jee 2010 bt

8/8/2019 Jee 2010 BT

1/61

SOLUTIONS TO IIT-JEE 2010Paper-I (Code: 8)

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

INSTRUCTIONSA. General:

1. This Question Paper contains 28 pages having 84 questions. The question paper CODE isprinted on the right hand top corner of this sheet and also on the back page of thisbooklet.

2. No additional sheets will be provided for rough work.3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and

electronic gadgets in any form are not allowed.

4. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is providedseparately.

5. Write your Registration No., Name and Name of centre and sign with pen in appropriateboxes. Do not write these anywhere else.

B. Question paper format and Marking scheme:6. The question paper consists of 3 parts (Chemistry, Mathematics and Physics). Each part

consists offour Sections.

7. For each question in Section I, you will be awarded 3 marks if you have darkened onlythe bubble corresponding to the correct answer and zero mark if no bubbles aredarkened. In all other cases, minus one (1) markwill be awarded.

8. For each question in Section II, you will be awarded 3 marks if you darken only thebubble(s) 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 beawarded in this Section.

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

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

8/8/2019 Jee 2010 BT

2/61

IITJEE 2010 SOLUTIONS

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

2

SOLUTIONS TO IIT-JEE 2010CHEMISTRY: Paper-I (Code: 8)

PARTI

Useful Data

Atomic numbers: Be = 4; C = 6; N = 7; O = 8; Al = 13; Si = 14; Cr = 24; Fe = 26; Zn = 30; Br = 35.

1 amu= 1.66 1027 kg R = 0.082 L atm K1 mol1

h = 6.626 1034 Js NA = 6.022 1023

me = 9.1 1031 kg e = 1.6 1019 C

c = 3.0 108 m s1 F = 96500 C mol1

RH = 2.18 1018 J 4o = 1.11 10

10 J1 C2 m1

SECTION I

Single Correct Choice Type

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which

ONLY ONE is correct.

Note: Questions with (*) mark are from syllabus of class XI.

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

(A) k

T

(B) k

T

(C) k

T

(D) k

T

Sol.: As per Arrhenius equation ),Aek(RT/Ea the rate constant increases exponentially with temperature.

Correct choice: (A)

2. In the reaction OCH3HBr the products are

(A) OCH3 and H2Br (B) Br and CH3Br

(C) Br and CH3OH (D) OH and CH3Br

Sol.: OCH3H+

OCH3

H

+ BrOH + CH3Br

Correct choice: (D)

3. The correct statement about the following disaccharide is

HO

H

OH

H

HO

H

H

CH2OH

OHOH

H

H

HO

HOH2C

OCH2CH2O

(a)

H

CH2OH

O

(b)

(A) Ring (a) is pyranose with -glycosidic link. (B) Ring (a) is furanose with -glycosidic link.

(C) Ring (b) is furanose with -glycosidic link. (D) Ring (b) is pyranose with -glycosidic link.

Sol.: Correct choice: (A)

*4. 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 CH3CH2CCH (B) BrCH2CH2CH3 and CH3CH2CH2CCH

(C) BrCH2CH2CH2CH2CH3 and CH3CCH (D) BrCH2CH2CH2CH3 and CH3CH2CCH

Sol.: H3CH2CCHNaNH2

NH3CH3CH2CC

Na+CH3CH2CH2CH2Br

(SN2)CH3CH2CCCH2CH2CH2CH3 + NaBr

3-Octyne

Correct choice: (D)

8/8/2019 Jee 2010 BT

3/61



3

5. 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

Sol.: Ionization isomer of [Cr(H2O)4Cl(NO2)]Cl is [Cr(H2O)4Cl2]NO2.

