copper impurity (D) removal of iron impurity Ans. Sol. (A, D)

Important ore of tin is cassiterite (SnO2). SnO2 is reduced to metal using carbon at 1200 1300C in an electric furnace. The product often contains traces of Fe, which is removed by blowing air through the molten mixture to oxidise FeO. Which then floats to the surface. SnO2 + 2C Sn + 2CO Fe + O2 FeO The correct statement(s) pertiaining to the adsorption of a gas on a solid surface is (are) (A) Adsorption is always exothermic (B) Physisorption may transform into chemisorption at high temperature (C) Physisorption increases with increasing temperature but chemisorption decreases with increasing temperature (D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation. Ans. (A, B, D)

9.

Sol.

(A) H = ve for adsorption (B) fact (D) chemical bonds are stronger than vander waals forces so chemical adsorption is more exothermic. According to kinetic theory gases (A) collisions are always elastic (B) heavier molecules transfer more momentum to the wall of the container (C) only a small number of molecules have very high velocity (D) between collisions, the molecules move in straight lines with constant velocities. Ans. (A, B, C, D)

10.

Sol.

(A) Fact (B) P = MV = M3RT = M3MRT

(C) Max well distribution (D) Fact

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CHEMISTRY11. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are)

(A) (C) H2C=C=O Ans. Sol. (B, C) In (B)

(B) HC (D) H2C=C=CH2

and (C) CH2 == C == O all atoms are always in same plane.

SECTION Ill (Total Marks : 15) (Paragraph Type)This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple choice questions and based on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Paragraph for Question Nos. 12 and 14 When a metal rod M is dipped into an aqueous colourless concentrated solution of compound N the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aqueous NH3 dissolves O and gives an intense blue solution. 12. The metal rod M is : (A) Fe Ans. 13. (B)

(B) Cu

(C) Ni

(D) Co

The compound N is : (A) AgNO3 Ans. (A)

(B) Zn(NO3)2

(C) Al(NO3)3

(D) Pb(NO3)2

14.

The final solution contains (A) [Pb(NH3)4]2+ and [CoCl4]2 (C) [Ag(NH3)2]+ and [Cu(NH3)4]2+ Ans. (C) (B) [Al(NH3)4]3+ and [Cu(NH3)4]2+ (D) [Ag(NH3)2]+ and [Ni(NH3)6]2+

Sol.

(12), (13) & (14) Cu(M) + AgNO3 (aqueous colorless solution) Resultant solution (Cu(NO3)2 + AgNO3(N) (blue solution)

Note : Here it is considered that complete AgNO3 is not utilized in the reaction. AgNO3 + NaCl AgCl + NaNO3(O) (white ppt.)

Solution containing white ppt. of AgCl also contains Cu(NO3)2.

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CHEMISTRYWhich developed deep blue colouration with aqueous NH3 solution AgCl (white) + 2NH3 (aq.) [Ag(NH3)2]+ Cu(NO3)2 (aq.) + 4NH3 (aq.) [Cu(NH3)4] (NO3)2(deep blue coloration)

So, Metal rod M is Cu. The compound N is AgNO3 and the final solution contains [Ag(NH3)2]+ and [Cu(NH3)42+ Paragraph for Question Nos. 15 and 16 An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic product through the following sequence of reactions, in which Q is an intermediate organic compound.

15.

The structure of compound P is (A) CH3CH2CH2CH2CCH

(B) H3CH2CCCCH2CH3

(C)

(D)

Ans. 16.

(D)

The structure of the compound Q is

(A)

(B)

(C)

(D)

Ans. Sol. (15 & 16)

(B)

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CHEMISTRYSECTION - IV (Total Marks : 28) (Integer Answer Type) This section contains 7 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS. 17. The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is Ans. 5

Sol.

18.

Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is Ans. 5

Sol. 19.

5NaBr + NaBrO3 + CO2 3Br2 + 3Na2CO3

The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number, ms = 1/2, is Ans. 9

Sol.1 will be 1 + 3 + 5 = 9. 2

So, electrons with spin quantum number =

20.

The work function () of some metas is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal isMetal f (eV) Li 2.4 Na 2.3 K 2.2 Mg 3.7 Cu 4.8 Ag 4.3 Fe 4.7 Pt 6.3 W 4.75

Ans. Sol. Ephoton =

412400 = 4.13 ev 3000

Photoelectric effect can take place only if Ephoton Thus, Li, Na, K, Mg can show photoectric effect.

