Solution Iit Jee

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IIT-JEE 2010 SOLUTION - 2 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

SOLUTIONS TO IIT–JEE 2010 (PAPER 1)

PART−1 : CHEMISTRY

SECTION−I Straight Objective Type

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

1. In the reaction HBr⎯⎯⎯→OCH3 the product are

(A) Br OCH3 and 2H

(B) Br and 3CH Br

(C) Br and 3CH OH

(D) OH and 3CH Br

1. (D)

HBrBr−−

⎯⎯⎯→O CH3 O CH3

H

: Br−

⎯⎯⎯→ OH 3CH Br+

2. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is

(A) k

T

(B) k

T

(C) k

T

(D)k

T

2. (A) From Arrhenius equation: Ea / RTK Ae−= , K will increase exponentially with temperature.

Hence (A).

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IIT-JEE 2010 SOLUTION - 3 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

3. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is

(A) Br2(g) (B) Cl2(g) (C) H2O(g) (D) CH4(g) 3. (B) The standard molar enthalpy of formation of an element is zero at 298K. Br2 although an

element, exists in the liquid state at 298K. Hence (B)

4. The ionization isomer of [Cr(H2O)4Cl(NO2)]Cl is (A) [Cr(H2O)4(O2N)]Cl2 (B) [Cr(H2O)4Cl2](NO2) (C) [Cr(H2O)4Cl(ONO)]Cl (D) [Cr(H2O)4Cl2(NO2)]H2O 4. (B) [Cr(H2O)4Cl(NO2)]Cl ⎯→ [Cr(H2O)4Cl(NO2)]+ + Cl– [Cr(H2O)4Cl2]NO2 ⎯→ [Cr(H2O)Cl2]+ + NO2

5. The correct structure of ethylenediaminetetraacetic acid (EDTA) is

(A) N CH CH NCH2

CH2CH2

CH2HOOC

HOOC

COOH

COOH

(B) N CH2CH2 NCOOH

COOHHOOC

HOOC

(C) N CH2 CH2 NCH2

CH2CH2

CH2HOOC

HOOC

COOH

COOH(D) N CH CH N

H

CH2H

CH2HOOC

COOHCH2HOOC

CH2COOH

5. (C) EDTA is

N CH2 CH2 NCH2

CH2CH2

CH2HOOC

HOOC

COOH

COOH Hence (C)

6. The bond energy (in kcal mol–1) of C – C single bond is approximately (A) 1 (B) 10 (C) 100 (D) 1000 6. (C) The bond energy of C – C single bond is ∼ 100 kcals mol–1. Hence (C). Other data are absurd.

7. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

(A) BrCH2CH2CH2CH2CH3 and CH3CH2C≡CH (B) BrCH2CH2CH3 and CH3CH2CH2C≡CH (C) BrCH2CH2CH2CH2CH3 and CH3C≡CH (D) BrCH2CH2CH2CH3 and CH3CH2C≡CH 7. (D)

CH3–CH2–C ≡C – H 3 2 2 22

3

–CH CH CH CH BrNaNH

3 2NH NaBrCH CH C C : Naδ+ δ−⊕

− − − −− −⎯⎯⎯→ − − ≡ ⎯⎯⎯⎯⎯⎯⎯⎯→

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IIT-JEE 2010 SOLUTION - 4 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

3 2 2 2 2 3

3 OctyneCH CH CH CH C C CH CH

−− − − − ≡ − −

Hence (D).

8. The correct statement about the following disaccharide is

O

CH2OH

HH

OHOH

H

H

OH

H

OCH2CH2O

OHOH2C

OH

H

H

OH

H

CH2OH

(a)(b)

(A) Ring (a) is pyranose with α-glycosidic link (B) Ring (a) is furanose with α-glycosidic link (C) Ring (b) is furanose with α-glycosidic link (D) Ring (b) is pyranose with β-glycosidic lilnk 8. (A) Ring (a) is pyranose with α-glycosidic link but ring (b) is furanose with β-glycosidic

linkage. Hence (A)

SECTION−II Multiple Correct Answer Type

This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.

9. In the Newman projection for 2,2-dimethylbuane

X

H HY

CH3CH3

X and Y can respectively be (A) H and H (B) H and C2H5 (C) C2H5 and H (D) CH3 and CH3 9. (B, D)

CH3 CH2 C

CH3

CH3

CH3

2,2 Dimethyl buane

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IIT-JEE 2010 SOLUTION - 5 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

(I) (II)

CH3

H HCH3

CH3CH3

H

H HC2H5

CH3CH3

I. Considering rotation about C1 – C2 bond one of the conformers will be as represented

by the Newman projection formula (I). Hence (B). Also II. Considering rotation about C2–C3 bond, one of the conformers can be represented by

the above Newman projection formula (II). Hence (D)

10. The reagent(s) used for softening the temporary hardness of water is (are) (A) Ca3(PO4)2 (B) Ca(OH)2 (C) Na2CO3 (D) NaOCl 10. (B, C) Temporary hardness is due to dissolved Ca(HCO3)2 Ca(HCO3)2 + Ca(OH)2 ⎯→ 2CaCO3↓ + 2H2O Ca(HCO3)2 + Na2CO3 ⎯→ CaCO3↓ + 2NaHCO3 Hence (B) and (C)

11. Among the following, the intensive property is (properties are) (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity 11. (A, B) By decreasing or increasing the amount of an electrolytic solution, neither molar

conductance nor EMF is going to change.

12. Aqueous solutions of HNO3, KOH, CH3COOH, and CH3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are)

(A) HNO3 and CH3COOH (B) KOH and CH3COONa (C) HNO3 and CH3COONa (D) CH3COOH and CH3COONa 12. (C, D)

13. In the reaction 2NaOH(aq)/ Br⎯⎯⎯⎯⎯→

OH

the intermediate(s) is(are)

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IIT-JEE 2010 SOLUTION - 6 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

(A)

Br

O

Br

(B)

O

Br Br

(C)

Br

O

(D)

O

Br

13. (A, C)

2

OHH O

−⎯⎯⎯→

OH O

Br Br−⎯⎯⎯→

O

H

Br ⎯⎯→

O

Br

Br Br−

O

H Br

Br

H+−←⎯⎯⎯

O

Br

Br

Br Br−←⎯⎯⎯

O

Br

Br

Br

HH+−←⎯⎯⎯

O

Br Br

Br

SECTION−III Linked Comprehension Type

This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct.

Paragraph for Questions 14 to 16 Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO4 5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.

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IIT-JEE 2010 SOLUTION - 7 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

14. Partial roasting of chalcopyrite produces (A) Cu2S and FeO (B) Cu2O and FeO (C) CuS and Fe2O3 (D) Cu2O and Fe2O3 14. (B)

15. Iron is removed from chalcopyrite as (A) FeO (B) FeS (C) Fe2O3 (D) FeSiO3 15. (D)

16. In self-reduction, the reducing species is (A) S (B) O2– (C) S2– (D) SO2 16. (C)

Paragraph for Questions 17 to 18 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is: M(s) | M+ (aq; 0.05 molar) ||M+ (aq; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV.

17. For the above cell (A) Ecell < 0; ΔG > 0 (B) Ecell > 0; ΔG < 0 (C) Ecell < 0; ΔG° > 0 (D) Ecell > 0; ΔG° < 0 17. (B) Ecell = –0.0591 log 0.05 = +ve as Ecell is +ve so ΔG is less than zero.

18. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be

(A) 35 mV (B) 70mV (C) 140mV (D) 700 mV 18. (C) As –0.0591 log 0.05 = 70 mV ∴ –0.0591 log 0.0025 = 140 mV

SECTION−IV Integer Type

This section contains TEN questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

19. The value of n in the molecular formula BenAl2Si6O is 19. 3 The value of n in the BenAl2Si6O18 is 3.

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IIT-JEE 2010 SOLUTION - 8 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

20. The total number of basic groups in the following form of lysine is

H3N CH2 CH2 CH2 CH2CH

NH2

CO

O

20. 2 In lysine – NH2 and –COO– are basic groups.

21. Based on VSEPR theory, the number of 90 degree F-Br-F angles in BrF5 is 21. 0 In BrF5 the shape is square pyramidal type but due to one lone pair no F-Br-F bond will be

of 90° angle, it is around 85°.

22. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is

KCN K2SO4 (NH4)2C2O4 NaCl Zn(NO3)2 FeCl3 K2CO3 NH4NO3 LiCN

22. 3 The aqueous solutions of KCN, K2CO3 and LiCN are alkaline in nature and hence will

trun red litmus blue..

23. Amongst the following, the total number of compounds soluble in aqueous NaOH is

NCH3CH3 COOH OCH2CH3

CH2OH

OH

NO2 OH

NCH3 CH3

CH2CH3

CH2CH3

COOH

23. 4 All carboxylic acids and phenols are more acidic than water. Therefore 4.

24. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is

24. 3

25. The number of neutrons emitted when 23592 U undergoes controlled nuclear fission to 142

54 Xe and 90

38Sr is

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IIT-JEE 2010 SOLUTION - 9 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

25. 4 235 1 142 90 1

92 0 54 38 0U n Xe Sr 4 n+ ⎯⎯→ + +

26. In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is

3

2

1. O 1. NaOH(aq.)2. Zn, H O 2. HeatY⎯⎯⎯⎯→ ⎯⎯⎯⎯⎯→

26. 1

OH−

⎯⎯⎯→

O

O

O

OH

2H OΔ

−⎯⎯⎯→

O

, unsaturated ketoneα β As this is [2.2.0] type bicycle compound not [2.2.1]type so dehydration at Bridge head is

possible, to form α, β-unsaturated ketone.

27. The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained:

[R] (molar) 1.0 0.75 0.40 0.10 t(min.) 0.0 0.05 0.12 0.18

The order of the reaction is 27. 0

This is zero order reaction as data satisfy zero order equation 0x C Ck or kt t

−= = .

0.25 0.60 0.90k 50.05 0.12 0.18

= = = =

28. The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is

28. 5 The total cyclic isomers possible for C4H6 is 5.

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IIT-JEE 2010 SOLUTION - 10 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

PART−2 : MATHEMATICS

SECTION – I Single Correct Choice Type

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

29. The number of 3 × 3 matrices A whose entries are either 0 or 1 and for which the system

⎥⎥⎥

⎢⎢⎢

⎡=

⎥⎥⎥

⎢⎢⎢

001

zyx

A has exactly two distinct solutions, is

(A) 0 (B) 29 − 1 (C) 168 (D) 2 29. (A) There are 3 possible cases only either 1 solution or infinite solutions or no solution Hence 2 distinct solutions are not possible.