Correct choice: (B)

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

(A)HOOCH2C

NCH=CHNHOOCH2C

CH2COOH

CH2COOH(B)

HOOCNCH2CH2N

COOH

COOHHOOC

(C) NCH2CH2NHOOCH2C

HOOCH2C

CH2COOH

CH2COOH(D) NCHCHN

HOOCH2C

H CH2COOH

CH2

COOH

CH2

HOOC

H

Sol.: The correct structure of EDTA is

NCH2CH2NHOOCH2C

HOOCH2C

CH2COOH

CH2COOH

Correct choice: (C)

*7. The bond energy (in kcal mol1) of a CC single bond is approximately

(A) 1 (B) 10 (C) 100 (D) 1000

Sol.: CC bond energy = 348 kJ/mol2.4

348kcal/mol = 82.85 kcal/mol 100 kcal/mol.

Correct choice: (C)

*8. 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)

Sol.: The species in its elemental form has zero standard molar enthalpy of formation at 298 K.

Br2(l) Br2(g) ;ofH 0

C(s) + 2H2(g) CH4(g) ;o

f

H 0

H2(g) + O2(g) H2O(g) ;ofH 0

Correct choice: (B)

SECTIONII

Multiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which

ONE OR MORE is/are correct.

9. In the reaction

OH

NaOH(aq)/Br2the intermediate(s) is(are)

(A)

Br

BrO

(B)

Br

O

Br

(C)

Br

O

(D)

Br

O

Sol.:

OH

NaOH

O

Br2

O

Br2Br

O

Br

Br2Br

O

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

Br

Br

H2O Br

Br Br

Product of reaction of phenol with NaOH/Br2 is sodium salt of 2,4,6-tribromophenol. Hence, species (I), (II), (III) are formed

as intermediate.

Correct choice: (A), (C)

8/8/2019 Jee 2010 BT

4/61



4

*10. Among the following, the intensive property is (properties are)

(A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity

Sol.: Mass independent properties are intensive properties. Resistance and heat capacity are extensive properties.

Correct choice: (A), (B)

*11. 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

Sol.: Temporary hardness is due to bicarbonates of calcium and magnesium. Temporary hardness can be removed by Clarksprocess, which involves the addition of slaked lime, Ca(OH) 2. Washing soda (Na2CO3) removes both the temporary and

permanent hardness by converting soluble calcium and magnesium compounds into insoluble carbonates.

Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O

Ca(HCO3)2 + Na2CO3 CaCO3 + 2NaHCO3

2OCl + 2H2O 2HOCl + 2OH

Ca(HCO3)2 + 2OH CaCO3 +

23CO + 2H2O

Correct choice: (B), (C), (D)

*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 CH3COONaSol.: Any solution of a weak acid and its salt with strong base acts as an acidic buffer solution.

If volume of HNO3 solution added is less as compared to that of CH3COONa solution, it results in the formation of an acidic

buffer solution.

CH3COONa + HNO3 CH3COOH + NaNO3

Excess limiting

reagent

MV MV

M(VV) 0 MV MV (V < V)

Correct choice: (C), (D)

*13. In the Newman projection for 2,2-dimethylbutane

X

H H

Y

H3C CH3

X and Y can respectively be

(A) H and H (B) H and C2H5 (C) C2H5 and H (D) CH3 and CH3

Sol.: Structural formula of 2, 2-dimethylbutane is

CH3CCH2CH3

CH3

CH3

1 2 3

(I) Newman projection using C1

C2

bond

H

H H

C2H5

H3C CH3

1

2

(II) Newman projection using C3C2 bond

H H

CH3

CH3 CH3

3

2

CH3

Correct choice: (B), (D)

8/8/2019 Jee 2010 BT

5/61



5

SECTION IIIParagraph 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) and D) out

of which ONLY ONE is correct.

Paragraph for Question 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.

14. Partial roasting of chalcopyrite produces

(A) Cu2S and FeO (B) Cu2O and FeO (C) CuS and Fe2O3 (D) Cu2O and Fe2O3

Sol.: 2CuFeS2 + 4O2 C14501400

Cu2S + 3SO2 + 2FeO

Partial roasting of FeS and Cu2S produces FeO and Cu2O respectively.