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CHEMISTRY21. To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1 mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm. at 0C) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0C is close to Ans. 7 PHe = 1 0.68 = 0.32 atm V=? n = 0.1 ; 22. V=nRT 0.1 0.0821 273 = =7 P 0.32

Sol.

The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is Ans. 5

Sol.

Total 5 products. 23. A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. Glycine contributes 47.0 % to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is Ans. Sol. 6

Molecular weight of decapeptide = 796 g/mol Total bonds to be hydrolysed = (10 1) = 9 per molecule Total weight of H2O added = 9 18 = 162 g/mol Total weight of hydrolysis product = 796 + 162 = 958 g Total weight % of glycine (given) = 47% Total weight of glycine in product =958 47 g = 450 g 100

Molecular weight of glycine = 75 g/mol Number of glycine molecule =450 = 6. 75

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PHYSICS PART - IISECTION - I (Total Marks : 21)(Single Correct Answer Type) This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 24. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is (A) 8.50 kHz Ans. Sol. (A) (B) 8.25 kHz (C) 7.75 kHz (D) 7.50 kHz

finisident = freflected =

320 8 kHz 320 10

fobserved =330 310

320 10 freflected 320

= = 25.

8

8.51 kHz 8.5 kHz

5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T1, the work done in the process is : (A)9 RT1 8

(B)

3 RT1 2

(C)

15 RT1 8

(D)

9 RT1 2

Ans. (A) Sol. Number of moles of He = Now

1 4

T1 (5.6) 1 = T2 (0.7) 1

1 T1 = T2 8

2/3

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PHYSICS4T1 = T2 Work done =

nR[T2 T1 ] 1

1 R [3T1 ] 4 = 2 39 = RT1 8

26.

, where E0 is a constant. The flux through the shaded area (as shown in Consider an electric field E E 0 xthe figure) due to this field is :

(A) 2E0a

2

(B)

2 E0a

2

(C) E0a

2

(D)

E 0a 2 2

Ans. (C)

Sol.

flux = (E0 cos 45) area) =

E0 2

a 2a

= E0a2

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PHYSICS27. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is : (A) 1215 Ans. (A) (B) 1640 (C) 2430 (D) 4687

Sol.

1 1 5 2 1 RZH 4 9 = R(1)2 H2 36

1 1 3 2 1 RZHe = R(4) He 4 16 16 He 1 16 5 5 H2 4 3 36 27 5 6561 = 1215 27

He = 28.

A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is :

(A) 9 Ans. (D)

(B) 18

(C) 27

(D) 36

Sol.

T sin = m Lsin2 324 = 0.5 0.5 2 2 =324 0 .5 0 .5

=

324 0.5 0.518 = 36 rad/sec. 0 .5

=

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MATHEMATICS29. A meter bridge is set-up as shown, to determine an unknown resistance X using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of X is

(A) 10.2 ohm Ans. (B) Sol. 1 = 52 + 1 = 53 cm 2 = 48 + 2 = 50 cm

(B) 10.6 ohm

(C) 10.8 ohm

(D) 11.1 ohm

1 x 53 x 2 R 50 10x = 10.6

30.

A 2F capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is (A) 0% (B) 20% (C) 75% (D) 80% Ans. (D)

Sol.

Ui =

V 1 (2)V 2 , Vcommon = 5 2

Uf =

V 1 (2 + 8) 2 5

2

V2 V2 Ui Uf 5 100 100 = Ui V24 100 = 80% Ans. 5

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MATHEMATICS SECTION II (Total Marks : 16)(Multiple Correct Answers Type)This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE may be correct. 31. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA) are kept far apart and each is given charge +Q. Now they are connected by a thin metal wire. Then (A) Einside 0 A Ans. (A), (B), (C), (D) (B) QA > QB

A RB (C) R B A

surface (D) E on A

on surface EB

Sol.

QA + QB = 2Q

...(i) ...(ii)

KQ A KQB RA RB

RA (i) and (ii) QA = QB R B RA QB 1 R B = 2Q 2Q R B = R R RA A B 1 R B 2Q

&

QB =

&

2Q RA QA = R R A B RB A Q A / 4R 2 A = 2 RA B QB / 4RB A EA = 0 EA < EB

QA > QB

using (ii)

B & EB = 0(at surface)

A < B

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PHYSICS32. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true? (A) They will never come out of the magnetic field region. (B) They will come out travelling along parallel paths. (C) They will come out at the same time. (D) They will come out at different times. Ans. (B), (D)

Sol.

tp =

2 RP 2 mp v 2mp v eBv eB( 2 2)m e v ( 2 2)m e (2 2) R e = eBv eB v

te =

te tp

RESONANCE

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PHYSICS33. A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state

(A) heat flow through A and E slabs are same (B) heat flow through slab E is maximum (C) temperature difference across slab E is smallest (D) heat flow through C = heat flow through B + heat flow through D. Ans. Sol. A: C: (A), (C), (D) At steady state, heat flow through A and E are same. T = i R i is same for A and E but R is smallest for E.