30. The value of ( )∫ +

+→

x

0430x 4t

t1lntx1lim dt is

(A) 0 (B) 121

(C) 241 (D)

641

30. (B)

( )∫ +

+→

x

0430x 4t

t1lntx1lim ⎟

⎠⎞

⎜⎝⎛ form

00

Using L−Hospital’s rule

( )( )4xx3

x1lnlim 40x ++

→ ⎟

⎠⎞

⎜⎝⎛ form

00

= ( ) ( ){ }340x x4x34x3x11lim

+++→=

121

31. Let p and q be real numbers such that p ≠ 0, p3 ≠ q and p3 ≠ −q. If α and β are nonzero

complex numbers satisfying α + β = −p and α3 + β3 = q, then a quadratic equation having

βα and

αβ as its roots is

(A) (p3 + q)x2 − (p3 + 2q)x + (p3 + q) = 0 (B) (p3 + q)x2 − (p3 − 2q)x + (p3 + q) = 0 (C) (p3 − q)x2 − (5p3 − 2q)x + (p3 − q) = 0 (D) (p3 − q)x2 − (5p3 + 2q)x + (p3− q) = 0 31. (B) α + β = −p, α3 + β3 = q ⇒ (α + β)(α2 + β2 − αβ) = q

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IIT-JEE 2010 SOLUTION - 11 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

⇒ (α + β)2 − 3αβ = −pq

⇒ 3αβ = p2 + pq

⇒ αβ = p3

qp3 +

and α2 + β2 = p3

q2pp3

qppq 33 −

=+

+−

∴ Sum of the roots = qpq2p

3

322

+−

=αβ

β+α

Product of roots = 1 ∴ Equation is (p3 + q)x2 − (p3 − 2q)x + (p3 + q) = 0

32. Equation of the plane containing the straight line 4z

3y

2x

== and perpendicular to the

plane containing the straight lines 2z

4y

3x

== and 3z

2y

4x

== is

(A) x + 2y − 2z = 0 (B) 3x + 2y − 2z = 0 (C) x − 2y + z = 0 (D) 5x + 2y − 4z = 0 32. (C) Vector along the plane is

k10ji8324243kji

−−=

∴ Direction of normal vector to the plane is

nk26j52i264321018kji

r=+−=−− (say)

∴ Equation of plane is 0n.r =rr

⇒ x − 2y + z = 0. 33. If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c

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

expression A2sinacC2sin

ca

+ is

(A) 21 (B)

23

(C) 1 (D) 3 33. (D) A + B + C = 180° ⇒ B = 60o and A + C = 120o

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IIT-JEE 2010 SOLUTION - 12 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

Now, AcosAsin2AsinCsinCcosCsin2

CsinAsinA2sin

acC2sin

ca

×+×=+

= 2sin(A + C) = 3 34. Let f g and h be real−valued functions defined on the interval [0, 1] by

( ) ( ) 2222 xxxx exexg,eexf −− +=+= and ( ) 22 xx2 eexxh −+= . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then

(A) a = b and c ≠ b (B) a = c and a ≠ b (C) a ≠ b and c ≠ b (D) a = b = c 34. (D) ( ) ( ) 2222 xxxx exexg,eexf −− +=+= ; ( ) 22 xx2 eexxh −+= , x ∈ [0, 1] Here ( ) ( ) ( )xhxgxf ≥≥ ∀ x ∈ [0, 1]

and maximum(f(x)) = e1e + at x = 1

But at x = 1, f(x) = g(x) = h(x) ⇒ a = b = c 35. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2

and r3 are the numbers obtained on the die, then the probability that 321 rrr ω+ω+ω = 0 is

(A) 181 (B)

91

(C) 92 (D)

361

35. (C) Three such groups are possible as 3p, 3p + 1, 3p + 2 for 3p: possible no. on dice are (3, 6) for 3p + 1: possible no. on dice (1, 4) for 3p + 2: possible no. on dice (2, 5)

∴ Probability = !362

62

62

×××

= 92

36. Let P, Q, R and S be the points on the plane with position vectors j3i3,i4,ji2 +−− and

j2i3 +− respectively. The quadrilateral PQRS must be a (A) parallelogram, which is neither a rhombus nor a rectangle (B) square (C) rectangle but not a square (D) rhombus, but not a square 36. (A) j3iQR;ji6PQ +−=+=

j3iSP;ji6RS +−=−−=

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IIT-JEE 2010 SOLUTION - 13 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

RS||PQQ and SP||QR ⇒ PQRS is a parallelogram

But 0QR.PQ ≠

and 0SQ.PR ≠ ⇒ Neither rectangle nor rhombus.

SECTION – II Multiple Correct Answer Type

This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.

37. Let z1 and z2 be two distinct complex numbers and let z = (1− t)z1 + tz2 for some real

numbers t with 0 < t < 1. If Arg(w) denotes the principal argument of a nonzero complex number w, then

(A) |zz||zz||zz| 2121 −=−+− (B) ( ) ( )21 zzArgzzArg −=−

(C) 0zzzzzzzz

1212

11 =−−−−

(D) ( ) ( )121 zzArgzzArg −=−

37. (A, C, D)

As ( )( ) tt1

tzzt1z 21

+−+−

=

⇒ z lies on line segment joining z1 & z2

t : 1– t

z1 z2z

⇒ 2121 zzzzzz −=−+− ⇒ 1212

11

zzzzzzzz

−−−−

= 0 and Arg(z – z1) = Arg(z2 – z1)

38. Let A and B be two distinct points on the parabola y2 = 4x. If the axis of the parabola

touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be

(A) r1

− (B) r1

(C) r2 (D)

r2

38. (C, D)

Centre C ≡ ⎟⎟⎠

⎞⎜⎜⎝

⎛+

+21

22

21 tt,

2tt

⇒ t1 + t2 = r

Slope of AB = 21 tt

2+

= r2

Similarly if circle is below x-axis then

⇒ slope of AB = r2−

( )121 t2,t

( )222 t2,t

A

B

C

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IIT-JEE 2010 SOLUTION - 14 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

39. Let f be a real-valued function defined in the interval (0, ∞) by f(x) = dttsin1nxx

0∫ ++l .

Then which of the following statement(s) is (are) true? (A) f ″(x) exists for all x ∈ (0, ∞) (B) f ′(x) exists for all x ∈ (0, ∞) and f′ is continuous on (0, ∞), but not differentiable on

(0, ∞) (C) there exists α > 1 such that ( ) ( )xfxf <′ for all x ∈ (α, ∞)

(D) there exists β > 0 such that ( ) ( ) β≤′+ xfxf for all x ∈ (0, ∞) 39. (B, C)

f(x) = lnx + dttsin1x

0∫ +

f′(x) = xsin1x1

++ ⇒ f ′(x) > 0 ∀ x > 0

f″(x) = xsin12

xcosx1

2 ++−

f″(x) do not exist when sinx = –1 i.e., when x = 2nπ – 2π .

So (A) is false and (B) is true. Now 0tsin1 ≥+

⇒ 0dttsin1x

0≥+∫ ∀ x ∈ (0, ∞)

And ln x > 0 ∀ x ∈ (1, ∞) ⇒ f(x) > 0 ∀ x ∈ (1, ∞) For x ≥ e3 f(x) = ln x + dttsin1

x

0∫ + ≥ 3

f′(x) = 2x1xsin1

x1

+≤++ ∀ x > 0

Now for x ≥ e3

⇒ 0 < f′(x) ≤ 2x1

+ < 2e1

3 + < 3 ∀ x ∈ (e3, ∞) ⇒ ( ) ( )xfxf <′

Hence (C) is correct. Note: ( )xflim

x ∞→ → ∞

Then ( ) ( )xfxf ′+ is not bounded. ⇒ (D) is wrong.

40. The value(s) of ( )∫ +

−1

02

44

x1x1x dx is (are)

(A) π−722 (B)

1052

(C) 0 (D) 2

31571 π

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IIT-JEE 2010 SOLUTION - 15 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

40. (A)

( )∫ +

−1

0 2

44

x1x1x

= ( )dxx1

1x4x6x4xx1

0 2

2344

∫ ++−+−

= ( ) ( )( )dx1x

1xx4x41xx1

0 2

2222

∫ ++−++4

= ( )∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

−−+−+

1

0 224

1x114x41xx dx

= ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+−+1

0 2

44546 dx

1xx4x4x4xx

= ( ) dx1x

11x4x4x5x1

0 22546∫ ⎟⎟

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

++−−−+

= dx41x

4x4x5x4x1

0 22456∫ ⎟

⎠⎞

⎜⎝⎛ +

+−−+−

= 44

434

515

614

71

+⎟⎠⎞

⎜⎝⎛ π

−−×+×−

= 4341

32

71

+π−−+−

= ⎟⎠⎞

⎜⎝⎛ π−

722

So, (A) is the correct option.

41. Let ABC be a triangle such that ∠ACB = 6π and let a, b and c denote the lengths of the

sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2 – 1 and c = 2x + 1 is (are)

(A) – ( )32 + (B) 1 + 3 (C) 2 + 3 (D) 34 41. (B)

cos6π = ( )( )1x1xx

1x2x3x2x2ab2

cba22

234222

−+++−−+

=−+

⇒ ( )( )( )( )1x1xx

1x1x2x222

22

−++−−+ = 3

⇒ ( ) ( ) ( )31x32x32 2 +−−+− = 0 ⇒ x = – ( )32 + , 1 + 3 But as b + c > a ⇒ x > 1 ⇒ x = 1 + 3

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IIT-JEE 2010 SOLUTION - 16 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

SECTION – III Linked Comprehension Type

This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct.

Paragraph for Questions 42 to 44

Let p be an odd prime number and Tp be the following set of 2 × 2 matrices:

Tp = { }⎭⎬⎫

⎩⎨⎧

−∈⎥⎦

⎤⎢⎣

⎡= 1p,...,2,1,0c,b,a:

acba

A

42. The number of A in Tp such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is

(A) (p – 1)2 (B) 2(p – 1) (C) (p – 1)2 + 1 (D) 2p – 1 42. (D) If A is symmetric then det(A) = a2 – b2 = (a – b)(a + b)

As it is divisible by p ⇒ either (a – b) = λ1p or (a + b) = λ2p ⇒ a = b or a + b = p hence a = b has p number of solution and a + b = p has (p – 1) number of solution Total solution = 2p – 1. 43. The number of A in Tp such that the trace of A is not divisible by p but det (A) is divisible

by p is [Note: The trace of a matrix is the sum of its diagonal entries.] (A) (p –1)(p2 – p + 1) (B) p3 – (p – 1)2 (C) (p – 1)2 (D) (p – 1)(p2 – 2) 43. (C) For the ordered pair (b, c) we have total (p – 1)2 choices.

In 2(p 1)

2− choices there exists exactly two values of a for which (a2 – bc) is a multiple of

p, and for remaining 2(p 1)

2− choices there exists no value of a for which (a2 – bc) is a

multiple of p. Hence total no. of choices for the triplet (a, b, c) such that a ≠ 0 and a2 – bc is multiple of

p, is 2

2(p 1) 2 (p 1)2−

× = − .

44. The number of A in Tp such that det (A) is not divisible by p is (A) 2p2 (B) p3 – 5p (C) p3 – 3p (D) p3 – p2 44. (D) Total no. of matrices = p3 total no. of matrices, whose det (A) is divisible by p but trace not zero = (p – 1)2

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IIT-JEE 2010 SOLUTION - 17 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

Total no. of matrices having trace zero and whose det(A) is divisible by p = 2p – 1 As trace (A) = 0 ⇒ 2a = 0 ⇒ a = 0 now a2 – bc = λp ⇒ –bc = λp ⇒ bc = 0 ⇒ b = 0 or c = 0 Note for b = 0, c has p choices (namely 0, 1, 2, … p-1) Similarly for c = 0, b has the choices 1, 2, …., p − 1. Total choices for b & c = p + p – 1 = 2p – 1 Hence, total number of required matrices p3 – (p – 1)2 – (2p – 1) = p3 – p2

Paragraph for Questions 45 to 46

The circle x2 + y2 – 8x = 0 and hyperbola 14y

9x 22

=− intersect at the points A and B.