2FeS + 3O2 2FeO + 2SO2

2Cu2S + 3O2 2Cu2O + 2SO2

Correct choice: (B)15. Iron is removed from chalcopyrite as

(A) FeO (B) FeS (C) Fe2O3 (D) FeSiO3

Sol.: FeO + SiO2 FeSiO3Correct choice: (D)

16. In self-reduction, the reducing species is

(A) S (B) O2 (C) S2 (D) SO2

Sol.: Cu2S + 2Cu2O 6Cu + SO2The reducing species is the one which gets oxidized. So, it is S2ion getting oxidized to S4+.

Correct choice: (C)

Paragraph for Question 17 to 18

The concentration of potassium ions inside a biological cell is atleast 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; 0G (B) 0G;0Ecell (C) 0G;0Ecell (D) 0G;0Ecell

Sol.: (A): M(s) e)aq(MA

(C): )s(Me)aq(MC

)aq(M)aq(M AC

]M[

]M[ln

F

RT

]M[

]M[ln

F

RTEE

A

C

C

Aocellcell

)0E( ocell

Ecell = 0.059 log05.0

]M[ C

0.07V = 0.05905.0

)M(log C

= 0.059 log 20 ]M[ C = 0.059 (log 20 + log ]M[ C

)

Since ]M[ C is greater than 0.05 M, so Ecell is +ve.

Ecell > 0 & .0G Correct choice: (B)

8/8/2019 Jee 2010 BT

6/61



6

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) 70 mV (C) 140 mV (D) 700 mV

Sol.: Ecell = 0.0590025.0

]M[log C

= 0.059 log 400 ]M[ C = 0.059 log (20)2 ]M[ C

= 0.059(2 log 20 + log ]M[ C )

= 0.059

59.0

07.02 = 0.14 V = 140 mV.

Correct choice: (C)

SECTIONIVInteger Answer Type

This section contains TEN questions. The answer to each 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.

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

1. O32. Zn, H2O Y

1. NaOH(aq)2. heat

Sol.:

1. O32. Zn, H2O

NaOH(aq)

O

O

(Y)

OH

O

heat

H2O

O

The number of intramolecular aldol condensation products (, -unsaturated carbonyl compound) formed from (Y) is 1.

The answer is 1.

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

NCH3H3C COOH OCH2CH3

CH2OH

OH

NO2 OH

CH3H3CN

CH2CH3

CH2CH3

COOH

Sol.: All carboxylic acids and phenols are soluble in aqueous NaOH. Four compounds are soluble in aqueous NaOH.

The answer is 4.

*21. 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

Sol.: KCN, K2CO3 and LiCN are the salts of weak acid and strong base. So, their aqueous solutions turns red litmus paper blue.

(NH4)2C2O4 is a salt of weak base (NH4OH ; Kb = 1.8 105) and weak acid (HOOC.COOH ;

1aK = 5.4 102 and

2aK = 5.2 105 ). As a result, the extent of hydrolysis of

4NH will be more than that of

242OC and hence the resulting

solution will be acidic.The answer is 3.

*22. Based on VSEPR theory, the number of 90 degree FBrF angles in BrF5 is

Sol.: According to VSEPR theory, number of electron pairs around central atom (Br) are 6.

2

N=

2

57 = 6. (Five are bond pairs and one is lone pair )

Its geometry is octahedral but due to lone pairbond pair repulsion, thefour fluorine atoms at corner are forced towards the upper fluorine atom

thus reducing FBrF angle from 90 to 84.8.

F

..F

F

F

Br

F

The answer is 0.

8/8/2019 Jee 2010 BT

7/61



7

*23. The value of n in the molecular formula BenAl2Si6O18 is

Sol.: It is the formula of Beryl (which has BeO, Al2O3 and SiO2).

The formula of Beryl should be

3BeO. Al2O3.6SiO2

Or

Total cationic charge = Total anionic charge

2n + 6 + 24 = 36

n = 3

The answer is 3.

*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

Sol.: The least significant figure in titre values is 3.

Average titre value = 1.253

4.75

3

2525.252.25

The number of significant figures in average titre value will also be 3.

The answer is 3.

