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Paper - I Code-1Time: 3 Hours Maximum Marks: 240Please read the instructions carefully. You are alloted 5 minutes specifically for this purpose.

INSTRUCTIONSC. Questions Paper Format:

1. The question paper consists of 3 parts (Mathematics, Physics and Chemistry). Each part has

4 sections.

2. Section-I contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C)

and (D) for its answer, out of which only one is correct.

3. Section-II contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C)

and (D) for its answer, out of which one or more is are correct.

4. Section III contains 2 groups of questions. Each group has 3 questions based on a paragraph.

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

correct.

5. Section IV contains 2 questions. Each question has four statements (A, B, C and D) given in

Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I

can have correct matching with one or more statement(s) given in Column II. For example,

if for a given question, statement B matches with the statements given in q and r, then for that

particular question, against statement B, darken the bubbles corresponding to q and r in the

ORS.

D. Marking Scheme:

6. For each question in Section-I, you will be awarded 3 marks if you have darken the bubble

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

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

7. For each question in Section-II, you will be awarded 4 marks if you darken the bubble(s)

corresponding to the correct choice(s) for the answer, and zero mark if no bubble is darkened.

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

8. For each question in Section-III, you will be awarded 4 marks if you darken only the

bubble corresponding to the correct answer, and zero mark if no bubble is darkened. In all

other cases, minus one (–1) mark will be awarded.

9. For each question in Section-IV, you will be awarded 2 marks for each row in which you

have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this

section carries a maximum of 8 marks. There is no negative marking for incorrect answer(s)

for this section.


PART I : CHEMISTRY

SECTION- ISingle Correct Choice Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and(D) for its answer, out of which ONLY ONE is correct.

1. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The molefraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 Kand 5 atm pressure is:

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

Sol: Ans [A]KH = 1 × 105 atm

2NX 0.8

2

2 2

NH

N H O

nP K

n n

2

2

N5

N

n5 0.8 1 10

n 10

2

2

N 5

N

n4 10

n 10

2Nn 10

2

5 4Nn 10 4 10 4 10

2. The correct acidity order of the following isOH OH

Cl

COOH COOH

CH3

(I) (II) (III) (IV)

(A) (III) (IV) (II) (I) (B) (IV) (III) (I) (II)(C) (III) (II) (I) (IV) (D) (II) (III) (IV) (I)

Sol: Ans [A]

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

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

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

Sol: Ans [B]

4 2 4 6P 3O P O , N2 provides inert medium hence, there is a controlled reaction of oxygen with

phosphorus to give selectively P4O6.


Chemistry (Paper-I) Code-1 IIT-JEE-2009

4. Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecularforce of attraction is weakest is

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

Sol: Ans [D]

5. Given that the abundances of isotopes 54Fe, 56Fe and 57Fe are 5%, 90% and 5%, respectively, theatomic mass of Fe is

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

Sol: Ans [B]

Average atomic mass of Fe = 54 5 56 90 57 5

100

= 55.95.

6. The IUPAC name of the following compound isOH

BrCN

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

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

Sol: Ans [B]

7. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent forSb2S3 sol is

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

Sol: Ans [C]

According to hardy-Schultze rule, cation having high magnitude of charge has greater coagulatingpower.

8. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is

(A) nb (B)2

2

anV

(C)2

2anV

(D) –nb

Sol: Ans [B]

2

2

anP V b nRTV



SECTION- IIMultiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and(D) for its answer, out of which ONE OR MORE is/are correct.

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

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

Sol: Ans [A, B]

10. The correct statement(s) about the compound H3C/(HO)HC – CH = CH – CH(OH)CH3 (X) is(are)(A) The total number of stereoisomers possible for X is 6(B) The total number of diastereomers possible for X is 3(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for

X is 4(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for

X is 2.

Sol: Ans [C, D]

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

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

Sol: Ans [C, D]

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

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

(B) Frenkel defect is a dislocation defect(C) Trapping of an electron in the lattice leads to the formation of F-center

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

Sol: Ans [B, C]

SECTION- IIIComprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice questions based ona paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLYONE is correct.

Paragraph for Question Nos. 13 to 15A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followedby dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoesintramolecular aldol reaction to give predominantly S.

3

222 4,

1. O1. MeMgBr 1. OH2. Zn, H O 2.2. H ,H O

3.H SO

P Q R S



13. The structure of the carbonyl compound P is

(A)O Me

(B)O Me

Me

(C)O Et

(D)

Me

O

Sol: Ans [B]

O Me

Me

22 4

1. MeMgBr2. H ,H O3.H SO ,

Me

Me Me

3

2

1. O2. Zn, H O H

O

Me

COMe

Me

1. OH2.

O

Me Me

P Q R

S14. The structures of the products Q and R, respectively, are

(A)Me

MeMe

H

O

Me

COMe

Me

,

(B)

MeMe

H

O

Me

CHO

Me

,

(C)

MeEt

H

O

Me

CHO

Et

,

(D)

Me

Me

CH3

O

Me

CHO

Et

,



Sol: Ans [A]

15. The structure of the product S is

(A)

O

Me

(B) OMe Me

(C)

Me Me

O

(D)

O

Me

Sol: Ans [B]

Paragraph for Question Nos. 16 to 18p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution istreated with a few drops of aqueous solution of Y to yield blue coloration due to the formation ofmethylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition ofthe reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of Z.