D:

iB = R B

T

iC = R C

T

T iD = R Dif ic = iB + iD Hence1 1 1 R C RB R D

8KA 3KA 5kA

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PHYSICS34. A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R ( Angular frequency for case B. (D) Angular frequency for case A < Angular frequency for case B. Ans. Sol. (A), (D)

torque is same for both the cases. T = 2 A > B A < B

mgd

SECTION III (Total Marks :15)(Paragraph Type)This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple choice questions and based on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

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PHYSICSParagraph for Question Nos. 35 to 37 Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum. upwards (or to right) is positive and downwards (or to left) is negative.

35.

The phase space diagram for a ball thrown vertically up from ground is :

Momentum

Momentum

(A)Position

(B)

Position

Momentum

Momentum

(C)

Position

(D)Position

Ans. (D)

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PHYSICS

Sol.

36.

The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the

two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the totalmechanical energies respectively. Then

(A) E1 2 E 2 Ans. (C) Sol.

(B) E1 2 E 2

(C) E1 4 E 2

(D) E1 16 E 2

In 1st case amplitude of SHM is a. In 2nd case amplitude of SHM is 2a Total energy =

1 k(amplitude)2 2

E1 =

1 k(2a)2 2 1 k(a)2 2

E2 =

E1 4 E 2

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PHYSICSAlternative : Linear momentum P = mv = m A 2 x 2 P2 = m22 (A2 x2) P2 + (m)2x2 = m22A2 Equation of circle (bigger) P2 + x2 = (2a)2 P2 + x2 = 4a2 Equation of circle (smaller) P2 + x2 = a2 Comparing (i) and (ii) Amplitude and (m)2 = 1 A = 2a m2 = ...(iii) ...(ii) ...(i)

1 m

1 m2 ( A )2 2So energy E1 =

1 m 2 (2a)2 2 1 1 ( 4a 2 ) 2m2a 2 m

=

=

Comparing (i) and (iii) A=a (m)2 = 1 m2 =

1 m

So E2 =

1 a2 1 a2 1 1 1 m 2 A 2 = a 2 = 2 2 m 2 m2 2 m

E1 So E 4 E1 = 4E2 2

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PHYSICS37. Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is :

Momentum

(A)

Position

(B)

(C)

(D)

Ans. (B) Sol. Linear momentum P = mv2 2 = m A x

P2 + m22x2 = m22A2

represents a circle on Px diagram with radius of circle R = A ( m22 = 1) of spring mass system remains constant and equal to

k m

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PHYSICSAmplitude of oscillation inside liquid will decrease due to viscous force So radius of circular arcs will decrease as position change Correctly shown in option B

Paragraph for Question Nos. 38 and 39 A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency p, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency , where a part of the energy is absorbed and a part of it is reflected. As approaches p all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. 38. Taking the electronic charge as e and the permitlivity as 0, use dimensional analysis to determine the correct expression for p.

(A) Ans. (C)

Ne m 0

(B)

m 0 Ne

(C)

Ne 2 m 0

(D)

m 0 Ne 2

1

Sol.

Ne 2 m 0

L

3

Q2 Q2 2

M

1 T

L F

So only (C) is dimensionally correct

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PHYSICS39. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons1 N 4 x 1027 m3. Take 0 1011 and m 1030, where these quantities are in proper SI units. s. (A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm

Ans. (B) Sol. For resonance

= P =

Ne 2 m 0

4 10 27 (1.6 10 19 )2 10 30 10 11

= 3.2 1015 f=

3.2 1015 1 1015 2 2 3.14 2

3 10 8 c = = 1 1015 f 2 600 nm

SECTION IV (Total Marks : 28)(Integer Answers Type)This sections contains 7 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS. 40. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is

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PHYSICSSol.Note : If net force applied by the rod is considered to be 2 N.

f '2 F 2 2FR f'R = 2mR2

...(i)

a R...(ii)

F f' = 2ma = 1.2 From (i) & (ii) (1.2 + f ')2 + f '2 = 22 2f '2 + 2.4f ' + 1.44 = 4 f '2 + 1.2f ' + 0.72 2 = 0 f '2 + 1.2f ' 1.28 = 0 f' =

1.2 1.44 4 (1.28 ) 2

= 0.6 0.36 1.28 = 0.6 = 0.68 From eq. (2) F = 1.88 =0.68 P = 1.88 10

1.64

P = 3.61 4

Ans.