25. The concentration of R in the reaction PR 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 reaction is

Sol.: The integrated form of a zero-order reaction is

[A0][At] = k0t

1.00.75 = k0 0.05, k0 = 5

1.00.4 = k0 0.12, k0 = 5

The answer is 0.

26. The number of neutrons emitted when U23592 undergoes controlled nuclear fission to Xe14254 and Sr

9038 is

Sol.: 1090

38142

541

0235

92 nySrXenU

235 + 1 = 142 + 90 + yy = 4. The number of neutrons emitted are (41) = 3.

The answer is 3.

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

CHC

H3NCH2CH2CH2CH2 O

H2N O

Sol.: The basic groups in the given form of lysine is NH2 and2CO . The answer is 2.

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

Sol.: The number of cyclic isomers for a hydrocarbon with molecular formula C4H6 is 5.

The structures are

, , and,

CH2CH3

CH3.

8/8/2019 Jee 2010 BT

8/61



8

SOLUTIONS TO IIT-JEE 2010MATHEMATICS: Paper-I (Code: 8)

PARTII

SECTION I


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

29. The value of

x

xdt

t

tt

x0

430 4

)1ln(1lim is

(A) 0 (B)12

1(C)

24

1(D)

64

1

Sol.:

dt

t

tt

x

x

x

0 430 4

1ln1lim

form

0

04

1ln

lim

3

0

4

0 x

dtt

ttx

x

Applying L. Hospital Rule

431ln

lim3

4

1ln

lim402

4

0

xx

x

x

x

xx

xx

12

1

4

1lim

1lnlim

3

1

400

xx

x

xx

Correct choice: (B)

30. The number of 33 matricesA whose entries are either 0 or 1 and for which the system A

0

0

1

z

y

x

has exactly two distinct

solutions, is

(A) 0 (B) 129 (C) 168 (D) 2

Sol.: For any matricesA it form three planes. But three planes can not intersect exactly at two distinct points.

Correct choice: (A)

31. Let P, Q, R and S be the points on the plane with position vectors ,2 ji jii 33,4 and ji 23 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

Sol.: IfO is origin jiOSjiORiOQjiOP 23;33;4;2

jiOQOSQS 27

jiPR 45

53449 QS ; 411625 PR

376 jiPQ ; 103 jiQR

376 jiRS ; 103 jiSP

Opposite sides are parallel and equal but diagonals are not equal.

Correct choice: (A)

8/8/2019 Jee 2010 BT

9/61



9

32. Let be a complex cube root of unity with 1 . A fair die is thrown three times. If 21, rr and 3r are the numbers obtained

on the die, then the probability that 321 rrr = 0 is

(A)18

1(B)

9

1(C)

9

2(D)

36

1

Sol.: 21663 Sn

Eis the event that 0321 rrr , is possible if one among 321 ,, rrr is of the form n3 ; another one is of the form

13 n and the remaining one is of the form 23 n , where Zn .

For n3 , two possible cases are 3, 6

For 13 n , two possible cases are 1, 4

For 23 n , two possible cases are 2, 5

48!3222 En

9

2

216

48

Sn

EnEP

Correct choice: (C)

33. Equation of the plane containing the straight line432

zyx and perpendicular to the plane containing the straight lines

243

zyx and

324

zyx is

(A) 022 zyx (B) 0223 zyx (C) 02 zyx (D) 0425 zyx

Sol.: Plane 2P contains the lineL 432

zyx

Plane 1P contains the lines 1L and 2L , where

2431

zyxL and

3242

zyxL

Let direction ratio of normal to the plane 1P is ,, , then

0243 and 0324

1018

Let direction ratio of normal of required plane is (a, b, c), then 0108 cba ; 0432 cba

265226

cba

121

cba

Required plane is 02 zyx

Correct choice: (C)

Alternative:

cbabcacban ..

kjikji 3242624326 kji 265226 direction ratio of normal of required plane is 1,2,1

Equation is 02 zyx

*34. If the angles BA, and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides

opposite to BA, and Crespectively, then the value of the expression Aa

cC

c

a2sin2sin is

(A)

2

1(B)

2

3(C) 1 (D) 3

8/8/2019 Jee 2010 BT

10/61



10

Sol.: SinceA, B, Care in A.P.