16. The compound X is

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

Sol: Ans [D]

2 2Na S 2HCl 2NaCl H S

NCH3CH3

NH2

2H S

NCH3CH3

NH2

3FeCl

N

S+

NCH3

CH3 Cl

N

CH3

CH3

17. The compound Y is

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

Sol: Ans [C]

3 4 6 4 6 3intense blue

4FeCl 3K [Fe(CN) ] Fe [Fe(CN) ] 12KCl



18. The compound Z is

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

Sol: Ans [B]

3 3 6 6Brown

FeCl K [Fe(CN) ] Fe[Fe(CN) ] 3KCl

SECTION- IVMatrix - Match Type

This section contains 2 questions. Each question contains statements given in two columns, whichhave to be matched. The statements in Column I are labelled A, B, C and D, while the statementsin Column II are labelled p, q, r, s and t. Any given statement in Column I can have correctmatching with ONE OR MORE statement(s) in Column II. The appropriate bubbles correspond-ing to the answers to these questions have to be darkened as illustrated in the following example:If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correctdarkening of bubbles will look like the following.

p q r s t

p q r s t

p q r s t

p q r s t

p q r s t

A

B

C

D

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

Column I Column II

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

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

(C) CH3–CH=CH–CH2OH (r) Development of foul smell on treatment withchloroform and alcoholic KOH

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

(t) Alkaline hydrolysisSol: Ans [A – p, q, s, t]

[B – s, t][C – p][D – r]



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

Column I Column II

(A) B2 (p) Paramagnetic

(B) N2 (q) Undergoes oxidation

(C) O2– (r) Undergoes reduction

(D) O2 (s) Bond order 2

(t) Mixing of ‘s’ and ‘p’ orbitalsSol: Ans [A – p, q, r, t]

[B – q, r, s, t][C – p, q, r, t][D – p, q, r, s, t]

PART II : MATHEMATICSSECTION- I

Single Correct Choice TypeThis section contains 8 multiple choice questions. Each question has 4 choice (A), (B), (C)and (D), for its answer, out of which ONLY-ONE is correct

21. Let z = x + iy be a complex number where x and y are integers. Then the area of the rectangle whosevertices are the roots of the equation is 3 3 350z z z z

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

Sol: Ans [A]

3 3 2 2350 ( ) 350zz zz zz z z

Let z = x + iy

(x2 + y2)((x + iy)2 + (x – iy)2) = 350

(x2 + y2)(x2 – y2) = 175 = 5 × 5 × 7 …(i)

So, x2 + y2 can have only 1, 5, 7, 25, 35 and 175 as values.

x2 + y2 = 1 |x| = 1, |y| = 0 does not satisfy (i)

x2 + y2 = 5 |x| = 2, |y| = 1 does not satisfy (i)

x2 + y2 can never be 7.

x2 + y2 = 25 either|x| = 5, |y| = 0 does not satisfy (i)

or |x| = 4, |y| = 3 satisfies (i)

So, (4, 3), (4, –3), (–4, 3) and (–4, –3) are 4 vertices.

So, Area of rectangle = 8 × 6 = 48.



22. If , , anda b c d

are unit vectors such that . 1a b c d

and 1. ,2

a c

then

(A) , ,a b c

are non coplanar (B) , ,b c d

are non-coplanar

(C) ,b d

are non-parallel (D) ,a d

are parallel and ,b c

are parallel

Sol: Ans [C]

( × ) . ( × ) = 1a b c d

| × | | × | cos = 1a b c d

| × | = | | | | | |sin |sin a b a b

... (i)

| × | = | | | | | |sin |sin c d c d

... (ii)

|cos | 1 ... (iii)

Multiplying (i), (ii) and (iii),

| × | | × | |cos a b c d

But | × | | × | cos a b c d

| × | =1, | × | a b c d

and cos = 1

|sin | = 1, |sin | = 1 and = 0º

cos = 0, cos = 0 and = 0º

= 90º, = 90º and = 0º

,a b c d

and × || ×a b c d

( × ) × ( × ) 0a b c d

Assume × = and × =a b e c d f

× ( × ) = 0e c d

( . ) – ( . ) = 0e d c e c d

a b d c

= a b c d

c and d are non-collinear [ ] = 0a b d and [ ] = 0a b c

... (iv)

Similarly, ( ) ( )a b c d = 0



( ) 0a b f = 0

( . ) [ ]a f b b c d a = 0

[ ] [ ]a c d b b c d a

a and b are non-collinear [ ] 0a c d and [ ] 0b c d

... (v)

(iv) and (v) , , ,a b c d are coplanar

1. | || | cos cos γ 60º2

a c a c

Angle between a and c is 60º, b a , d

c and , , ,a b c d are coplanar ,b d

are non-parallel.