45. Equation of a common tangent with positive slope to the circle as well as to the hyperbola

is (A) 2x – 5 y – 20 = 0 (B) 2x – 5 y + 4 = 0 (C) 3x – 4y + 8 = 0 (D) 4x – 3y + 4 = 0 45. (B) Let the tangent be ( ) 4sinycos4x =θ+θ− ⇒ ( ) θ+−θ−= eccos44xcoty ⇒ ( ) ( )θ+θ+θ−= eccoscot4xcoty Now, c2 = a2m2 − b2 ⇒ 16(cotθ + cosecθ)2 = 9cot2θ − 4 ⇒ 16(cosθ + 1)2 = 9cos2θ − 4sin2θ= 13cos2θ − 4

⇒ 3cos2θ + 32cosθ + 20 = 0 ⇒ cosθ = −32

So, slope = 5

2cot =θ−

⇒ cosecθ = 5

3

So, equation of tangent is

( )5

124x5

2y +−=

⇒ 2x − 5 y + 4 = 0 46. Equation of the circle with AB as its diameter is (A) x2 + y2 – 12x + 24 = 0 (B) x2 + y2 + 12x + 24 = 0 (C) x2 + y2 + 24x – 12 = 0 (D) x2 + y2 – 24x – 12 = 0 46. (A)

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IIT-JEE 2010 SOLUTION - 18 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

Solving x2 + y2 − 8x = 0 and 14y

9x 22

=−

⇒ 0x84x94x 22 =−−+

⇒ 04x8x9

13 2 =−−

⇒ 6

913

3144

913

952164

x =+

=⎟⎠⎞

⎜⎝⎛

++=

So, y = 32xx8 2 ±=−± So, A and B are (6, 32± )

O 3 4 6

A

B

Centre of circle is the mid point of AB, which is (6, 0) and radius = 32 , so its equation is

(x − 6)2 + y2 = 12 ⇒ x2 + y2 − 12x + 24 = 0

SECTION IV Integer Type

This section contains TEN questions. The answer to each question is single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

47. Let ω be the complex number 3

2sini3

2cos π+

π . Then the number of distinct complex

numbers z satisfying ω+ω

ω+ωωω+

z11z

1z

2

2

2

= 0 is equal to

47. 1

3

2sini3

2cos π+

π=ω , so ω3 = 1 and 1 + ω + ω2 = 0.

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IIT-JEE 2010 SOLUTION - 19 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

ω+ω

ω+ωωω+

z11z

1z

2

2

2

= 0

⇒ (z + 1) (z2 − z) − ω(zω) + ω2(−zω2) = 0 ⇒ z3 −z − zω2 − zω = 0 ⇒ z3 = 0 ⇒ z = 0 So number of distinct solution is 1.

48. The number of values of θin the interval ⎟⎠⎞

⎜⎝⎛ ππ−

2,

2such that θ ≠

5nπ for n = 0, ±1, ±2 and

tanθ = cot5θ as well as sin2θ = cos4θ is 48. 3

tanθ = cot5θ, θ ≠ 5

cosθcos5θ − sin5θsinθ = 0 cos6θ = 0

⇒ 6θ = −2

5,2

3,2

,2

,2

3,2

5 ππππ−

π−

π

⇒ 125,

412,

12,

4,

125 ππππ

−π

−π

−=θ

Again sin2θ = cos4θ = 1 − 2sin22θ ⇒ 2sin22θ + sin2θ − 1 = 0

⇒ sin2θ = −1, 21

⇒ 6

5,6

,2

2 πππ−=θ

⇒ 125,

12,

4πππ

−=θ

So, common solution are θ = 12

,4

ππ− and

125π

So number of solution is 3. 49. For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a

real valued function defined on the interval [–10, 10] by

f(x) = x [x] if[x]is odd,

1 [x] x if[x]is even−⎧

⎨ + −⎩

Then the value of ( ) dxxcosxf10

10

10

2

ππ

∫−is

49. 4

( ) { } [ ]{ } [ ]⎩

⎨⎧

=−=

=evenxifx1oddxifx

xf

Clearly f(x) is an even function and periodic with period 2.

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IIT-JEE 2010 SOLUTION - 20 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

Also cosπx is periodic with period 2 So, f(x).cosπx is periodic with period 2

So, ( ) dxxcosxf10

10

10

2

ππ

∫−

= ( )1

2

1

f x cos xdx−

π π∫

= ( ) xdxcosxf21

0

2 π×π ∫ (As f(x) is even also)

= ( ) dxxcosx121

0

2 π−π ∫

= ⎟⎟⎠

⎞⎜⎜⎝

⎛π−×π ∫

1

0

2 xdxcosx2 (Replace x by (1 − x))

= ⎟⎠⎞

⎜⎝⎛

π−π−=

⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ π

π+⎟

⎠⎞

⎜⎝⎛ π

ππ− 2

21

02

1

0

2 22xcos1xsinx2 = 4

50. If the distance between the plane Ax – 2y + z = d and the plane containing the lines

43z

32y

21x −

=−

=− and

54z

43y

32x −

=−

=− is 6 , then d is

50. 6

( ) ( ) ( ))98k1210j1615i543432kji

−+−−−=

= kj2i −+− (normal vector) Equation of plane is −1(x − 1) + 2(y − 2) − 1(z − 3) = 0 −x + 1 + 2y − 4 − z + 3 = 0 −x + 2y − z = 0 x − 2y + z = 0 and Ax − 2y + z − d = 0

∴ distance = 6141

|d|=

++

|d| = 6.

51. The line 2x + y = 1 is tangent to the hyperbola 2

2

2

2

by

ax

− = 1. If this line passes through the

point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is

51. 2

y = 0, x = 21 ,

21

ea

= , 2ea =

1 = 4a2 − (a2(e2 − 1)) = 4a2 − a2e2 + a2 1 = 5a2 − a2e2

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IIT-JEE 2010 SOLUTION - 21 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

4 = 5e2 − e4 ⇒ e4 − 5e2 + 4 = 0 e2 = 1, e2 = 4 but e2 ≠ 1 ∴ e = 2. 52. Let Sk, k = 1, 2, …, 100, denote the sum of the infinite geometric series whose first term is

!k1k − and the common ratio is

k1 . Then the value of ( )∑

=

+−+100

1kk

22

S1k3k!100

100 is

52. 4

( )100

2k

k 1k 3k 1 S

=

− +∑ (where ( )k

k 1 1 1S / 1k! k k 1 !− ⎛ ⎞= − =⎜ ⎟ −⎝ ⎠

=( )

( )( )( )

( )

2100 100

k 1 k 3

k 3k 1 k 1 k 2 11 1

k 1 ! k 1 != =

− + − − −= + +

− −∑ ∑ (for k ≥ 3, k2 − 3k + 1 > 0)

= ( ) ( )

100

k 3

1 12k 3 ! k 1 !=

⎛ ⎞+ −⎜ ⎟⎜ ⎟− −⎝ ⎠∑

= !99

1!98

14 −−

∴ !99

1!98

14!100

1002

−−+

= !98

14!99

1!99

100−+−

= !98

1!98

14 −+ = 4.

53. Let f be a real-valued differentiable function on R (the set of all real numbers) such that

f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to

53. 9

The equation of tangent : Y − y = ( )dy X xdx

y intercept = 3xdxdyxy =−

2xxy

dxdy

−=−

I.F. = x1e

dxx1

=∫−

c2x

xy 2

+−=

f(1) = 1, c = 23

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IIT-JEE 2010 SOLUTION - 22 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

23

2x

xy 2

+−= ⇒ 3x 3xy

2 2= − +

f(−3) = 9.

54. If ar and br

are vectors in space given by 5

j2ia −=

r and 14

k3ji2b ++=

r, then the value of

( ) ( ) ( )[ ]b2ababa2rrrrrr

−××⋅+ is 54. 5 ( ) ( ) ( )( )b2aba.ba2

rrrrrr−××+

1a.a =rr 1b.b =

rr 0b.a =

rr ( ) ( ) ( ) ( ) bba2abab2aba

rrrrrrrrrr××−××=−××

= ( ) ( )baabab2rrrrrr

××−×× = ( ) ( )( )bb.aab.b2

rrrrrr− − ( ) ( )ba.aab.a

rrrrrr+

= ( ) ba2rr

+ ( ) ( ) ( )( )b2aba.ba2

rrrrrr−××+

= ( )( )ba2.ba2rrrr

++ = b.ba.b2b.a2a.a4

rrrrrrrr+++

= 4 + 1 = 5 55. The number of all possible values of θ, where 0 < θ < π, for which the system of equations

(y + z) cos3θ = (xyz) sin3θ

x sin3θ = z

3sin2y

3cos2 θ+

θ

(xyz) sin3θ = (y + 2z) cos3θ + y sin3θ have a solution (x0, y0, z0) with y0 z0 ≠ 0, is 55. 3 xyz sin3θ = (y + z)cos3θ (1) xyz sin3θ = 2zcos3θ + 2ysin3θ (2) xyz sin3θ = (y + 2z)cos3θ + ysin3θ (3) ⇒ (cos3θ − 2sin3θ)y − (cos3θ)z = 0 ((1) − (2)) and (sin3θ)y + (cos3θ)z = 0 ((3) − (1)) for y ≠ 0, z ≠ 0

⇒ θθ

−=θ

θ−θ3cos3cos

3sin3sin23cos

cos3θ − 2sin3θ = −sin3θ tan3θ = 1

3θ = nπ + 4π , n ∈ I

θ = 123

n π+

π , n ∈ I

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IIT-JEE 2010 SOLUTION - 23 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

∴ θ = 129,

125,

12πππ

∴ number of values of θ = 3.

56. The maximum value of the expression θ+θθ+θ 22 cos5cossin3sin

1 is

56. 2 ( ) θ+θθ+θ=θ 22 cos5cossin3sinf

= 1 + 4cos2θ + 23 sin2θ

= 3 + 2cos2θ + 23 sin2θ

f(θ)min = 3 − 25 = 1

2

∴ maximum value of ( )θf1 = 2.