CAB 2

180CBA

1803B 60B

Aa

c

Cc

a

2sin2sin

AAA

CCC

C

Acossin2

sin

sincossin2.

sin

sin 120sin22sin2sin2cossinsincos2 BCAACAC

32

32

Correct choice: (D)

35. Let f, g and h be real-valued functions defined on the interval [0, 1] by22

)( xx eexf ,22

)( xx exexg and

222)( xx eexxh . Ifa, b and c denote, respectively, the absolute maximum off, g and h on [0, 1], then

(A) ba and bc (B) ca and ba (C) ba and bc (D) cba

Sol.: 22 xx

eexf

22

2 xx eexxf

22 xx exexg

222

2 xxx eexexxg

222 xx eexxh

222

223 xxx eexexxh

for ]1,0[x

xgxf , and 0 xh

xgxf , and xh will be increasing functions and their maximum values are

e

ec

e

eb

e

ea1

,1

,1

at 1x

cba

Correct choice: (D)

Alternative:

],1[]1,0[as1 2

2

2

eex

e

exf x

x

x

],1[,1

ett

ttf ; e

etf1

max

max e

exf1

xfxgxh

e

exg1

but at 1x , e

exg1

and similarly at 1x , xh e

e1

So, cba

*36. Let p and q be real numbers such that qpp 3,0 and qp 3 . If and are nonzero complex numbers satisfying

p and q 33 , then a quadratic equation having

and

as its roots is

(A) 0)()2()( 3323 qpxqpxqp (B) 0)()2()( 3323 qpxqpxqp

(C) 0)()25()( 3323 qpxqpxqp (D) 0)()25()( 3323 qpxqpxqp

8/8/2019 Jee 2010 BT

11/61



11

Sol.: p (i)

q 33 (ii)

Sum of roots

2

222

S (iii)

Product of roots P = 1

From (i) and (ii) 22pq

32pq p

pq

3

3

3

3

3

33

3

32

2223

3

3

2

pq

qp

pq

pqp

p

pq

p

pqp

S

Required quadratic equation is 02 PSxx

02 3323 pqxqpxpq Correct choice: (B)

SECTIONIIMultiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) for its answer, out of

which ONE OR MORE may be correct.

*37. Let ABC be a triangle such that6

ACB and let a, b and c denote the lengths of the sides opposite to A, B and C

respectively. The value(s) ofx for which 1,1 22 xbxxa and 12 xc is (are)

(A) 32 (B) 31 (C) 32 (D) 34

Sol.:

1121211

6cos

22

22222

xxx

xxxx

112

123

2

3

22

2222

xxx

xxxxx

1213 22 xxxxx 0313232 2 xx

322

3132432322

x 31,32

But 32 x (as 0,, cba 1x )Correct choice: (B)

38. Letfbe a real-valued function defined on the interval ),0( by dttxxf

x

0

sin1ln)( . Then which of the following

statement(s) is (are) true?

(A) )('' xf exists for all ),0( x

(B) )(' xf exists for all ),0( x and xf' is continuous on ),0( , but not differentiable on ),0(

(C) there exists > 1 such that |)(||)('| xfxf for all ),( x

(D) there exists > 0 such that |)('||)(| xfxf for all ),0( x

8/8/2019 Jee 2010 BT

12/61



12

Sol.: xx

xf sin11

Clearly xf exists and continuous for all 0x .

Also xf is not differentiable at point where 1sin x .

xgx

xxfxf 1ln , where xdttxgx

sin1sin1

0

Now dtt

x

0

sin1 is always continuous, increasing and has range ,0 .