23. The line passing through the extremity A of the major axis and extremity B of the minor axis of theellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M. Then the area of the triangle with verticesat A, M and the origin O is

(A)3110

(B)2910

(C)2110

(D)2710

Sol: Ans [D]

Equation of ellipse is2 2

19 1x y

Equation of AB is1 0 10 3 0

yx

or 3y – 3 = –x

x = 3 – 3y ... (i)

Equation of auxiliary circle is

x2 + y2 = 9 ... (ii)

Solving (i) and (ii), we get

(3 – 3y)2 + y2 = 9

10 y2 – 18y = 0

2y(5y – 9) = 0

y = 0, 95

y

Then x = 3, 125

x

Co-ordinate of M is 12 9,5 5



Area of triangle OAM is

0 0 11 3 0 12

12 9 15 5

= 1 9 1 27 273 02 5 2 5 10

square unit.

24. Let z = cos + i sin . Then the value of 15

2 1

1Im( )m

mz

at = 2° is

(A)1

sin 2(B)

13sin 2 (C)

12 sin 2 (D)

14 sin 2

Sol: Ans [D]

z2m – 1 = (ei)2m – 1 = ei(2m – 1)

Im(z2m – 1) = sin(2m – 1)

15 152 1

1 1Im( ) sin(2 1)θm

m mz m

= sin + sin 3 + sin 5 + ... + sin 29

= 2sin (sin sin 3 sin 5 ... sin 29 )

2sin

= (1 cos2 ) (cos2 cos4 ) (cos4 cos6 ) ... (cos 28 cos30 )

2sin

= 1 cos30 1 cos60º 1

2sin 2sin 2º 4sin 2º

.

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

Then the value of µ for which the vector PQ

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

(A)14

(B)14

(C)18

(D)18

Sol: Ans [A]

Given line, ˆ ˆˆ ˆ ˆ ˆ2 3 5r i j k i j k

1 1 3 2

3 1 5x y

Co-ordinate of point Q be (1 – 3, – 1, 5 + 2)

Hence direction ratio of line PQ be (3 + 2, 3 – , 4 – 5)

Since line PQ

parallel to x – 4y + 3z = 1

(3 + 2) . 1 + (3 – )(–4) + (4 – 5)3 = 0



3 + 2 – 12 + 4 + 12 – 15 = 0

2 – 8 = 0

14

.

26. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only, is

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

Sol: Ans [C]

Let digits be x1, x2, x3, ..., x7

Then x1 + x2 + x3 + x4 + x5 + x6 + x7 = 10

1 xi 3

No. of ways = coefficient of x10 in (x + x2 + x3)7

= coefficient of x10 in x7(1 + x + x2)7

= coefficient of x3 in (1 + x + x2)7

= 2 1 1 5

2 0 3 47!( ) ( ) (1) 7! ( ) ( ) (1)1!1!5! 0!3!4!

x x x x = 42 + 35 = 77.

27. Let f be a non-negative function defined on the interval [0, 1].

If 2

0 0

1 ( ( )) ( ) , 0 1,x x

f t dt f t dt x and f (0) = 0, then

(A)1 1 1and

2 2 3 3f f

(B)1 1 1and

2 2 3 3f f

(C)1 1 1and

2 2 3 3f f

(D)1 1 1and

2 2 3 3f f

Sol: Ans [C]

Given f (x) 0 and x [0, 1]

and 2

0 0

1 ( ( )) ( ) , 0 1x x

f t dt f t dt x Differentiating both sides w.r.t. ‘x’ using leibnitz rule

21 ( ( )) ( )f x f x

2 21 [ ( )] ( )f x f x

Let f (x) = y ( ) dyf xdx



221 dy y

dx

221dy y

dx

21dy ydx

21.

1dy dx

y

sin–1 y = ± x + c

Given f (0) = 0

0 = + c c = 0

sin–1 y = ± x

y = ± sin x

f (x) = ± sin x

But f (x) 0, x [0, 1]

f (x) = + sin x

1 1 1sin2 2 2

f

( sin , [0,1])x x x

1 1 1sin3 3 3

f

Answer (C).

28. Tangents drawn from the point P(1, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0 touch the circle at thepoints A and B. The equation of the circumcircle of the triangle PAB is

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

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

Sol: Ans []

Let S = x2 + y2 – 6x – 4y – 11

S(1, 8) = 1 + 64 – 6 – 32 – 11 > 0

Point P(1, 8) is outside of circle circumcircle of DPABwill pass through the centre O of given circle, with OPas diameter.

(As OAPB is cyclic quadrilateral and PAO = 90°)

A

O(3, 2)

BP(1, 8)



Equation of circumcircle of PAB is

(x – 1)(x – 3) + (y – 8)(y – 2) = 0

x2 – 4x + 3 + y2 – 10y + 16 = 0

x2 + y2 – 4x – 10y + 19 = 0.