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IIT-JEE 2010 SOLUTION - 24 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

PART−3 : PHYSICS

SECTION - I

This Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

57. An AC voltage source of variable angular frequency ω and fixed amplitude V0 is

connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased

(A) the bulb glows dimmer (B) the bulb glows brighter (C) total impedance of the circuit is unchanged (D) total impedance of the circuit increases 57. (B)

orms 2

2

VI12 RC

=⎛ ⎞+ ⎜ ⎟ω⎝ ⎠

58. A thin flexible wire of length L is connected to two adjacent fixed points and carries a

current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× ×××××××××××××

(A) IBL (B) π

IBL

(C) π2

IBL (D) π4

IBL

58. (C) 2Tsinθ = Fm

For small θ 2Tθ = B × I × Δ l = B × I × 2θR

T = BIR = BIL2π

θ θθ

θ

Tsinθ Tsinθ

Fm

T cosθ T cosθ

T T

Δ l

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IIT-JEE 2010 SOLUTION - 25 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

59. A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is μ and tanθ > μ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sinθ – μ cosθ) to P2 = mg(sinθ + μcosθ), the frictional force f versus P graph will look like

θ

P

(A)

f

P1 P2

P (B)

f

P1 P2 P

(C)

f

P1 P2

P (D)

f

P1 P2 P

59. (A) P f mgsin± = θ = constant 60. A real gas behaves like an ideal gas if its (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high 60. (D) 61. Consider a thin square sheet of side L

and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is

L

t

(A) directly proportional to L (B) directly proportional to t (C) independent of L (D) independent of t

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IIT-JEE 2010 SOLUTION - 26 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

61. (C)

LRL t tρ× ρ

= =×

62. A thin uniform annular disc (see figure)

of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is

4R

3R

4R

P

(A) ( )524

R7GM2

− (B) ( )524R7

GM2−−

(C) R4

GM (D) ( )12R5

GM2−

62. (A)

4R

p 2 23R

rdrV 2 G16R r

= − π σ+

2

2GM 4 2 5 R7R

− ⎡ ⎤= −⎣ ⎦

ext P2GMW (0 V ) 4 2 57R

⎡ ⎤= − = −⎣ ⎦

63. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistance R100, R60 and R40, respectively, the relation between these resistance is

(A) 6040100 R

1R1

R1

+= (B) R100 = R40 + R60

(C) R100 > R60 > R40 (D) 4060100 R

1R1

R1

>>

63. (D)

2VR

P=

100 60 40R R R< <

100 60 40

1 1 1R R R

> >

64. To verify Ohm’s law, a student is provided with a test resistor RT, a high resistance R1, a

small resistance R2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is

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IIT-JEE 2010 SOLUTION - 27 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

(A)

G1

G2 R1

R2

RT

V

(B)

G1

G2 R2

R1

RT

V

(C)

G1

R2

R1

RT

V

G2 (D)

G1

R1

R2

RT

V

G2

64. (C) When high resistance R1 is connected in series with G1, it acts as a voltmeter.

SECTION – II Multiple Correct Answer Type

This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.

65. A point mass of 1 kg collides elastically with a stationary point mass of 5kg. After their

collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms–1. Which of the following statement(s) is (are) correct for the system of these two masses?

(A) total momentum of the system is 3kg ms–1 (B) momentum of 5kg mass after collision is 4 kg ms–1 (C) kinetic energy of the centre of mass is 0.75 J (D) total kinetic energy of the system is 4 J 65. (A, C) u = 5 v – 2 (1) v + 2 = u (2) Solving equations (1) and (2), we get v = 1m/s, u = 3 ms−1

vcm = 1 3 16 2×

=

cm1 1 3KE 6 J2 4 4

= × × =

1 9TE 1 9 3.5J2 2

= × × = =

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IIT-JEE 2010 SOLUTION - 28 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

66. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that

(A) 21 QQ > (B) 21 QQ < (C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero. 66. (A, D) 67. A ray OP of monochromatic light is incident on the face

AB of prism ABCD near vertex B at an incident angle of 60° (see figure). If the refractive index of the material of the prism is 3 , which of the following is (are) correct?

(A) the ray gets totally internally reflected at face CD (B) the ray comes out through face AD (C) the angle between the incident ray and the emergent

ray is 90° (D) the angle between the incident ray and the emergent

ray is 120°

90o 75o

135o

60o

A D

C

B

67. (A, B, C)

sin 60º 3sin r

=

1sin r ; r 30º2

= ⇒

∴ Ray PQ is || to BC TIR will take place at the face CD since i > θc

where i = 45o and θc = sin−1 13

sinθ < 1μ

1 o1sin 453

− ⎛ ⎞θ < <⎜ ⎟

⎝ ⎠

90o 75o

135o

60o

A D

C

B

60o

30o

45o

45o

60o

30o

68. One mole of an ideal gas in initial state A undergoes a

cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the following

(A) Internal energies at A and B are the same (B) Work done by the gas in process AB is P0V0 nl 4

(C) Pressure at C is 4P0

(D) Temperature at C is 4

T0

4V0

V0

T0

A

B

C

T

V

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IIT-JEE 2010 SOLUTION - 29 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

68. (A, B) AB→ Isothermal and AC→ Isobaric ∴ UA = UB

WAB = PV ln 2

1

VV

⎛ ⎞⎜ ⎟⎝ ⎠

= PoVoln4

4V0

V0

T0

A

B

C

T

V

69. A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec. for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is are true?

(A) Error ΔT in measuring T, the time period, is 0.05 seconds (B) Error ΔT in measuring T, the time period, is 1 second (C) Percentage error in the determination of g is 5% (D) Percentage error in the determination of g is 2.5% 69. (A, C)

Error in time period, ΔT = least count 1 0.05sn 20

= =

T = tn

= 2πgl

2 2

2

4 ngt

π=

l

g 2 t 2 1100 100 100 5g t 40

Δ Δ − ×× = − × = × = −

SECTION−III Linked Comprehension Type

This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct.

Paragraph for Questions 70 to 72 When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The

corresponding time period is proportional to mk

, as can be seen

easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential

0V

0Xx

V(x)

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IIT-JEE 2010 SOLUTION - 30 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

energy is V(x) = αx4 (α > 0) for |x| near the origin and becomes a constant equal to V0 for |x| ≥ X0 (see figure). 70. If the total energy of the particle is E, it will perform periodic motion only if (A) E < 0 (B) E > 0 (C) V0 > E > 0 (D) E > V0 70. (C)

71. For periodic motion of small amplitude A, the time period T of this particle is proportional to

(A) mAα

(B) 1 mA α

(C) Amα (D) 1

A mα

71. (B) For small amplitude (mω2A2) ∝ (α A4)

T2 ∝ 2

mA

⎛ ⎞⎜ ⎟α⎝ ⎠

T∝1 mA α

72. The acceleration of this particle for |x| > X0 is

(A) proportional to V0 (B) proportional to 0

0

VmX

(C) proportional to 0

0

VmX

(D) zero

72. (D) For, |x| ≥ xo V(x) = constant ∴ F = 0, acceleration of particle = zero

Paragraph for Questions 73 to 74 Electrical resistance of certain materials, known as superconductors, changes abruptly from a non zero value to zero as their temperature is lowered below a critical temperature TC(0). An interesting property of superconductors is that their critical temperature becomes smaller than TC(0) if they are placed in a magnetic field, i.e., the critical temperature TC(B) is a function of the magnetic field strength B. The dependence of TC(B) on B is shown in the figure.

CT (0)

CT (B)

O B

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IIT-JEE 2010 SOLUTION - 31 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

73. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B2 (dashed line). If B2 is larger than B1, which of the following graphs shows the correct variation of R with T in these fields?

(A)

B2B1

TO

R

(B)

B2

B1

TO

R

(C)

B2B1TO

R

(D)B2

B1

TO

R

73. (A) 2 1B B>Q ∴ (Tc)2 < ( Tc)1

74. A superconductor has TC(0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its TC decreases to 75K. For this material one can definitely say that when

(A) B = 5 Tesla, TC(B) = 80 K (B) B = 5 Tesla, 75 K < TC(B) < 100 K (C) B = 10 Tesla, 75 K < TC (B) < 100 K (D) B = 10 Tesla, TC(B) = 70K 74. (B) As shown in graph for B = 5 75 K < Tc (B) < 100 K

T75

7.5 5 O B

Tc (B)

Tc (0)

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IIT-JEE 2010 SOLUTION - 32 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

SECTION−IV Integer Type

This section contains TEN questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

75. A binary star consists of two stars A (mass 2.2MS) and B (mass 11MS), where MS is the

mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is

75. 6 Total angular momentum of binary star, Ltotal = Itotalω Itotal = MA (5d/6)2 + MB(d/6)2 angular momentum of star B LB = IBω

CM

5d/6 d/6

MA MB

IB = MB (d/6)2

( )( )2 2

s stotal2

B s

2.2M 5d / 6 11M (d / 6)L 6L 11M (d / 6)

+ ω= =

ω

76. The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance

of 25 cm in front of it to50 cm, the magnification of its image changes from m25 to m50.

The ratio 25

50

mm

is

76. 6

m = ff u

⎛ ⎞⎜ ⎟+⎝ ⎠

25 5020 20m and m

20 25 20 50= =

− −

25

50

m 6m

=

77. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m

and its cross-sectional area is 4.9 × 10–7m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s–1. If the Young’s modulus of the material of the wire is n × 109 Nm–2, the value of n is

77. 4 By Hooke's law

F xYA L

= −

YAF xL

⎛ ⎞= −⎜ ⎟⎝ ⎠

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IIT-JEE 2010 SOLUTION - 33 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

YALm

ω =

( ) ( ) 22

7

0.1 1 (140)mLYA 4.9 10−

× ×ω= =

×

9 9Y 4 10 n 10= × = × n = 4

78. When two progressive waves y1 = 4 sin (2x – 6t) and y2 = 3 sin 2x 6t2π⎛ ⎞− −⎜ ⎟

⎝ ⎠ are

superimposed, the amplitude of the resultant wave is 78. 5 2 2 2

1 2 1 2A A A 2A A cos= + + φ given that A1 = 4, A2 = 3 and φ = - π/2 A = 5 79. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1and T2

respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?

79. 9 λmT = const

1 2

2 1

T 3T

λ= =

λ

1

44 21 41 1 1

42 2 2 2 2

A TE r T 6 (3)E A T r T 18

2σ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = × = ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟σ ⎝ ⎠⎝ ⎠ ⎝ ⎠=9

80. Gravitational acceleration on the surface of a planet is 6 g11

, where g is the gravitational

acceleration on the surface of the earth. The average mass density of the planet is 23

times

that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1 will be

80. 3

3

2 2

4G RGM 43g GRR R 3

×ρ× π= = = πρ

p

p p p

e e e e

2 Rg R 63g R R 11

×ρ ×= = =

ρ ×

⇒ Rp = e3 6 R22

p p p

e e e

v 2g R 6 3 6 3v 2g R 11 22 11

= = × =

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IIT-JEE 2010 SOLUTION - 34 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

p e3v v

11=

vp = 3 km/sec 81. A stationary source is emitting sound at a fixed frequency f0, which is reflected by two

cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms–1.

81. 7 Frequency received by car is

c1 o

v vf fv+⎛ ⎞= ⎜ ⎟

⎝ ⎠

Reflected frequency received by observer

2

c2 o

v vf fv+⎛ ⎞= ⎜ ⎟

⎝ ⎠=

2c

ovf 1v

⎛ ⎞+⎜ ⎟⎝ ⎠

(since vc << v)

c2 o

vf f 1 2v

⎛ ⎞= +⎜ ⎟⎝ ⎠

o c2

2f ( v )fvΔ

Δ =

C1.2 330v2 100

×Δ =

×ms-1 = 7 km/hr

82. When two identical batteries of internal resistance 1Ω each are connected in series across a

resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J2 then the value of R in Ω is

82. 4

In series I1 = 2R 2r

ε+

and J1 = 21I R

and in parallel I2 = R r / 2

ε+

and J2 = 22I R

given J1 = 2.25 J2

( )

2 2

22

4 R 2.25(R 2r) R r / 2

ε ε= ×

+ +

R r / 2 1.5R 2r 2

+⎛ ⎞ =⎜ ⎟+⎝ ⎠

⇒ R = 4Ω 83. A piece of ice (heat capacity = 2100 J kg–1J°C–1 and latent heat = 3.36 × 105 J kg–1) of

mass m grams is at –5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is

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IIT-JEE 2010 SOLUTION - 35 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

83. 8 mC(0ºC - (-5ºC)) + (1× 10-3kg) (3.36×105 Jkg-1 ) = 420 J

m × (2100 Jkg-1ºC-1) = 420 3365−

⇒ m = 8 gm 84. An α-particle and a proton are accelerated from rest by a potential difference of 100V.