So 0xg for some ,0k as 2sin1 x for ,0x

Also 01

ln x

x for ,kx

0 xfxf for all ,kkx

xfxf for all ,x , where kk Correct choice: (B), (C)

k1

y = lnx

y = 1/x

*39. LetA andB be two distinct points on the parabola xy 42 . If the axis of the parabola touches a circle of radius rhavingABas its diameter, then the slope of the line joiningA andB can be

(A)r

1 (B)

r

1(C)

r

2(D)

r

2

Sol.: Let mid-point ofAB is rh ,

Equation ofAB is

sincos

ryhx

SoA andB can be rh sin,cos

But these will satisfy xy 42

C(h, r)

y2 = 4x

B

A

xO

hr cos4sin 2

04cos2sin2sin 222 hrr Sum of roots = 0 CBAC

cos2sinr r

2tan

Correct choice: (C), (D)

Alternative:

Slope of21

2tt

AB

y-coordinate of mid-point of rttAB 21

slope ofAB can ber

2or

r

2

222 2, tt B

A

x

y

121 2, tt r

40. The value(s) of

1

0

2

44

1

)1(dx

x

xxis (are)

(A) 7

22(B)

105

2(C) 0 (D)

2

3

15

71

8/8/2019 Jee 2010 BT

13/61



13

Sol.: dx

x

xdxxdxxxdx

x

xxxI

1

0

2

61

0

5

1

0

24

1

0

2

224

1

441

1

21

7

224

3

4

5

4

3

2

7

1

5

1

1

114

6

4

7

1

5

11

0

2

24dx

xxx

Correct choice: (A)

*41. Let1Z

and 2z be two distinct complex numbers and let 21)1( tzztz for some real number twith 10 t . If Arg(w)

denotes the principal argument of a nonzero complex number w, then

(A) |||||| 2121 zzzzzz (B) )(Arg)(Arg 21 zzzz

(C) 01212

11

zzzz

zzzz(D) )(Arg)(Arg 121 zzzz

Sol.: 21 ,, zCzBzA are collinear

ACBCAB

2121 zzzzzz

21 zzArgzzArg

121 zzArgzzArg

Equation of line CBA ,, is 01212

11

zzzz

zzzz

Correct choice: (A), (C), (D)

SECTION IIIComprehension Type

This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the secondparagraph 2 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out


Paragraph for Questions 42 to 44

Letp be an odd prime number and pT be the following set of 2 2 matrices :

}1.....,,2,1,0{,,: pcba

ac

baATp

42. The number ofA in pT such thatA is either symmetric or skew-symmetric or both, and det (A) divisible byp is

(A)2)1( p (B) )1(2 p (C) 1)1( 2 p (D) 12 p

Sol.: ForA to be symmetric cb

22 ba should be multiple ofp

)()( baba should be multiple ofp

Either )(or)( baba should be multiple ofp

ba can be multiple ofp in )1( p ways (excluding 0 ba )

)( ba can be multiple ofp inp ways (including 0 ba )

Total number of ways = 2p1

Correct choice: (D)

8/8/2019 Jee 2010 BT

14/61



14

43. The number ofA in pT such that the trace ofA is not divisible byp but det (A) is divisible byp is

[Note: The trace of a matrix is the sum of its diagonal entries.]

(A) )1)(1( 2 ppp (B) 23 )1( pp (C) 2)1( p (D) )2)(1( 2 pp

Sol.: As Atr is not amultiple ofp

0a

bca 2 will be a multiple ofp

if remainder ofp

a2

andp

bcis same.

Let when 2a divided byp remainder is r

when bc is divided byp for different values of }1...,1,0{ pc we get distinct remainder }1...,1,0{ p

there exist only one value ofb for which remainder ofbc divided byp gives r

a and c can be taken in )1( p ways each and b can be choose in only one way.

Total ways 1)1(2

p

Correct choice: (C)

44. The number ofA in pT such that det (A) is not divisible byp is

(A) 22p (B) pp 53 (C) pp 33 (D) 23 pp

Sol.: The total number ofAin3

pTp (i)

Now theA for which det A is divisible byp

Case-I: 0a

bc should be divisible byp

either 0b or 0c or both

total ways 12 p

Case-II: 0a

Total ways2)1( p [as in previous question]

Total number of ways = 121 2 pp 2p (ii)

Required ways = 23 pp

Correct choice: (D)


The circle 0822 xyx and hyperbola 149

22

yx

intersect at the pointsA andB.