SECTION- IIMultiple Correct Choice Type

This section contains 4 multiple choice questions. Each question has 4 choice (A), (B), (C)and (D) for its answer, out of which ONE OR MORE is/are correct.

29. In a triangle ABC with fixed base BC, the vertex A moves such that 2cos cos 4sin .2AB C

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

(A) b + c = 4a

(B) b + c = 2a

(C) locus of point A is an ellipse

(D) locus of point A is a pair of straight lines

Sol: Ans [B, C]

cos B + cos C = 4 sin2A/2

22cos cos 4sin2 2 2

B C B C A

2sin cos 2sin2 2 2A B C A

sin cos 2sin 02 2 2A B C A

sin 02A A = 2n which is not possible.

So, cos 2sin 2cos2 2 2

B C A B C

cos cos sin sin 2 cos cos sin sin2 2 2 2 2 2 2 2B C B C B C B C

cos cos 3sin sin2 2 2 2B C B C

1tan tan

2 2 3B C



( )( ) ( )( ) 1

( ) ( ) 3s a s c s a s b

s s b s s c

13

s as

3s – 3a = s

2s = 3a a + b + c = 3a b + c = 2a

Locus of point A is an ellipse.

30. If 4 4sin cos 1 ,

2 3 5x x then

(A) 2 2tan3

x (B)8 8sin cos 1

8 27 125x x

(C) 2 1tan3

x (D)8 8sin cos 2

8 27 125x x

Sol: Ans [A, B]

Given 4 4sin cos 1

2 3 5x x

4 2 2sin (1 sin ) 1

2 3 5x x

25 sin4x – 20 sin2x + 4 = 0

(5 sin2x – 2)2 = 0

sin2x = 25

2 3cos5

x

... (i)

2 2tan3

x

Now from equation (i)

8 16sin625

x and 8 81cos625

x

8 8sin cos 1 16 1 81

8 27 8 625 27 625x x = 125.

31. Let

22 2

40

4lim , 0.x

xa a xL a

x

If L is finite, then

(A) a = 2 (B) a = 1 (C)164

L (D)1

32L



Sol: Ans [A, C]

Given,

22 2

40

4lim , 0x

xa a xL a

x

L is in 00

form, so apply L Hospital Rule

2 2

30

1 ( 2 )22lim

4x

xxa xL

x

2 2

30

2lim

4x

x xa xL

x

=

2 2 2 2

2 2 2 2 20

2 2lim8 2x

z x a xx a x a x

L =

2 2

0 2 2 2 2 2

4lim8 (2 )x

a x

x a x a x

For L to be finite a = 2 to form 00

form.

2

2 2 20

4 4lim8 4 (2 4 )x

xLx x x

= 2

2 2 20lim

8 4 (2 4 )x

xx x x

= 1 1

8 2 (2 2) 64

(A), (C)

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

(A) e – 1 (B)1

ln ( 1 )e

e y dy (C)1

0

xe e dx (D)1

lne

y dy

Sol: Ans [B, C, D]

Required area 1

e

x dy

1

lne

I y dy …(1)

y

y e =

xO

(0, 1)

(0, 0)

Also, 1

ln (1 )e

I e y dy …(2)

[Using property ( ) ( )b b

a a

f x dx f a b x dx From (1),

1 1

ln ln .1e e

I y dy y dy



11

1[ln . ]e

ey y y dyy

1

( 0) 1e

e dy

1[ ]ee y

= e – e + 1 = 1

And option (c) is,1

10

0

[ ]x xxe e d e e

= e – (e – 1) = 1

So ans (c) is also correct.

(B), (C), (D) are answers.

SECTION- IIIComprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice questionsbased on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONLY ONE is correct.

Paragraph for Question Nos. 33 to 35

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

33. The probability that X = 3 equals

(A)25

216(B)

2536

(C)536

(D)125216

Sol: Ans [A]

5 5 1 25( 3)6 6 6 216

P X .

34. The probability that X 3 equals

(A)125216

(B)2536

(C)536

(D)25

216

Sol: Ans [B]

P(X 3) = 1 – P(X < 3) = 1 – [P(X = 1) + P(X = 2)]

= 1 5 1 11 251 16 6 6 36 36

.



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

(A)125216

(B)25

216(C)

536

(D)2536

Sol: Ans [D]

6 ( 6 3) ( 6)3 ( 3) ( 3)

X P X X P XPx P X P X

= 1 ( 5)1 ( 3)

P XP X

=

2 4

2

1 5 1 5 5 11 . ...6 6 6 6 6 6

1 5 1 5 11 .6 6 6 6 6

=

5

2

3

1 516 6

1 51 5 2566 361 51

6 61 51

6

.


Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of theseentries are 1 and four of them are 0.

36. The number of matrices in A is

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

Sol: Ans [A]

Let the symmetric matrix be

x a ba y cb c z

Among a, b, c, number of 1s will always be even. (Since number of a’s, b’s and c’s is 2)

So number of 1s among x, y, z will be odd. (Since total number of 1s is 5)

So number of 1s among x, y, z is 1 or 3.