After this, their de Broglie wavelengths are λα and λp respectively. The ratio p

a

λ

λ, to the

nearest integer, is 84. 3 qα = 2qp = 2e mα = 4 mP

KE = 2p qV

2m= p 2mqV=

λ = hp

h

2mqVλ =

ph

2mqVλ =

ph

2 4m 2e V 8α

λλ = =

× × ×

p 8α

λ=

λ

nearest integer = 3

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IIT-JEE 2010 SOLUTION - 36 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

IIT-JEE 2010 PAPER – I (KEY) CHEMISTRY MATHEMATICS PHYSICS

1. (D) 2. (A) 3. (B) 4. (B) 5. (C) 6. (C) 7. (D) 8. (A) 9. (B, D) 10. (B, C) 11. (A, B) 12. (C, D) 13. (A, C) 14. (B) 15. (D) 16. (C) 17. (B) 18. (C) 19. 3 20. 2 21. 0 22. 3 23. 4 24. 3 25. 4 26. 1 27. 0 28. 5

29. (A) 30. (B) 31. (B) 32. (C) 33. (D) 34. (D) 35. (C) 36. (A) 37. (A, C, D) 38. (C, D) 39. (B, C) 40. (A) 41. (B) 42. (D) 43. (C) 44. (D) 45. (B) 46. (A) 47. 1 48. 3 49. 4 50. 6 51. 2 52. 4 53. 9 54. 5 55. 3 56. 2

57. (B) 58. (C) 59. (A) 60. (D) 61. (C) 62. (A) 63. (D) 64. (C) 65. (A, C) 66. (A, D) 67. (A, B, C) 68. (A, B) 69. (A, C) 70. (C) 71. (B) 72. (D) 73. (A) 74. (B) 75. 6 76. 6 77. 4 78. 5 79. 9 80. 3 81. 7 82. 4 83. 8 84. 3

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IIT-JEE 2010 SOLUTION - 37 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

SOLUTIONS TO IIT–JEE 2010 (PAPER 2)

PART−1 : CHEMISTRY Useful Data : Atomic Mass : H = 1, C = 12, N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, K = 39, Ca=40, Mn = 55, Fe = 56, Cu = 63.5, Br=80, I = 127.

SECTION−I Straight Objective Type

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. The compound P, Q and S

OH

COOH

CH3

OCH3 CO

O

P Q S

where separately subjected to nitration using HNO3/H2SO4 mixture. The major product formed in each case respectively, is

(A)

OH

COOH

NO2

CH3

OCH3

NO2

CO

O

O2N

(B)

OH

COOH

NO2 CH3

OCH3

NO2

CO

O

NO2

(C)

OH

COOH

NO2

CH3

OCH3

NO2

CO

ONO2

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IIT-JEE 2010 SOLUTION - 38 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

(D)

OH

COOH

NO2

CH3

OCH3

NO2

CO

O

NO2

1. (C) In P, –NO2 group goes ortho- to – OH group in Q, –NO2 group goes ortho – to methoxy

group and in S, para substitution occurs on oxygen bonded ring.

2. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is

(A) 1 and diamagnetic (B) 0 and diamagnetic (C) 1 and paramagnetic (D) 0 and paramagnetic 2. (A)

2x2 * 2 2 * 2oy

2p1s 1s 2s 2s

2p⎧π⎪σ σ σ σ ⎨π⎪⎩

∴ BO = 6 4 12−

= and diamagnetic

3. The packing efficiency of the two-dimensional square unit cell shown below is

L

(A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54% 3. (D)

2

2

2 rP.F.8r 4π π

= =

4. The complex showing a spin only magnetic moment of 2.82 B.M. is (A) Ni(CO)4 (B) [NiCl4]2– (C) Ni(PPh3)4 (D) [Ni(CN)4]2– 4. (B) In [NiCl4]2–

2Ni + ⇒

3d 4s

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IIT-JEE 2010 SOLUTION - 39 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

5. In the reaction 2(1) NaOH / Br T⎯⎯⎯⎯⎯→CH3 CNH2

O

Cl

O the structure of the Product T is

(A)

CH3 CO

O

CO

(B) NH

CO

CH3

(C) NH

CO

CH3 (D)

CH3 CNH

O

CO

5. (C)

2(1) NaOH/ Br⎯⎯⎯⎯⎯→CH3 C

NH2

OMe NH2 Cl−−

⎯⎯⎯⎯⎯→Ph C

O

ClPh C

O

NH Me

6. The species having pyramidal shape is (A) SO3 (B) BrF3 (C) SiO3

2– (D) OSF2 6. (D) In OSF2, sulphur is central atom with sp3 hybridisation and pyramidal shape.

SECTION−II Integer Type

This section contains 5 questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

7. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12m2 can be expressed in scientific notation as y × 10x. The value of x is

7. 7

Radius of Ag – atom = 13

23

108 310.5 6.022 4 3.14 10

×⎛ ⎞⎜ ⎟× × × ×⎝ ⎠

= 1.59 × 10–8

Now Ag atoms in given area = 8

8 2

103.14 (1.59 10 )

−× ×

= 1.25 × 107 ∴ x = 7

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IIT-JEE 2010 SOLUTION - 40 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

8. Among the following, the number of elements showing only one non-zero oxidation state is

O, Cl, F, N, P, Sn, Tl, Na, Ti 8. 2 Flourine and sodium show only one non zero state.

9. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is ws and that along the dotted line path wd, then the integer closest to the ratio wd/ws is

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

0.0

b

a

P (atm.)

V (lit.)

9. 2

8.9 24.25

=

10. The total number of diprotic acids among the following is H3PO4 H2SO4 H3PO3 H2CO3 H2S2O7 H3BO3 H3PO2 H2CrO4 H2SO3 10. 6 H2SO4, H3PO3, H2CO3, H2S2O7, H2CrO4, H2SO3; all six are diprotic acids.

11. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3) (NH3)] is 11. 3 It is M abcd type square planar complex.

SECTION−III Linked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

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IIT-JEE 2010 SOLUTION - 41 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

Paragraph for Questions 12 to 14 Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, whicih upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below:

O

CH3 OH

OCH3

12. The compounds P and Q respectively are

(A) C

O

CHCH3

CH3

H andC

O

CH3 H

(B) C

O

CHCH3

CH3

H andC

O

H H

(C) C

O

CH2 HCH

CH3

CH3

and C

O

CH3 H(D)

C

O

CH2 HCH

CH3

CH3

and C

O

H H

12. (B)

13. The compound R is

(A) HC

OHH2C

OCH3

CH3

(B) HC

OHCH

OCH3

CH3

CH3

(C) HC

OHH2C

O

CH

CH3

CH3

(D) HC

OH

O

CH

CH3

CH3

CH3 13. (A)

14. The compound S is (A)

HC

CNH2C

O

CH

CH3

CH3

(B)

HC

CNH2C

OCH3

CH3

(C)

OHCH

OHH2C

CN

CH

CH3

CH3

(D)

OHCH

OHH2C

CNCH3

CH3

14. (D)

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IIT-JEE 2010 SOLUTION - 42 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

12-14 OH

2CH O−

+ = ⎯⎯⎯→

Δ←⎯⎯ ←⎯⎯

CH3 CH

CH3

CHO CH3 C

CH3

CH2OH

CHO

HCN

CH3 C

CH3

CH2OH

HCOH

CNCH3 C

CH3

CH2OH

HCOH

COOHO O

OHMe

Me

(P) (Q) (R)

(S)

Paragraph for Questions 15 to 16 The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. 15. The state S1 is (A) 1s (B) 2s (C) 2p (D) 3s 15. (B)

16. Energy of the state S1 in units of the hydrogen atom ground state energy is (A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50 16. (C)

17. The orbital angular momentum quantum number of the state S2 is (A) 0 (B) 1 (C) 2 (D) 3 17. (B) for p orbital 1=l 15-17. The spherically symmetrical state with one radial node is 2s while S2 is 3p as it also has

one radial node and transition is bound to satisfy the principle of conservation of angular momentum.

SECTION−IV Linked Comprehension 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 ORS.

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IIT-JEE 2010 SOLUTION - 43 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

18. Match the reactions in Column I with appropriate options in Column II. Column I Column II (A)

2o

NaOH / H O0 C

⎯⎯⎯⎯→+N2Cl OH N N OH

(p) Racemic mixture

(B)

2 4H SO⎯⎯⎯→CH3 C

CH3

OH

C

CH3

OH

CH3 C

CH3

CCH3

CH3

CH3

O (q) Addition reaction

(C) 4

3

1. LiAlH2. H O+⎯⎯⎯⎯→C

CH3

O

HCCH3

OH

(r) Substitution reaction

(D) Base⎯⎯⎯→HS Cl S

(s) Coupling reaction

(t) Carbocation intermediate 18. (A – r, s), (B – t), (C – p, q), (D – r)

19. All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II.

Column I Column II (A) (CH3)2SiCl2 (p) Hydrogen halide formation (B) XeF4 (q) Redox reaction (C) Cl2 (r) Reacts with glass (D) VCl5 (s) Polymerization (t) O2 formation 19. (A – p, s), B – p, q, r, t), (C – p, q, t), (D – p)

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IIT-JEE 2010 SOLUTION - 44 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

PART−2 : MATHEMATICS SECTION – I (Single Correct Choice Type)

This section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

20. If the distance of the point P(1, −2, 1) from the plane x + 2y − 2z = α, where α > 0, is 5,

then the foot of the perpendicular from P to the plane is

(A) ⎟⎠⎞

⎜⎝⎛ −

37,

34,

38 (B) ⎟

⎠⎞

⎜⎝⎛ −

31,

34,

34

(C) ⎟⎠⎞

⎜⎝⎛

310,

32,

31 (D) ⎟

⎠⎞

⎜⎝⎛ −

25,

31,

32

20. (A) x + 2y − 2z − α = 0 Perpendicular distance from P(1, −2, 1) is

5441

|241|=

++α−−−

α = 10

Equation of PQ is 21z

22y

11x

−−

=+

=− = k

Q

P(1, −2, 1)

Coordinates of Q is (k + 1, 2k − 2, −2k + 1) satisfying this, in equation of plane

We get, k = 35

⇒ Q is ⎟⎠⎞

⎜⎝⎛ −

37,

34,

38

21. A signal which can be green or red with probability 54 and

51 respectively, is received by

station A and then transmitted to station B. The probability of each station receiving the

signal correctly is 43 . If the signal received at station B is green, then the probability that

the original signal was green is

(A) 53 (B)

76

(C) 2320 (D)