Note : In Hindi friction force is aksed, so the answer is P = 6.8. (for Hindi)

Note : But if only normal reaction applied by the rod is considered to be 2 N.

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PHYSICS Law a = R 0.3 = [0.5] =3 rad/s 5

2 f = 2 [0.3] f = 2 0.6 f - 1.4 Nx ...(i)

....(ii)

c = c fR 2R = mR2 f 2 = mR 1.4 2 =

2 3 2 5

1.4 0.6 = 2 0.8 = 2 = 0.4 =P 10

P=4

Ans.

Note : In Hindi friction force is aksed, so the answer is P = 8. (for Hindi) 41. A block is moving on an inclined plane making an angle 45 with the horizontal and the coefficient of friction is . The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 , then N is

Sol.

mgF1 =

2

mg 2

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PHYSICSmgF2 =

2

mg 2

F1 = 3F2 1 + = 3 3 4 = 2 =

1 2Ans.

N = 10 N=5 42.

Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side a.The surface tension of the soap film is . The system of charges and planar film are in equilibrium, and q2 a =k 1/ N

, where k is a constant. Then N is

Sol.

Surface Tension =

force length

2kq2 kq2 2 2 = 2 a 2 2a a 1

q2 a = (Some constant) 43.

3

So

N=3

Ans.

Steel wire of length L at 40C is suspended from the ceiling and then a mass m is hung from its free end. The wire is cooled down from 40C to 30C to regain its original length L. The coefficient of linear thermal expansion of the steel is 105 /C, Youngs modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of m in kg is nearly.

Sol.

F L =y A L mg y ( ) A

m= Ans. 3

Ay( ) r 2 y( ) (10 3 )2 1011 10 5 10 = = 3 g g 10

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PHYSICS44. The activity of a freshly prepared radioactive sample is 10 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 1025 kg. The mass (in mg) of the radioactive sample is Sol. N = N0 etdN 9 = 1010 = N0 () e 10 t dt10

at (t = 0) 1010 = N0 109 N0 = 1019 mass of sample = N0 1025 = N0 (mass of the atom) = 106 kgm = 106 103 gm = 103 gm = 1 mg Ans. 45. 1

A long circular tube of length 10 m and radius 0.3 m carries a current along its curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as = 0 cos (300 t) where 0 is constant. If the magnetic moment of the loop is N 0 0 sin (300 t), then N is

Sol.

Flux through circular ring = (0 ni) r2 =

0 2 r 0 cos 300 t Ld Rdt

i=

i=

0 r 2 0 . sin 300 t 300 RL

=

r 2 .300 0 0 sin 300 t RL

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PHYSICSM = . r2

=

2r 4 .300 0 0 sin 300 t RL 10 10 4 300 100 10N=6 Ans.

(Take 2 = 10)

=

46.

Four solid spheres each of diameter N 104 kg-m2, then N is

5 cm and mass 0.5 kg are placed with their centers at the corners of

a square of side 4cm. The moment of inertia of the system about the diagonal of the square is

Sol.

2 2 2 2 2 = MR 2 + MR Mx 2 5 5 2 2 2 2 = MR 2 + MR 2 + (Mx2) 2 5 5 2 2 = 4 MR + 2mx2 5 8 MR 2 + 2mx2 5

=

2 8 0.5 5 2 (0.5) ( 4 2)10 4 = 5 2

5 = 8 104 5

= 9 104 = N 104 So, N = 9 Ans.

RESONANCE

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MATHEMATICS

PART - IIISECTION - I (Total Marks : 21)(Single Correct Answer Type)This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

47.

Let (x0, y0) be the solution of the following equations (2x)n2 = (3y)n3 3nx = 2ny . Then x0 is (A)1 6

(B)

1 3

(C)

1 2

(D) 6

Ans. Sol.

(C)

(2x)n2 = (3y)n3 also n2 n(2x) = n3 n(3y) = n3 (n3 + ny) 3nx = 2ny nx n3 = ny n2 ......... (2)nxn3 n 2 . n (2x) = n3 n3 n2

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

by (1) n2 n(2x) = n3 (n3 + ny) n22 n2x = n23 (n2 + nx)

n 2 n 3 (n2x) = 0n2x = 0 x =

2

2

1 2

n 3

48.