Case I : If all of x, y, z are 1s, then among a, b, c, we want 2 1s which will be either a or b or c i.e., 3matrices.

Case II : If only one among x, y, z is 1, then one among x, y, z can be chosen to be 1 in 3C1 ways.

Now among a, b, c, we want 4 1s which will be either a and b or b and c or c and a i.e., 3 ways.

Number of such matrices = 3 × 3 = 9.

So total number of matrices = 3 + 9 = 12.

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

100

xA y

z

has a unique solution is

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

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

Sol: Ans [B]

Let symmetric matrix be

a h gA h b f

g f c

For unique solution, |A| 0.

|A| = abc + 2fgh – af 2 – bg2 – ch2 0

Among f, g, h, number of zeroes will always be even. (Since no. of fs, gs, hs is 2)

So number of zeroes among a, b, c will also be even. (Since total 4 zeroes are there)

Case I : None of a, b, c is 0.

a = b = c = 1

Either f = g = 0 or g = h = 0 or f = h = 0

h = 1 f = 1 g = 1

|A| = 0 |A| = 0 |A| = 0

Case II : Number of zeroes among a, b, c is 2.

Either a = b = 0 or b = c = 0 or c = a = 0

c = 1 a = 1 b = 1

In all these 3 cases, one among f, g, h has to be zero.

fgh = 0



|A| = –ch2 0 |A| = –af 2 0 |A| = –bg2 0

h 0 f 0 g 0

h = 1 f = 1 g = 1

Either f = 0 or g = 0 Either g = 0 or h = 0 Either f = 0 or h = 0

So no. of matrices = 2 No. of matrices = 2 No. of matrices = 2

Total number of matrices = 6.

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

100

xA y

z

is inconsistent, is

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

Sol: Ans [B]

Total matrices = 12

Matrices with (|A| 0) = 6

So, matrices with (|A| = 0) = 12 – 6 = 6

For inconsistency, (adj. A)B 0

2 1(adj ) 0

0

bc fA B fg ch

fh bg

2

0bc ffg chfh bg

... (i)

We have obtained all 6 matrices with |A| = 0 in Q. NO. 37

Out of those 6, only 4 satisfy (i)

So, No. of matrices = 4.



SECTION- IVMatrix – Match Type


p q r s t

p q r s t

p q r s t

p q r s t

p q r s t

A

B

C

D

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

Column I Column II

(A) Circle (p) The locus of the point (h, k) for which the line hx + ky = 1touches the circle x2 + y2 = 4.

(B) Parabola (q) Points z in the complex plane satisfying |z + 2| – |z – 2| = ± 3.

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

(D) Hyperbola2

2 2

1 23 , .1 1

t tx yt t

(s) The eccentricity of the conic lies in the interval 1 x < .

(t) Points z in the complex plane satisfying Re(z + 1)2 = |z|2 + 1.

Sol: Ans [A-(p); B-(r, s, t); C-(s); D-(q, s)

(p) hx + ky = 1 touches the circle x2 + y2 = 4 then

hx ky + = 1

(0, 0)

2

perpendicular distance of (0, 0) from line hx + ky = 1 is

2 2

1 2h k



22 2 1

2h k

2 2 14

x y Locus is circle.

(q) |z + 2| – |z – 2| = ±3

(i) |z + 2| – |z – 2| = 3–2( )Q 2( )R

z P( )

PQ – PR = 3

Hyperbola.

(ii) |z + 2| – |z – 2| = –3

|z – 2| – |z + 2| = 3–2 2

R

P z( )

Q

PR – PQ = 3 Hyperbola.

(r)2

2 2

1 231 1

t tx yt t

2

2 2

1 2,1 13

x t tyt t

2 2 2 2 2

2 2 2 2

(1 ) 43 1 (1 ) (1 )x y t t

t t

2 2

13 1x y

Ellipse.

(s) 1 x <

If eccentricity of conic is in between 1 and means that can be

Conic section can be Parabola or Hyperbola.

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

z = (x + iy)

Re((x + 1) + iy)2 = x2 + y2 + 1

Re((x + 1)2 – y2 + 2i(x + y)y) = x2 + y2 + 1

(x + 1)2 – y2 = x2 + y2 + 1

x2 + 1 + 2x – y2 = x2 + y2 + 1

2x = 2y2

y2 = x Parabola.



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

Column I Column II

(A) Interval contained in the domain of definition of non-zero (p) ,2 2

solutions of the differential equation (x – 3)2 y + y = 0.

(B) Interval containing the value of the integral (q) 0,2

5

1

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

(C) Interval in which at least one of the points of local (r)5,

8 4

maximum of cos2 x + sin x lies.

(D) Interval in which tan–1(sin x + cos x) is increasing. (s) 0,8

(t) (–, )

Sol: Ans [A-(r, t); B-(p, t); C-(p, q, r, t); D-(s)

(A) (x – 3)2 y + y = 0

2 0( 3)

yyx

2 0( 3)

dy ydx x

2 0( 3)

dy dxy x

2( 3)dy dy cy x

ln y – (x – 3)–1 = c

1ln( 3)

y cx

Domain : R – {3}

Answer {r, t}.