209

21. (C) Original A B R ⎯⎯→ R ⎯⎯→ G R ⎯⎯→ G ⎯⎯→ G G ⎯⎯→ G ⎯⎯→ G G ⎯⎯→ R ⎯⎯→ G ∴ Required probability

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IIT-JEE 2010 SOLUTION - 45 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

=

41

41

54

43

43

54

43

41

51

41

43

51

41

41

54

43

43

54

××+××+××+××

××+××

= 2320

4640

=

22. Two adjacent sides of a parallelogram ABCD are given by AB 2i 10 j 11k= + +

uuur ) ) ) and

k2j2iAD ++−= The sides AB is rotated by an acute angle α in the plane of the parallelogram so that AD

becomes AD′. If AD′ makes a right angle with the side AB, then the cosine of the angle α is given by

(A) 98 (B)

917

(C) 91 (D)

954

22. (B) α + θ = 90o α = 90o − θ cosα = sinθ

sinθ = |b||a|

barr

rr×

k14j15i2ba +−−=×rr

∴ sinθ= 917

A B

C D

θ

α

D′

23. For r = 0, 1, …., 10, let Ar, Br and Cr denote, respectively, the coefficient of xr in the

expansions of (1 + x)10, (1 + x)20 and (1 + x)30. Then

( )r10r10

10

1rr ACBBA −∑

=

is equal to (A) B10 − C10 (B) ( )1010

21010 ACBA −

(C) 0 (D) C10 − B10 23. (D) (1 + x)10 (1 + x)20 (1 + x)30 10 20 30

r r r r r rA C B C C C= = =

( )r10

1030

r20

1020

10

1rr

10 CCCCC −∑=

= ∑∑==

−2 −

10

1r

2r

1010

30r

2010

1rr10

1010

0 CCCCC

= ( ) ( )20 30 30 20 30 2010 10 10 10 10 10 10 10C C 1 C C 1 C C C B− − − = − = −

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IIT-JEE 2010 SOLUTION - 46 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

24. Let f be a real valued function defined on the interval (−1, 1) such that

e−xf(x) = 2 + dt1tx

0

4∫ + , for all x ∈ (−1, 1) and let f−1 be the inverse function of f. Then

(f−1)′(2) is equal to

(A) 1 (B) 31

(C) 21 (D)

e1

24. (B)

( ) dtt12xfex

0

4x ∫ ++=− ∀ x ∈ (−1, 1)

At x = 0, f(0) = 2 Now −e−xf(x) + e−x f ′(x) = 0 + 4x1+ ⇒ − e0f(0) + e0f ′(0) = 01+ ⇒ −2 + f ′(0) = 1 ⇒ f ′(0) = 3 Now f-1(f(x)) = x ⇒ (f−1f(x))′ f ′(x) = 1 ⇒ (f−1f(0))′f ′(0) = 1

⇒ (f−1(2))′ = ( ) 31

0f1

=′

25. Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to (A) 25 (B) 34 (C) 42 (D) 41 25. (D) S = {1, 2, 3, 4} Let P and Q be disjoint subsets of S ai ∈ P and ai ∉ Q ai ∉ P and ai ∈ Q ai ∉ P ad ai ∉ Q for every element there are three option hence total option are 34 = 81 which includes the

case in which P and Q are φ

Now after removing the order in P and Q we get the total Number = 12

181+

− = 41.

SECTION II (Integer Type)

This section contains 5 questions. The answer to each question is single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

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IIT-JEE 2010 SOLUTION - 47 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

26. Let a1, a2, a3, ….., a11 be real numbers satisfying a1 = 15, 27 − 2a2 > 0 and ak =2ak−1 − ak−2 for k = 3, 4, …., 11.

If 11

a.....aa 211

22

21 +++ = 90, then the value of

11a...aa 1121 +++ is equal to

26. 0 a1 = 15

27 − 2a2 > 0 ⇒ a2 < 2113

227

=

ak = 2ak−1 − ak−2 ⇒ ak − ak−1 = ak−1 − ak−2 ⇒ a1, a2, a3, …. a11 are in A.P.

Let d be the common difference as a2 < 2113 and a1 = 15 , so, d < − 3

2

( ) ( ) ( ) 9011

d10a......d2adaa 211

21

21 =

+++++++

⇒ ( ) ( )10.....321da2d10......21a11 122222

1 +++++++++ = 11 × 90 ⇒ 7d2 + 30d + 27 = 0 ⇒ (7d + 9) (d + 3) = 0

⇒ d = −3, 79

But d < 23

− ⇒ d = −3

So, 1 2 3 11 1a a a ..... a 2a 10d1111 2 11

+ + + + +⎛ ⎞= ×⎜ ⎟⎝ ⎠

= a1 + 5d = 15 + 5(−3) = 0 27. Let f be a function defined on R (the set of all real numbers) such that f ′(x) = 2010(x − 2009) (x − 2010)2(x − 2011)3(x − 2012)4, for all x ∈ R. If g is a function defined on R with values in the interval (0, ∞) such that f(x) = ln(g(x)), for all x ∈ R, then the number of points in R at which g has a local maximum is 27. 1 ( ) ( )( ) ( ) ( )432 2012x2011x2010x2009x2010xf −−−−=′ f(x) = ln g(x) ⇒ g(x) = ef(x)

⇒ g′(x) = ef(x) . f ′(x)

+ − − + +

2009 2010 2011 2012 Local maximum at x = 2009 Local minimum at x = 2011

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IIT-JEE 2010 SOLUTION - 48 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

28. Let k be a positive real number and let

⎥⎥⎥

⎢⎢⎢

−−−

−=

1k2k2k21k2k2k21k2

A and ⎥⎥⎥

⎢⎢⎢

−−−

−=

0k2kk20k21

k1k20B .

If det(adj A) + det(adj B) = 106, then [k] is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer

less than or equal to k]. 28. 4

1k2k2k21k2k2k21k2

|A|−−

−−

= = 11k2k2k2k21k2k201k2

−+−−+

=

2k 1 0 2 k

2 k 1 2k 2k

4 k 0 2k 1

+ −

− −

= (1 + 2k) ((2k − 1)2 + 8k)

= (1 + 2k) ( )2

k81k4k4 2 ++−

⇒ |A| = (1 + 2k) (2k + 1)2 = (2k + 1)3 |B| = 0 (as B is skew symmetric of odd order) det (adj A) + det(adjB) = |A|2 + |B|2 = 106 ⇒ (2k + 1)6 = 106 ⇒ 2k + 1 = 10 ⇒ k = 4.5 [k] = 4. 29. Two parallel chords of a circle of radius 2 are at a distance 13 + apart. If the chords

subtend at the centre, angles of kπ and

k2π , where k > 0, then the value of [k] is

[Note [k] denotes the largest integer less than or equal to k] 29. 3

( )13k2

cosk

cos2 +=⎟⎠⎞

⎜⎝⎛ π

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=−

π+

π2

131k2

cosk2

cos2 2

2

13,23

434131

k2cos +−

=+±−

2π/k

π/k

6

cosk2

cos π=

π

⇒ k = 3.

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IIT-JEE 2010 SOLUTION - 49 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

30. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is

315 . If ∠ACB is obtuse and if r denotes the radius of the incircle of triangle, then r2 is equal to

30. (3)

Δ = Csinab21

⇒ 115 3 6 10 sin C2

= × × ×

⇒ 23Csin = ⇒ C = 120o

So, c2 = a2 + b2 − 2ab cosC

A

B C

c

6

10 Δ = 15 3

= 36 + 100 − 120 × 19621

=⎟⎠⎞

⎜⎝⎛−

⇒ c = 14

So, r = 3

214106

315s

=⎟⎠⎞

⎜⎝⎛ ++

⇒ r2 = 3.

SECTION – III (Paragraph Type) This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice

questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct.

Paragraph for Questions 31 to 33 Consider the polynomial f(x) = 1 + 2x + 3x2 + 4x3 Let s be the sum of all distinct real roots of f(x) and let t = |s|. 31. The real numbers s lies in the interval

(A) ⎟⎠⎞

⎜⎝⎛− 0,

41 (B) ⎟

⎠⎞

⎜⎝⎛ −−

43,11

(C) ⎟⎠⎞

⎜⎝⎛ −−

21,

43 (D) ⎟

⎠⎞

⎜⎝⎛

41,0

31. (C) f(x) = 4x3 + 3x2 + 2x + 1 f ′(x) = 12x2 + 6x + 2 > 0 ∀ x ∈ R ⇒ Only one root is real

Also 1 1f2 4

⎛ ⎞− =⎜ ⎟⎝ ⎠

> 0 and 21

43f −=⎟

⎠⎞

⎜⎝⎛− < 0

⇒ s ∈ ⎟⎠⎞

⎜⎝⎛ −−

21,

43

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IIT-JEE 2010 SOLUTION - 50 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

32. The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

(A) ⎟⎠⎞

⎜⎝⎛ 3,

43 (B) ⎟

⎠⎞

⎜⎝⎛

1611,

6421

(C) (9, 10) (D) ⎟⎠⎞

⎜⎝⎛

6421,0

32. (A) Required area, A

= ( )∫ +++t

0

22 dx1x2x3x4

⇒ A = t4 + t2 + t2 + t

As 143t

21

<<< , 4A1615

<< x=3/4

x = t x=1/2 x=−1/2x=−3/4

33. The function f ′(x) is

(A) increasing in ⎟⎠⎞

⎜⎝⎛ −−

41,t and decreasing in ⎟

⎠⎞

⎜⎝⎛− t,

41

(B) decreasing in ⎟⎠⎞

⎜⎝⎛ −−

41,t and increasing in ⎟

⎠⎞

⎜⎝⎛− t,

41

(C) increasing in (−t, t) (D) decreasing in (−t, t) 33. (B) ( ) 2x6x12xf 2 ++=′

⇒ f″(x) = 24x + 6 = 24 1x4

⎛ ⎞+⎜ ⎟⎝ ⎠

14

+ −

⇒ decreasing in ⎟⎠⎞

⎜⎝⎛ −−

41,t and increasing in ⎟

⎠⎞

⎜⎝⎛− t,

41

Paragraph for Questions 34 to 36

Tangents are drawn from the point P(3, 4) to the ellipse 14y

9x2

=+2

touching the ellipse at

points A and B.

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IIT-JEE 2010 SOLUTION - 51 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

34. The coordinates of A and B are

(A) (3, 0) and (0, 2) (B) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

151612,

58 and ⎟

⎠⎞

⎜⎝⎛−

58,

59

(C) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

151612,

58 and (0, 2) (D) (3, 0) and ⎟

⎠⎞

⎜⎝⎛−

58,

59

34. (D) y − 4 = m(x − 3) ⇒ y = mx + 4 − 3m Now, c2 = a2m2 + b2 gives (4 − 3m)2 =

9m2 + 4 ⇒ 9m2 − 24m + 16 = 9m2 + 4 ⇒ 0 m2 − 24 m + 12 = 0

⇒ 1m2

=

From the other value of m tangent is vertical

So, A ≡ (3, 0) and the other tangent is

25

2xy +=

P (3, 4)

B

Y

O A X

D2

3

i.e,, 15y2

5x

=+−

i.e., 154

y59

9x

=⎟⎠⎞

⎜⎝⎛ 8

+⎟⎠⎞

⎜⎝⎛−

So, B ≡ ⎟⎠⎞

⎜⎝⎛−

58,

59

35. The orthocenter of the triangle PAB is

(A) ⎟⎠⎞

⎜⎝⎛

78,5 (B) ⎟

⎠⎞

⎜⎝⎛

825,

57

(C) ⎟⎠⎞

⎜⎝⎛

58,

511 (D) ⎟

⎠⎞

⎜⎝⎛

57,

258

35. (C)

Clearly the orthocenter will lie on the line BD (in the figure) whose equation is y = 58 .