The value of

n 2

x sin x 2 dx is sin x sin(n6 x 2 )2

(A)

1 3 n 4 2

(B)

1 3 n 2 2

(C) n

3 2

(D)

1 3 n 6 2

Ans. Sol. Put

(A) x2 =t x dx =

dt 2

RESONANCE

J10411Page # 31

MATHEMATICS

n 3

sin t.

I =

n 2

sin t sin (n6 t)b

dt ......(1) 2

b

apply

a

f ( x )dx =

f (a b x)dxa

1 I= 2

n 3

n 2

sin(n6 t) sin tn 3

sin(n6 t )

dt .......(2)

adding (1) and (2)

1 2I = 2

n 2

1.dt1 3 n 4 2

I=

49.

,b and c Let a i jk i jk i j k be three vectors. A vector in the plane of a and b , whose 1 projection on c is , is given by 3 (A) i 3 j 3kAns. (C)

(B) 3 i 3 jk

(C) 3 i j 3k

(D) i 3 j 3k

Sol.

Let

= a b i + ( ) j + ( )k = ( )

Now

1 = . c

3 1=

( ) ( ) ( )

For =

3

3

j + (2) k = (2+ 1) i 1, = 3 l j 3k

RESONANCE

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MATHEMATICS50. Let P = { : sin cos = (A) P Q and Q P (C) P Q Ans. (D)2 cos } and Q = { : sin + cos = 2 sin } be two sets. Then

(B) Q P (D) P = Q

Sol.

P = {: sin cos = sin = ( 2 + 1) cos tan = = n+2 +1

2 cos }

3 ; n I 82 sin }

Q = { : sin + cos = cos = ( 2 1) sin

1 tan =

2 1

=

2 +1

51.

= n + P=Q

3 ; n I 8

Let the straight line x = b divide the area enclosed by y = (1 x)2, y = 0, and x = 0 into two parts R1 (0 x b) and R2(b x 1) such that R1 R2 =

1 . Then b equals 41 3

(A)

3 4

(B)

1 2

(C)

(D)

1 4

Ans.b

(B)( x 1)2 dx =

Sol.

R1 =

0

( x 1)3 3

b

=0

(b 1)3 1 31

1

also R2 =

b

( x 1)2 dx =

( x 1)3 3

=b

(b 1)3 3

R1 R2 =

2(b 1)3 1 3 31 8

1 2(b 1)3 1 = 4 3 3

(b 1)3 =

b=

1 2J10411Page # 33

RESONANCE

MATHEMATICS52. Let and be the roots of x2 6x 2 = 0, with > . If an = n n for n 1, then the value of is (A) 1 Ans. Sol. (C) (B) 2 (C) 3 (D) 4

a10 2a 8 2a 9

x2 6x 2 = 0 having roots and 2 6 2 = 0 10 69 28 = 0 10 28 = 69 similarly by (i) and (ii) (10 10) 2(8 8) = 6 (9 9) a10 2a8 = 6a910

.... (i)8 9

2 = 6

.... (ii)

Aliter

a10 2a8 =3 2a9

10 10 2( 8 8 ) 10 10 ( 8 8 ) 9 ( ) 9 ( ) = = 2( 9 9 ) 2( 9 9 ) 2( 9 9 )

= 53.

6 = =3 2 2 3 x y 1. If L also

A straight line L through the point (3, 2) is inclined at an angle 60 to the line intersects the x-axis, then the equation of L is (A) y + (C) Ans.

3x+23 3=0

(B) y (D)

3x+2+3 3 =0

3yx+3+2 3 =0(B)

3y+x3+2 3 =0

Sol.

Let slope of line L = m

m ( 3 )

1 m( 3 )

= tan 60 =

m 3

3

1 3m

=

3

taking positive sign,

m+ m=0

3 =

3 3m

taking negative sign m+ m= As L cuts x-axis so L is y + 2 = m=

3 + 3 3

3 3m = 0

3 (x 3)J10411Page # 34

RESONANCE

MATHEMATICSSECTION - II (Total Marks : 16)(Multiple Correct Answers Type)The section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE may be correct.

54.

and , and perpendicular to the vector The vector(s) which is/are coplanar with vectors i j 2k i 2 jk is/are i jk (A) j kAns. (A, D) (B) i j (C) i j

(D) jk

Sol.

j + 2k a = i+ j+k i +2 b= j +k i+ c =

Required vector is c ( a b )

[( c . b ) a ( c . a ) b ] ) (1+1+2) ( )] [(1+2+1) ( i+ j +2 k j +k i +2 ] j +4 k [4 so our vector in parallel

j + k

55.