(B)5

1

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

Let x – 3 = t



2

2

( ) ( 2)( 1) ( 1)( 2)f t t t t t t dt

f(t) is odd function

f(t) = 0

Answer {p, t}

(C) f (x) = cos2 x + sin x

( ) 2cos .sin cosdf x x x xdx

( ) sin 2 cos 0df x x xdx

sin 2x – cos x = 0

sin x = 1/2 cos x = 02 ( ) 2cos2 sind f x x xdx

22 2

sin 1/ 2

( ) 2(cos sin ) sinx

d f x x x xdx

3 1 124 4 2

= – 1 –1/2 = –ve (maxima)

sin x = 1/2 gives us maxima

5, , 2 , 3 , , ,

6 6 6 6 6 6x

(p, q, r, t).

(D) f (x) = tan–1 (sin x + cos x)

2

( ) 1 .(cos sin )1 (sin cos )

df x x xdx x x

( ) 0df xdx

r = –3 /4 r = /4

/2

–/2–

cos x – sin x > 0

cos x > sin x

3 ,4 4

x



PART III : PHYSICS

SECTION- I

Single Correct Choice Type

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

and (D) for its answer, out of which ONLY-ONE is correct

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

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

mass of the ink used to draw the outer circle is 6 m. The coordinates of the centres of the different

parts are: outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and

horizontal line (0, –a). The y-coordinate of the centre of mass of the ink in this drawing is

(A)10a

(B)8a

x

y

(C)12a

(D)3a

Sol: Ans [A]

ycm = 6 0 0 (– )

10m m a m a m m a

m

= 10a

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

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

field increases with time. I1 and I2 are the currents in the segments ab and cd. Then,

(A) I1 > I2

(B) I1 < I2

(C) I1 is in the direction ba and I2 is in the direction cd

× × × × × ××××××

×××××

×××××

×××××

×××××

×××××

a b

c d

(D) I1 is in the direction ab and I2 is in the direction dc

Sol: Ans [D]

Since B increases ; hence flux increases.

× × × × × ××××××

×××××

×××××

×××××

×××××

×××××

a b

c d

According to Lenz’s law direction of induced current

is anticlockwise.



43. Two small particles of equal masses start moving in opposite directionsfrom a point A in a horizontal circular orbit. Their tangential velocities arev and 2v, respectively, as shown in the figure. Between collisions, theparticles move with constant speeds. After making how many elasticcollisions, other than that at A, these two particles will again reach thepoint A?

(A) 4 (B) 3

AV 2V

(C) 2 (D) 1

Sol: Ans [C]

1st time the particles collide at B after moving by 240° with

2v. The speed gets exchanged. The 2nd time they collide at

C when the other particle moves with speed 2 V and reachs

to C, then again speed gets exchanged and they will meet

at A.

A

B C 21

v´´2v´´

2v

v´ 2v´´

v

O

S = 2344. A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre

at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis fromx = a/4 to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/4, 3a/4, 0),respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ± a/2. Theelectric flux through this cubical surface is

(A)o

C 2

(B)o

C2

y

x

(C)o

C

10(D)

o

C

12

Sol: Ans [A]

net = 0

enq =

0

6 8 72 4C C a C

a

y

x

3C6C

8C

–7C

a

= 0 0

(3 2 7) 2C C

45. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively.It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratioof the charges given to the shells, Q1 : Q2 : Q3, is

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



Sol: Ans [B]

Charge on the outer surfaces of shell will be Q1, Q1 + Q2 and Q1 + Q2 + Q3 respectively.

= 124

QR

= 1 2

24 (2 )Q Q

R

= 1 2 3

24 (3 )Q Q Q

R

1 2 31 21 4 9

Q Q QQ QQ

CB

A

R

2R3RQ1

Q2

Q3

Q1 : Q2 : Q3 = 1 : 3 : 5

46. The x–t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of theparticle at t = 4/3s is

(A) 22 scm32

3 (B)

22cm/ s

32

0

1

–1

4 8 12 t(s)x(

cm)

(C) 22

scm32

(D) 22 scm32

3

Sol: Ans [D]

x = 1 sin (t)

= 1 sin 28

t

x = sin 4t

When t = 43

s, we have x = sin 4 3

4 3 2

cm.

a = – 2x = 2 22 3 3

8 2 32

cm/s2

47. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index ofwater is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, whenthe ball is 12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2]

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

Sol: Ans [C]



2 1 0v u

413 0

v ( )x

12.8

m

20 mv

43

x = –v

43

dxdt

= vd

dt

v 43

d dxdt dt

Applying v2 = u2 + 2gh; have,

v = 0 2 10 7.2 = 12 m/s v 4 12

3ddt

= 16 m/s

48. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of frictionbetween them is 3. The inclination of this inclined plane from the horizontal plane is graduallyincreased from 0°. Then

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

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

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

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

Sol: Ans [B]For sliding, µ mg cos = mg sin tan = µ

N

C

Amg

B

10 cm

15 cm

= tan–1 ( 3 ) = 60°

For toppling at but not sliding mg must pass through A and c 0

f × 152

= N × 102

mg sin × 152

= mg cos × 102

tan = 2015

= 23

As angle of toppling is less than angle of sliding.

first it will topple



SECTION- IIMultiple correct choice Type

This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C)and (D), for its answer out of which ONE OR MORE is/are correct.