36. The equation of the locus of the point whose distances from the point P and the line AB

are equal, is (A) 9x2 + y2 − 6xy − 54x − 62y + 241 = 0 (B) x2 + 9y2 + 6xy − 54x + 62y − 241 = 0 (C) 9x2 + 9y2 − 6xy − 54x − 62y − 241 = 0 (D) x2 + y2 − 2xy + 27x + 31y − 120 = 0 36. (A)

Equation of line AB is T = 0 i.e., 14y4

9x3

=+

⇒ x + 3y − 3 = 0

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IIT-JEE 2010 SOLUTION - 52 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

The (x − 3)2 + (y − 4)2 = ( )10

3y3x 2−+

⇒ 10(x2 + y2 − 6x − 8y + 25) = x2 + 9y2 + 6xy − 6x − 18y + 9 ⇒ 9x2 + y2 − 6xy − 54x − 62y + 241 = 0

SECTION – IV (Paragraph Type) This section contains 2 questions. Each questions 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 II. For example, if for a given questions, 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.

37. Match the statements in Column I with the values in Column II. [Note : Here z takes values in the complex plane and Im z and Re z denote, respectively,

the imaginary part and the real part of z.] Column I Column I

(A) The set of points z satisfying |z − i|z|| = |z + i |z|| is contained in or equal to

(p) an ellipse with eccentricity

54

(B) The set of points z satisfying |z + 4| + |z − 4| = 10 is contained in or equal to

(q) the set of points z satisfying Im z = 0

(C) If |w| = 2, then the set of points

w1wz −= is contained in or equal to

(r) the set of points z satisfying |Im z| ≤ 1

(D) If |w| = 1, then the set of points

w1wz += is contained in or equal to

(s) the set of points z satisfying |Re z| ≤ 2

(t) the set of point z satisfying |z| ≤ 3 37. (A) − (q), (B) − (p), (C)−(s, t), (D) − (q, s, t) (A) − (q) |z − i|z|| = |z + i|z|| for any z; |z − i|z|| represents distance of

z form i|z| and |z + i|z|| represent distance of z

from −i|z| ∴ distances are equal ⇒ z should lie on the perpendicular

bisector of (0, |z|) and (0, |z|).

(0, +|z|)

Im −axis

Re −axis

(0, −|z|)

(B) − (p)

|z + 4| + |z − 4| = 10, represents an ellipse whose eccentricity = 54

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IIT-JEE 2010 SOLUTION - 53 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

(C)−(p, s, t)

|z| = 1 1 5w | w | 3w | w | 2

− ≤ + ≤ ≤

Let w = 2(cosθ + isinθ)

⇒ z = 2(cosθ + isinθ) − ( )1 cos isin2

θ − θ

⇒ 3 5z cos i sin2 2

= θ + θ

⇒ 2 2x y 1

3 / 2 5 / 2⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ z represents an ellipse with eccentricity 45

⇒ |Re(z)| = 3 cos2

θ ≤ 2

(D) − (q, s, t) |w| = 1 ⇒ |z| ≤ 2 ⇒ |z| ≤ 3 and (Re(z)) ≤ 2 Let x2 + y2 = 1 for |w| = 1 z = x + iy + x − iy = 2x ⇒ Im(z) = 0 38. Match the statements in Column – I with the values in Column – II.

Column I Column II

(A) A line from the origin meets the lines 1

1z21y

12x +

=−−

=−

and 1

1z13y

238x −

=−+

=−

at P and Q respectively. If

Length PQ = d then d2 is

(p) –4

(B) The values of x satisfying

tan–1(x + 3) – tan–1(x – 3) = sin–1 ⎟⎠⎞

⎜⎝⎛

53 are

(q) 0

(C) Non-zero vectors b,arr and cr satisfy ba

rr⋅ = 0,

( ) ( )cbab rrrr+⋅− = 0 and 2 abcb rrrr

−=+ . If c4ba rrr+μ= ,

then the possible values of μ are

(r) 4

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IIT-JEE 2010 SOLUTION - 54 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

(D) let f be the function on [ –π, π] given by f(0) = 9 and f(x) =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

2xsin

2x9sin for x ≠ 0.

(s) 5

The value of ( )dxxf2∫π

π−πis

(t) 6

38. (A) − (t), (B)−(p), (r), (C)−(q), (s), (D)−(r) From given conditions, we have

11

123

2382

−λ+μ

=−λ

+μ=

⇒ λ = 3, μ = 31

⇒ P ≡ (5, −5, 2), Q ≡ ⎟⎠⎞

⎜⎝⎛ −

34,

310,

310

⇒ 222

2

342

3105

3105PQ ⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ +−+⎟

⎠⎞

⎜⎝⎛ −=

= 6

(0, 0, 0)

x 2 y 1 z 11 2 1− − +

= =−

Q

(λ + 2, 1−2λ, λ−1) P

11z

13y

23/8x −

=−+

=−

⎟⎠⎞

⎜⎝⎛ +μ−μ−+μ 1,3,

282

(B)−(p), (r)

( ) ( )1 1 1 13 3tan x 3 tan x 3 sin tan5 4

− − − −⎛ ⎞ ⎛ ⎞+ − − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ ( ) ( )( )433xtan3xtantan 11 =−−+ −−

⇒ ( ) ( )( )( )

x 3 x 3 31 x 3 x 3 4

+ − −=

+ + −

⇒ x2 = 16 ⇒ x = ± 4 For x = 4, we have

1 1 1 17 1 3tan 7 tan 1 tan tan1 7 4

− − − −−⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

for x = −4, we have

( ) ( ) ( ) ( )1tan7tan7tan1tan 1111 −−−− −=−−− = ⎟⎠⎞

⎜⎝⎛−

43tan 1

(C)−(q), (s)

4

bac μ−= and 0b.a =

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IIT-JEE 2010 SOLUTION - 55 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

Now ( ) ( )b a . b c 0− + = ⇒ ( ) 04

bab.ab =⎟⎟⎠

⎞⎜⎜⎝

⎛ μ−+−

⇒ ( ) 04ab

44ab =⎟

⎠⎞

⎜⎝⎛ +

μ−−

⇒ ( ) 04ab

44 2

2 =−μ− ⇒ (4 − μ)b2 − a2 = 0

2

2 4 a2 b | b a |4 4− μ

+ = − 2

⇒ 222

22

ab4ab

444 +=+⎟

⎠⎞

⎜⎝⎛ μ−

⇒ ( )2 2 2 2 24 b a 4b 4a− μ + = + ⇒ ((4 − μ)2 − 4) b2 − 3a2 = 0

⇒ ( )μ−

−μ−4

44 2

= 13

⇒ μ2 − 8μ + 12 = 12 − 3 μ ⇒ μ2 − 5μ = 0 μ = 0 or 5 (D)−(r)

Let In = ∫π

π−dx

2xsin

2nxsin

, n = odd natural number.

then In – In – 2 =

( ) x n 1n 2 xnx 2sin cos xsin sin 2 22 2 dx dxx xsin sin2 2

π π

−π −π

−⎛ ⎞⎛ ⎞−− ⎜ ⎟⎜ ⎟ ⎝ ⎠=⎜ ⎟

⎜ ⎟⎜ ⎟⎝ ⎠

∫ ∫

= n 12 cos x dx2

π

−π

−⎛ ⎞⎜ ⎟⎝ ⎠∫

= 0x2

1nsin1n

4=

π−π

⎟⎠⎞

⎜⎝⎛ −

− as n = odd

⇒ In = In – 2 So, I9 = I7 = I5 = I3 = I1 = ∫

π

π−π= 2dx1

So, ( )∫π

π− π=

π2dxxf2 I9 = 4

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IIT-JEE 2010 SOLUTION - 56 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

PART−3 : PHYSICS

SECTION – I (Single Correct Choice Type) This section contains 6 multiple choice questions. Each question has four choices (A),

(B), (C) and (D) out of which ONLY ONE is correct. 39. A block of mass 2 kg is free to move along the x-

axis. It is at rest and from t = 0 onwards it is subjected to a time-dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 second is

(A) 4.50 J (B) 7.50 J (C) 5.06 J (D) 14.06 J

4.5s

F(t)

4 N

O 3s t

39. (C) Area of F – t graph = change in linear momentum Δp = +6 – 1.5 = 4.5 Qpi = 0 ∴ pf = 4.5 kg m/s.

∴ KE = m2

p2

= 5.06 J

40. A uniformly charged thin spherical shell of

radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to

FF

(A) 22

0

R1σ

ε (B) R1 2

0

σε

(C) R

1 2

0

σε

(D) 2

2

0 R1 σε

40. (A)

∴ F = ( )2 2 2

2

0 0

RR2 2

⎛ ⎞σ σπ =⎜ ⎟ε ε⎝ ⎠

41. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical

uniform electric field of strength 5107

81×

π Vm–1. When the field is switched off, the drop

is observed to fall; with terminal velocity 2 × 10–3 ms–1. Given g = 9.8 ms–2, viscosity of the air = 1.8 × 10–5 Ns m–2 and the density of oil = 900 kg m–3, the magnitude of q is

(A) 1.6 × 10–19 C (B) 3.2 × 10–19 C (C) 4.8 × 10–19 C (D) 8.0 × 10–19 C

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IIT-JEE 2010 SOLUTION - 57 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

41. (D)

For balanced position, qE = mg = 3r34

π ρoil g (1)

for tiny drop Fb ≅ 0 (Buoyancy force) ∴ After acquiring terminal velocity while (E = 0) Fv = mg ∴ 6πηrv = mg = qE [by equation (1)]

∴ q3E3 = (6πηv)3r3 = (6πηv)3⎥⎦

⎤⎢⎣

⎡πρ g4

qE3

oil

On solving and by putting all given values q = 8 × 10–19 C 42. A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the

vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is

(A) 0.02 mm (B) 0.05 mm (C) 0.1 mm (D) 0.2 mm 42. (D) Least count = 1 MSD – 1 VSD

= aNN1a

NNa ⎟

⎠⎞

⎜⎝⎛

′−=⎟

⎠⎞

⎜⎝⎛

′−

Where a : one main scale division N : Number of divisions in main scale which concide with N divisions of vernier

scale

L.C = ⎥⎦⎤

⎢⎣⎡ −

20161 (1 mm) = 0.2 mm

43. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between

the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

(A) virtual and at a distance of 16 cm from the mirror (B) real and at a distance of 16 cm from the mirror (C) virtual and at a distance of 20 cm from the mirror (D) real and at a distance of 20 cm from the mirror 43. (B) After refraction from the lens position of image = 30 cm. (from lens) ∴ for the mirror object position = 20 cm (right of the mirror) Now for the second refraction from the lens after getting reflected from the mirror, the

position of incident ray = 10 cm (left of the lens) ∴final position of image = 6 cm from left of the lens and 16 cm from the plane mirror.