Let M and N be two 3 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes the transpose of P, then M2 N2 (MT N)1 (MN1)T is equal to (A) M2 Ans. (C) (B) N2 (C) M2 (D) MN

Sol.

Data inconsistent A 3 3 non-singular matrix cannot be skew-symmetric However considering M, N matrices as even order, we obtain correct answer. M2 N2 (MT N)1 (MN1)T = M2N2 N1 (MT)1 (N1)T MT M2 N2 N1 M1 N1 M M2 NM1 N1 M MNN1 M M2

(So the question is wrong)

RESONANCE

J10411Page # 35

MATHEMATICSx2 y2 = 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the 2 a b2

56.

Let the eccentricity of the hyperbola

hyperbola passes through a focus of the ellipse, then (A) (B) (C) (D) Ans. Sol. the equation of the hyperbola is a focus of the hyperbola is (2, 0) the eccentricity of the hyperbola is5 3

x2 y2 =1 3 2

the equation of the hyperbola is x2 3y2 = 3 (B, D)

Eccentricity of ellipse =

1

1 3 = 4 2

1

b2

2 3

= a2

b = a

1 3

focus of ellipse 3 , 0

3

( 3 )2

a2

=1 a=

b = 1 & focus of hyperbola (2, 0)

Hence equation of hyperbolax2 y2 =1 3 1

57.

Let f : R R be a function such that f(x + y) = f(x) + f(y), x, y R. If f(x) is differentiable at x = 0, then (A) f(x) is differentiable only in a finite interval containing zero (B) f(x) is continuous x R (C) f(x) is constant x R (D) f(x) is differentiable except at finitely many points Ans. (B, C)

Sol.

f(x) = kx Hence f(x) is continuous & differentiable at x R & f (x) = k (constant)

RESONANCE

J10411Page # 36

MATHEMATICSSECTION - III (Total Marks : 15)(Paragraph Type) This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple choice questions and based on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choice (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions Nos. 58 to 60 Let a, b and c be three real numbers satisfying

1 9 7 [a b c] 8 2 7 = [0 0 0] 7 3 7

...........(E)

58.

If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is (A) 0 Ans. (D) (B) 12 (C) 7 (D) 6

59.

Let be a solution of x3 1 = 0 with m () > 0. if a = 2 with b and c satisfying (E), then the value of3 1 3 a + b + c

is equal to (B) 2 (C) 3 (D) 3

(A) 2 Ans. 60. (A)

Let b = 6, with a and c satisfying (E). If and are the roots of the quadratic equation ax2 + bx + c = 0,

then

n0

1 1

n

is

(A) 6

(B) 7

(C)

6 7

(D)

Ans. Sol.

(B)

(Q.No. 58 to 60) a + 8b + 7c = 0 9a + 2b + 3c = 0 a+b+c=0 ........... (i) ........... (ii) ........... (iii)

1 8 7 = 9 2 3 = 1.(1) 8(6) + 7(7) = 0 1 1 1

RESONANCE

J10411Page # 37

MATHEMATICSLet C= a + 8b = 7 a + b = b=

6 & a= 7 7

6 , , where R (a, b, c) 7 7

58.

P(a, b, c) lies on the plane 2x + y + z = 1

2 6 = 1 7 7

=1 7

= 7

7a + b +c = 7 + 6 7 = 6 59. a=2 = 14

b = 12 & c = 14 Now3 1 3 3 1 b c = 2 12 3.14 = 3 + 1 + 32 = 3( +2) + 1 = 2 a

60.

b=6 now

= 7 a=1 & c=7 ax2 + bx + c = 0 x = 7 , 1 x2 + 6x 7 = 0

1 1 6 = 7 n 0 n 0

n

n

=

1+

6 6 + 7 7

2

+ ....... =

1 1 6 7

=7

Paragraph for Question Nos. 61 and 62 Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2. 61. The probability of the drawn ball from U2 being white is (A)13 30

(B*) (B)

23 30

(C)

19 30

(D)

11 30

Ans. Sol. P(white)

= P (H white) + P(T white)2 C 1 3 2 1 1 3C 1 3 C 2 C 2 1 5 2 1 5 2 51 1 2 5 5 2 2 C2 3 C2 C2 3

=

RESONANCE

J10411Page # 38

MATHEMATICS=

1 8 1 3 1 12 2 10 2 10 30 30 4 1 22 10 2 30 23 30

=

=

62.