49. For the circuit shown in the figure

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

24V 6 k

2 k

R2

R1

1.5kRL

(B) the potential difference across RL is 18V

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

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

Sol: Ans [A, D]

Req = 2 + 1.5 66 1.5

= 2 + 9 215

= 165

k

I = 24V

16 / 5k = 7.5 mA

LRV = 65

× 103 × 7.5 mA = 9 volt

1RP = (7.5 × 10–3)2 × 2 × 103

2RP = 237.5 10 1.5

6 1.5

× 6 × 103

1

2

R

R

PP =

253

LRP =

237.5 10 67.5

× 1.5

= 36 × 10–6 × 1.5



if R1 and R2 are interchanged, then

6247667

LRV

= 3 volt

2(3) 9'

1.5 1.5LRP

' 1

9L

L

R

R

PP

50. CV and CP denote the molar specific heat capacities of a gas at constant volume and constant pressure,respectively. Then

(A) CP – CV is larger for a diatomic ideal gas than for a monoatomic ideal gas

(B) CP+CV is larger for a diatomic ideal gas than for a monoatomic ideal gas

(C) CP/CV is larger for a diatomic ideal gas than for a monoatomic ideal gas

(D) CP.CV is larger for a diatomic ideal gas than for a monoatomic ideal gas

Sol: Ans [B, D]

CP – CV = R

CP + CV = 2CV+ R

as f is larger for diatomic molecule.

21p

v

CC f

hence p

v

CC is smaller for diatomic

Cp. Cv = (R + Cv). Cv

= .2 2f fR R R

= 2

2

2 4f fR

as f is large for diatomic.

Cp. Cv is large for data



51. A student performed the experiment of determination of focal length of a concave mirror by u–vmethod using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. Themaximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by thestudent (in cm) are: (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come fromexperiment and is (are incorrectly recorded, is (are)

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

Sol: Ans [C, D]For set (42, 56)

(a) 1 1 1v u f

1 1 142 24v

1 1 124 42v

42 24 124 42 56

v = –56data is correct

(b) If lens is placed at origin, then at u = 48, v = 48as u= 2f data is correct.

(c) For set (66, 33)

1 1 1v 66 24

1 1 1v 66 24

1 24 66v 66 24

v –38so v – 33

Data is incorrect(d) For set (78, 39)

1 1 178 24v

1 1 178 24v

v = –64 21cm3

Data is incorrect



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

frame, one can surely say that

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

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

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

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

Sol: Ans [A]

If Fext = 0

Linear momentum of system remains constant

Kinetic energy of system can change due to workdone by internal force.

Angular momentum of system can change due to couple of forces which produce internal torque.

Potential energy of system can change since internal forces can change their relative configuration.

SECTION- III

Comprehension Type

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

based on a paragraph. Each question has 4 choices (A), (B), (C) and (D), for its answer, out

of which ONLY ONE is correct.


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

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

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

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

according to the de Broglie relation. The energy of the particle of mass m is related to its linear

momentum as m

pE2

2

. Thus, the energy of the particle can be denoted by a quantum number ‘n’

taking values 1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the

standing wave.

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

x = 0 to x = a. Take h = 6.6 ×10–34 Js and e = 1.6 ×10–19C.

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

(A) a–2 (B) a–3/2 (C) a–1 (D) a2



Sol: Ans [A]

for n loops

n 2

= a = 2an

– = x ax = 0

n loops

since p = h

p = 2han

; p = 2nh

a

En =

2

22

nham

; En = 2 2

28n hma

E a–2

54. If the mass of the particle is m = 1.0 ×10–30 kg and a = 6.6 nm, the energy of the particle in its groundstate is closest to

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

Sol: Ans [B]

For ground state n = 1

E = 2

28hma

= 34 2

30 9 2

(6.6 10 )8 1 10 (6.6 10 )

J

= 34 2

19 30 9 2

(6.6 10 )1.6 10 8 1 10 (6.6 10 )

eV = 8 meV

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

(A) n–3/2 (B) n–1 (C) n1/2 (D) n

Sol: Ans [D]

En = 2 2

28h nma



En = 12

m Vn2

2 2

2n 2

1 mV2 8

h nma

Vn2 =

2 2

2 24h nm a

Vn = 2hnma

Vn n


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

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

energynHeHH 32

21

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

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

nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to

take place. Usually, the temperatures in the reactor core are too high and no material wall can be used

to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the

particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt0 is

called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is

greater than 5 × 1014 s/cm3.