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IIT-JEE 2010 SOLUTION - 58 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

I1I I2 O 6cm 20 cm

30 cm 10 cm 20 cm 44. A hollow pipe of length 0.8 m is closed at one end. At its open end a 0.5 m long uniform

string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is 50 N and the speed of sound is 320 ms–1, the mass of the string is

(A) 5 grams (B) 10 grams (C) 20 grams (D) 40 grams 44. (B)

Fundamental frequency for closed pipe = L4v

f0 = 100 Hz. And second harmonic for string

f0 = μT

L1

∴ By putting f0 = 100 Hz, T = 50N, L = 0.5 M ∴ μ = 20 gram/meter. ∴mass of the string = μl = 20 × 0.5 = 10 grams.

SECTION – II (Integer Type) This section contains 5 questions. The answer to each of the questions is a single digit

integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

45. A large glass slab (μ = 5/3) of thickness 8 cm is placed over a point source of a light on a

plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?

45. 6

35 sinθ = 1 ⇒ sinθ =

53

tanθ = 5/45/3

8R

= ⇒ R = 6 cm

θ θ 8 cm

R

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IIT-JEE 2010 SOLUTION - 59 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

46. Image of an object approaching a convex mirror of radius of curvature 20 m along its

optical axis is observed to move from 325 m to

750 m in 30 seconds. What is the speed of

the object in km per hour? 46. 3

1u

1253

+ = 101 &

101

u1

507

2

=+

u1 = – 50 m & u2 = –25 m Δu = 25 cm

Average speed = 5

183025

× = 3 km / hr.

47. To determine the half life of

a radioactive element, a student plots a graph of

( )dt

tdNnl versus t. Here

( )dt

tdN is the rate of

radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor p after 4.16 years, the value of p is

1 2 3 4 5 6 7 8

2

3

4

5

6

Years

( )dt

tdNnl

47. 8

λ = 21 from the given graph

t1/2 = 2/12ln = 2 × 0.693 = 1.386

⇒ Number of half lives n = 4.16 31.386

=

So p = 8 as N = N0 3

21

⎟⎠⎞

⎜⎝⎛ =

8N0

48. A diatomic ideal gas is compressed adiabatically to 321 of its initial volume. In the initial

temperature of the gas is Ti (in Kelvin) and the final temperature is αTi the value of α is 48. 4 Ti 1

ff1

i VTV −γ−γ =

if aTT =Q and f i1V V32

=

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IIT-JEE 2010 SOLUTION - 60 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

a = ( ) 157

32 − = 22 = 4 49. At time t = 0, a battery of 10 V is connected

across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V?

[take : nl 5 = 1.6, nl 3 = 1.1]

B

2MΩ

2MΩ

2μF

2μF

A

49. 2

V = ( )RC/t0 e1Cq −−

4 = ( )4/te1440 −−

⇒ et/4 = 35 ⇒ t = 4(1.6 – 1.1) = 2 sec.

Section - III Paragraph for Question 50 to 52.

When liquid medicine of density ρ is to be put in the eye. It is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

50. If the radius of the opening of the dropper is r, the vertical force due to the surface tension

on the drop of radius R (assuming r << R) is (A) 2πrT (B) 2πRT

(C) 22 r T

Rπ (D)

22 r Tr

π

50. (C) Vertical force due to surface tension = Tsinθ. 2πr

= T r2.Rr

π

= R

Tr2 2π

90–θ

θ

r

T

R

51. If r = 5 × 10-4m, ρ = 103 kgm–3, g = 10 ms-2, T = 0.11Nm–1, the radius of the drop when it

detaches from the dropper is approximately (A) 1.4 × 10-3m (B) 3.3 ×10-3m (C) 2.0 × 10-3m (D) 4.1 × 10-3m

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IIT-JEE 2010 SOLUTION - 61 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

51. (A)

gR34

RTr2 3

2

ρπ=π ⇒ R =

4/12

gTr

23

⎟⎟⎠

⎞⎜⎜⎝

⎛ρ

= 1.4 × 10–3 m

52. After the drop detaches, its surface energy is (A) 1.4 × 10-6 J (B) 2.7 × 10-6J (C) 5.4 × 10-6J (D) 8.1 × 10-6J 52. (B) Surface energy = TA = T × 4πR2 = 2.7 × 10–6 J

Paragraph for question 53 to 55 The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of

angular-momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.

53. A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotation

energy in the nth level (n = 0 is not allowed) is

(A) 2

2 2

1 hn 8 I

⎛ ⎞⎜ ⎟π⎝ ⎠

(B) 2

2

1 hn 8 I

⎛ ⎞⎜ ⎟π⎝ ⎠

(C) 2

2

hn8 I

⎛ ⎞⎜ ⎟π⎝ ⎠

(D) 2

22

hn8 I

⎛ ⎞⎜ ⎟π⎝ ⎠

53. (D)

π2

nh = Iω ⇒ ω = I2

nhπ

∴ K = 21 Iω2 = ⎟⎟

⎞⎜⎜⎝

⎛π I8hn 2

22

54. It is found that the excitation frequency from ground to the first excited state of rotation

for the CO molecule is close to 4π

× 1011 Hz. Then the moment of inertia of CO molecule

about its center of mass is close to (Take h = 2π × 10-34 J s) (A) 2.76 × 10–46 kg m2 (B) 1.87 × 10–46 kg m2 (C) 4.67 × 10–47 kg m2 (D) 1.17 × 10–47 kg m2 54. (B)

ΔE = hν = I8

h2

2

π(22 – 12) =

I8h3

2

2

π

⇒ I = νπ h8

h32

2

= 112

34

1048

1023

×π

×π

×π× −

= 1.87 × 10–46 kg m2.

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IIT-JEE 2010 SOLUTION - 62 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

55. In a CO molecule, the distance between C (mass = 12 a.m.u) and O (mass = 16 a.m.u),

where 1 a.m.u = 271035 −× kg, is close to

(A) 2.4 × 10–10 m (B) 1.9 × 10–10 m (C) 1.3 × 10–10 m (D) 4.4 × 10–11 m 55. (C)

I = 2

21

21 dmm

mm⋅

+ ⇒ d =

( )21

21

mmmmI +⋅ = 1.3 × 10–10 m.

SECTION - IV (Matrix 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 statements(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.

56. Two transparent media of refractive indices μ1 and μ3 have a solid lens shaped transparent

material of refractive index μ2 between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationship between μ1, μ2 and μ3 are given. Match them to the ray diagrams shown in Column II.

Column I Column II

(A) μ1 < μ2 (p)

μ3

μ2 μ1

(B) μ1 > μ2

μ3 μ2 μ1

(C) μ2 = μ3

μ3

μ2

μ1

(D) μ2 > μ3 μ3

μ2

μ1

μ3

μ2

μ1

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

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IIT-JEE 2010 SOLUTION - 63 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

57. You are given many resistances, capacitors and inductors. These are connected to a-variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2. (Indicated in circuits) are related as shown in Column I. Match the two

Column-I Column -II

(A) I ≠ 0, V1 is proportional to I (p) V1 V2

3μF 6 mH

V (B) I ≠ 0, V2 > V1 (q) V1 V2

2Ω 6 mH

V (C) V1= 0, V2 = V (r) V1 V2

2Ω 6 mH

V (D) I ≠ 0, V2 is proportional to I (t) V1 V2

3μF 6 mH

V (t) V1 V2

3μF 1kΩ

V 57. (A – r, s, t), (B – q, r, s, t), (C – q), (D – q, r, s, t)

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IIT-JEE 2010 SOLUTION - 64 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

IIT-JEE 2010 PAPER – II (KEY) CHEMISTRY MATHEMATICS PHYSICS

1. (C) 2. (A) 3. (D) 4. (B) 5. (C) 6. (D) 7. 7 8. 2 9. 2 10. 6 11. 3 12. (B) 13. (A) 14. (D) 15. (B) 16. (C) 17. (B) 18. (A – r, s) (B – t) (C – p, q) (D – r) 19. (A – p, s) (B – p, q, r, t) (C – p, q, t) (D – p)

20. (A) 21. (C) 22. (B) 23. (D) 24. (B) 25. (D) 26. 0 27. 1 28. 4 29. 3 30. 3 31. (C) 32. (A) 33. (B) 34. (D) 35. (C) 36. (A) 37. (A) − (q) (B) − (p) (C)−(p, s, t) (D) − (q), (s), (t) 38. (A) − (t) (B)−(p), (r) (C)−(q), (s) (D)−(r)

39. (C) 40. (A) 41. (D) 42. (D) 43. (B) 44. (B) 45. 6 46. 3 47. 8 48. 4 49. 2 50. (C) 51. (A) 52. (B) 53. (D) 54. (B) 55. (C) 56. (A – p, r) (B – q, s, t) (C – p, r, t) (D – q, s) 57. (A – r, s, t) (B – q, r, s, t) (C – q) (D – q, r, s, t)

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IIT-JEE 2010 SOLUTION - 65 PIE IIT JEE PAPER – I & II (SOLUTION – 2010) π

IIT-JEE 2010 MODEL PAPER

INSTRUCTION A. General: 1. This Question Paper contains 32 pages having 84 questions. 2. The questions paper CODE is printed on the right hand top corner of this sheet and also on the back page (page no. 32) of this

booklet. 3. No additional sheets will be provided for rough work. 4. Blank papes, clipboards, log tables, slide rules, calculators, cellular phones, pagers and electronic gadgets in any form are not

allowed. 5. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is provided separately. 6. Do not Tamper/mutilate the ORS or this booklet. 7. Do not break the seals of the question-paper booklet before instructed to do so by the invigilators. B. Filling the bottom-half of the ORS: 8. The ORS has CODE printed on its lower and upper Parts. 9. 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. 10. Write your Registration Nno., Name and Name of centre and sign with pen in appropriate boxes. Do not write these anywhere

else. 11. Darken the appropriate bubbles below your registration number with HB Pencil.

IIT-JEE 2010 MODEL PAPER (0)

MARKING SCHEME PAPER – I 12. The question paper consists of 3 parts (Chemistry, mathematics and Physics). Each part consists of four Sections. 13. For each question in Section I, you will be awarded 3 marks if you have darkened only the bubble corresponding to the correct

answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. 14. For each question in Section II, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and

zero mark if no bubbles are darkened. Partial marks will be awarded for partially correct answers. No negative marks will be awarded in this section.

15. For each question in Section III, you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.

16. For each question in Section IV, you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for in this Section.

PAPER – II 13. The question paper consists of 3 Parts (Chemistry, Mathematics and Physics), and each Part consists of four Sections. 14. For each question in Section I : you will be awarded 5 marks if you have darkened only the bubble corresponding to the correct answer

and zero mark if no bubbles are darkened. In all other cases minus two (–2) mark will be awarded. 15. For each question in Section II: you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero

mark if no bubble is darkened. No negative marks will be awarded for incorrect answers in this Section. 16. For each question in Section III: you will be awarded 3 marks if you darken only the bubble corresponding to the correct answer and

zero mark if no bubbles are darkened. IN all other cases, minus one (–1) mark will be awarded. 17. For each question in Section IV: you will be awarded 2 marks for each row in which you have darkened in bubble(s) corresponding to

the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this Section.