Given that the drawn ball from U2 is white, the probability that head appeared on the coin is (A)17 23

(B) (D)

11 23

(C)

15 23

(D)

12 23

Ans. Sol.

P Head

White =

P(Head white) P( white)

1 3 2 1 4 1 12 2 5 5 2 10 = = = 23 23 23 30 30

SECTION - IV (Total Marks : 28)(Integer Answer Type)This section contains 7 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9). The boubble corresponding to the correct answer is to be darkened in the ORS.

63.

Consider the parabola y2 = 8x. Let 1 be the area of the triangle formed by the end points of its latus rectum1 and the point P , 2 on the parabola, and 2 be the area of the triangle formed by drawing tangents at P 2

1 and at the end points of the latus rectum. Then is 2Ans. Sol. (2) 2 =

1 2

(by property)

1 2 = 2

RESONANCE

J10411Page # 39

MATHEMATICSp

64.

Let a1, a2, a3,....., a100 be an arithmetic progression with a1 = 3 and Sp =

a , 1 p 100.i i 1

Sm For any integer n with 1 n 20, let m = 5n. If S does not depend on n, then a2 is nAns. (9)

Sol.

5n [6 (5n 1)d] 2 S 5n Sm 5[( 6 d) 5nd] = = = n Sn Sn [( 6 d) nd] [6 (n 1)d] 2d = 6 or d = 0 a2 = 3 + 6 = 9 Now if d = 0 then a2 = 3 else a2 = 9

for single choice more appropriate choice is 9, but in principal, question seems to have an error.

65.

The positive integer value of n > 3 satisfying the equation1 1 1 is 2 3 sin sin sin n n n

Ans. Sol. 1 sin n

(n = 7) 1 1 = 3 2 sin sin n n

2 cos

2 sin 1 n n 3 = sin 2 sin sin n n n 4 3 = sin n n

sin

4 3 = (1)k + k , k n nIf k = 2m

= 2m n

1 = 2m , not possible nIf Ans. k = 2m + 1 n = 7, m = 0 n=7

7 = (2m + 1) n

RESONANCE

J10411Page # 40

MATHEMATICS66. Let f : [1, ) [2, ) be a differentiable function such that f(1) = 2. Ifx

6 f ( t ) dt 3 xf ( x ) x 3

1

for all x 1, then the value of f(2) is Ans. Sol. (6)

Data inconsistent. Putting x = 1 , in given integral equation f(1) = 1/3 , a contradiction (given that f(1) = 2).x

However if considering integral equation as 6 f ( t ) dt 3 xf ( x ) x 3 5

1

we obtain correct answer. Differentiating the integral equation 6f(x) = 3f(x) + 3xf(x) 3x2 f(x) put

1 f(x) = x x

y = f(x)dy 1 y=x dx x

I.F. =

1 x 1 =x+c x c=1

General solution is y Put x = 1, y = 2 y=x +x f(x) = x2 + x2

f(2) = 4 + 2 = 6

67.

If z is any complex number satisfying |z 3 2i| 2, then the minimum value of |2z 6 + 5i| is Ans. (5)

Sol.

5i |2z 6 + 5i| = 2 z 3 2 5 =5 2

for minimum = 2

RESONANCE

J10411Page # 41

MATHEMATICS68. The minimum value of the sum of real numbers a5, a4, 3a3, 1, a8 and a10 where a > 0 is Ans. Sol. (8)

A.M. G.M.

1 1 1 1 1 1/ 8 4 3 3 3 1 a 8 a10 5 1 1 1 1 1 8 10 a a a a a 5 . 4 . 3 . 3 . 3 .1.a .a 8 a a a a a 1 1 3 1 a8 a10 8(1)1/ 8 a5 a 4 a 31 1 3 4 3 1 a 3 a10 = 8, at a = 1 5 a a a

minimum value of

69.

d sin ( f ()) is tan 1 Let f() = sin cos 2 , where 4 < < 4 . Then the value of d (tan ) Ans. (1)

Sol.

sin sin = sin1 tan 1 cos cos 2 f() = tandf =1 d tan

RESONANCE

J10411Page # 42

Name of the Candidate

Roll Number

I have read all the instructions and shall abide by them. -------------------------------Signature of the Candidate

I have verified all the information filled in by the Candidate. -------------------------------Signature of the Invigilator

RESONANCE

J10411Page # 43

QUESTIONS AND SOLUTIONS OF IIT-JEE 2011Date : 11-04-2011 Duration : 3 Hours Max. Marks : 240