It may be helpful to use the following: Boltzmann constant k = 8.6 × 10–5 eV/K; eVm1044.14

92

o

e

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

(A) strong nuclear force acting between the deuterons

(B) Coulomb force acting between the deuterons

(C) Coulomb force acting between deuteron electron pairs

(D) the high temperature maintained inside the reactor core

Sol: Ans [D]

Due to high temperature, havy Hydrogen collides strongly and get ionized.



57. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towardseach other, each with kinetic energy 1.5 kT, when the separation between them is large enough toneglect Coulomb potential energy. Also neglect any interaction from other particles in the core. Theminimum temperature T required for them to reach a separation of 4 × 10–15 m is in the range.

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

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

Sol: Ans [A]

From conservation of energy

kEi + kEi + 0 = 0

1 ( )( )4

e er

+ 0

3kT = 2

0

14

er

T = 2

0

14 3

er k

= 9

15 5

1.44 104 10 3 8.6 10

T = 1.39 × 109 k

58. Results of calculations for four different designs of a fusion reactor using D-D reaction are givenbelow. Which of these is most promising based on Lawson criterion?

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

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

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

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

Sol: Ans [B]

n to 5 × 1014

For, A nt0 = 10 × 109 s/cm3

For B, nt0 = 72 × 1013 s/cm3

For C, nt0 = 4 × 1012 s/cm3

For D, nt0 = 4 × 1012 s/cm3



SECTION- IVMatrix - Match Type


p q r s t

p q r s t

p q r s t

p q r s t

p q r s t

A

B

C

D

59. Column II shows five systems in which two objects are labelled as X and Y. Also in each casea point P is shown. Col.umn I gives some statements about X and/or Y. Match thesestatements to the appropriate system (s) from Column II.

Column-I Column-IIA. The force exerted by X on Y has (p) Block Y of mass M left on a fixed inclined plane X,

a magnitude Mg slides on it with a constant velocity.

P

X

Y

B. The gravitational potential energy (q) Two ring magnets Y and Z, each of mass M, areof X is continuously increasing. kept in frictionless vertical plastic stand so that they

re pel each other. Y rests on the base X and Zhangs in air in equilibrium. P is the topmost point ofthe stand on the common axis of the two rings.The whole system is in a lift that is going up witha constant velocity

PZ

Y

X



C. Mechanical energy of the system (r) A pulley Y of mass m0 is field to a table through aX+ Y is continuously decreasing. clamp X. A block of mass M hangs from a string

that goes over the pulley and is fixed at point P ofthe table. The whole system is kept in a lift that isgoing down with a constant velocity.

X

PY

D. The torque of the weight of Y about (s) A sphere Y of mass M is put in a nonviscous liquid point P is zero X kept in a container at rest. The sphere is released

and it moves down in the liquid

X

Y

P(t) A sphere Y of mass M is falling with its terminal

velocity in a viscous liquid X kept in a container

X

Y

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

(p)

P

N = Mg cosP.E decreasesM.E = decreases 0

(q)

P

F > MgP.E increasingM.E. = increasing = 0

(r)X

P Ym0

mF

Mg

F = mg2P.E decreasingM.E = decreasing 0



(s)

X

Y M

P

F < MgP.E increasingM.E = constant 0

(t)X

Y F = MgP.E increasingM.E decreasing 0

60. Six point charges, each of the same magnitude q, are arranged in different manners as shown incolumn-II. In each case, a point M and a line PQ passing through M are shown. Let E be the electricfield and V be the electric potential at M (potential at infinity is zero) due to the given charge distributionwhen it is at rest. Now, the whole system is set into rotation with a constant angular velocity about theline PQ. Let b be the magnetic field at M and be the magnetic moment of the system in this condition.Assume each rotating charge to be equivalent to a steady current.

Column -I Column-IIA. E = 0 (p) Charges are at the corners of a regular hexagon. M is at

the centre of the hexagon. PQ is perpendicular to the planeof the hexagon.

B. V 0 (q) Charges are on a line perpendicular two PQ at equalintervals. M is the mid-point between the two innermostcharges.

C. B = 0 (r) Charges are placed on two coplanar insulating rings at equalintervals M is the common centre of the rings. PQ isperpendicular to the plane of the rings.



D. µ 0 (s) Charges are placed at the corners of a rest angle of sides aand 2a and at the mid points of the longer sides. M is at thecentre of the rectangle. PQ is parallel to the longer sides.

(t) Charges are placed on two coplanar, identical insulating ringsat equal intervals. M is the mid point between the centres ofthe rings. PQ is perpendicular to the line joining the centresand coplanar to the rings.

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

(p)

+

+

+

Ma

a

aV = 0, E = 0, B = 0, = 0

(q)M

+ + + V = 0, E 0, B = 0, = 0

(r)

+ +

+

M

P

Qi1

i2

r1

r2

V 0, E = 0. M 0, B 0



(s)

+

+

P QME = 0. B 0, V 0, 0

(t)

+

+ +

P

Q

M V = 0, E 0, = 0, B = 0

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