PAPER - 2Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

INSTRUCTIONSA. General : 1. The question paper CODE is printed on the right hand top corner of this sheet and on the back page of this booklet. 2. 3. 4. 5. 6. 7. 8. 9. No additional sheets will be provided for rough work. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and electronic gadgets are NOT allowed. Write your name and registration number in the space provided on the back page of this booklet. The answer sheet, a machine-gradable Optical Response Sheet (ORS), is provided separately. DO NOT TAMPER WITH/MUTILATE THE ORS OR THE BOOKLET. Do not break the seals of the question-paper booklet before being instructed to do so by the invigilators. This Question Paper contains having 60 questions. On breaking the seals, please check that all the questions are legible.

B. Filling the Right Part of the ORS: 10. 11. 12. The ORS also has a CODE printed on its Left and Right parts. Make sure the CODE on the ORS is the same as that on this booklet. If the codes do not match, ask for a change of the booklet. Write your Name, Registration No. and the name of centre and sign with pen in the boxes provided. Do not write them anywhere else. Darken the appropriate bubble UNDER each digit of your Registration No. with a good quality HB pencil.

C. Question paper format and Marking Scheme: 13. 14. The question paper consists of 3 parts (Chemistry, Physics and Mathematics). Each part consists of four sections. In Section I (Total Marks: 24), for each question you will be awarded 3 marks it you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one (-1) mark will be awarded. In Section II (Total Marks: 16), for each queshon you will be awarded 4 marks if you darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. There are no negative marks in this section. In Section Ill (Total Marks: 24), for each question you will be awarded 4 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in this section. In Section IV (Total Marks: 16), for each question you will be awarded 2 marks for each row in which you have darkened ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. Thus, each question in this section carries a maximum of 8 marks. There are no negative marks in this section. Write your Name registration number and sign in the space provided on the back page of this booklet.

15.

16.

17.

DO NOT BREAK THE SEALS WITHOUT BEING INSTRUCTED TO DO SO BY THE INVIGILATOR

Useful Data : R = 8.314 JK1 mol1 or 8.206 102 L atm K1 mol1 1 F = 96500 C mol1 h = 6.626 1034 Js 1 eV = 1.602 1019 J c = 3.0 108 m s1 NA = 6.022 1023

RESONANCE

J10411Page # 2

CHEMISTRY PART-IISECTION I (Total Marks : 24)(Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 1. Oxidation states of the metal in the minerals haematite and magnetite, respectively, are : (A) II, III in haematite and III in magnetite (B) II, III in haematite and II in magnetite (C) II in haematite and II, III in magnetite (D) III in haematite and II, III in magnetite Ans. Sol. (D)

In haematite(Fe2O3), Fe is present (III) oxidation state and in magnetite (Fe3O4) Fe is present in (II) and (III) oxidation state. Among the following complexes (KP), K3[Fe(CN)6] (K), [Co(NH3)6]Cl3 (L), Na3[Co(oxalate)3] (M), [Ni(H2O)6]Cl2 (N), K2[Pt(CN)4] (O) and [Zn(H2O)6](NO3)2 (P) the diamagnetic complexes are : (A) K, L, M, N Ans. (C) (B) K, M, O, P (C) L, M, O, P (D) L, M, N, O

2.

Sol.

K[Fe(CN)6]3 : 3d5 electron configuration after pairing of electrons for d2sp3 hybridisation it contains one unapaired electrons. L[Co(NH3)6]3+ : 3d6 electron configuration, d2sp3, diamagnetic. M[Co(ox)3]3 : 3d6 electron configuration, d2sp3, diamagnetic. N[Ni(H2O)6]2+ : 3d8 electron configuration, sp3d2, with two unpared electrons ; paramagnetic. O[Pt(CN)4]2 : 5d8 electron configuration, dsp2 diamagnetic. P[Zn(H2O)6]2+ : 3d10 electron configuration, sp3d2 diamagnetic. Passing H2S gas into a mixture of Mn2+, Ni2+, Cu2+ and Hg2+ ions in an acidified aqueous solution precipitates: (A) CuS and HgS (B) MnS and CuS (C) MnS and NiS (D) NiS and HgS Ans. (A)

3.

Sol.

In presence of acidic medium, ionisation of H2S is supressed so less number of S2 ions are produced. So only those sulphides are precipitated which have low solubility product (KSP) value, For example CuS and HgS. Consider the following cell reaction : 2

iitjeesol_2011

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