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

(A) 02052 yx (B) 0452 yx (C) 0843 yx (D) 0434 yx

Sol.: Let sin4,cos14 be a point on the circle.

Equation of tangent to the circle is cos14sincos yx = 0

sin

cos14

sin

cosxy (i)

Condition of tangency for hyperbola is

2222

bmac

8/8/2019 Jee 2010 BT

15/61



15

4sin

cos9

sin

cos116

2

2

2

2

020cos32cos3 2

3

2cos and 10 (rejected)

35sin

Slope of tangent from (i) is m =5

2

35

32

sin

cos

5

4

5

2;

35

3214

5

2

xyxy

425 xy 0452 yx

Correct choice: (B)

*46. Equation of the circle withAB as its diameter is

(A) 0241222 xyx (B) 0241222 xyx

(C) 0122422 xyx (D) 0122422 xyx

Sol.:22

8 xxy

14

8

9

22

xxx

369724 22 xxx 0367213 2 xx 03667813 2 xxx

0)6(6613 xxx 6,13

6 xx

Clearly 0,6 is the centre.

if 6x , 36482 y 122 y 32y

End points of diameter are 32,6,32,6 32r and 0,6C

126 22 yx 123612 22 yxx 0241222 xyx

Correct choice: (A)

SECTION IVInteger Answer Type

This section contains TEN questions. The answer to each of the questions is a single-digitinteger, ranging from 0 to 9. Thecorrect digit below the question numbers in the ORS is to be bubbled.

47. If a

and b

are vectors in space given by5

2 jia

and

14

32 kjib

, then the value of bababa 22 . is

Sol.: 1,1 ba

and 0. ba

]2..2[.2 22 abbabbaaba

541.220.201 2

2

baabba

Ans. 5

8/8/2019 Jee 2010 BT

16/61



16

*48. The line 12 yx is tangent to the hyperbola 12

2

2

2

b

y

a

x. If this line passes through the point of intersection of the

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

Sol.: 12 xy (i)

222

bmamxy (ii)

(i) and (ii) identical 2m and 14 22 ba (iii)

Satisfying 12 yx by 12

e

a

224 ea (iv)

Also 1222 eab 2222 aeab (v)From (iii) and (iv) 122 eb 11 222 eae 1a ae 2 2e

Ans. 2

49. If the distance between the plane dzyAx 2 and the plane containing the lines4

3

3

2

2

1

zyxand

5

4

4

3

3

2

zyxis 6 , then || d is

Sol.: Let d.c. of normal to the plane is nml ,, :

0432 nml

0543 nml

121

nml

D.R. of normal to the plane is 1,2,1

Equation of plane is dzyx 2

Satisfying by 3,2,1 0d

Required plane is 02 zyx (i)

Given plane is dzyAx 2 (ii)

From (i) and (ii) we get 1A because planes are parallel.

Now, 6121

0

222

d 6d

Ans. 6

50. For any real numberx, let [x] denote the largest integer less than or equal to x. Letfbe a real valued function defined on the

interval [10, 10] by

evenis[x]if][1

odd,is[x]if][)(

xx

xxxf . Then the value of

10

10

2

cos)(10

dxxxf is

8/8/2019 Jee 2010 BT

17/61



17

Sol.: Period of 2xf

Period of 2cos x

So, period of 2cos xxf

dxxxfdxxxf 2.5

2.5

10

10

coscos

dxxxf cos102

0

1 0235 1 2 3 5

Graph off(x)

dxxxfdxxxf

2

1

1

0

coscos10

dxxxdxxx

2

1

1

0

coscos110

dxxxdxxx

1

0

2

1

cos110cos110

0

22010coscoscoscos10

22

1

0

2

1

2

1

1

0

dxxdxxxdxxxxdx

2

40

So, 4cos10

10

10

2

dxxxf

Ans. 4

51. Let be the complex number3

2sin

3

2cos

i . Then the number of distinct complex numbers z satisfying

z

z

z

1

1

1

2

2

2

= 0 is equal to

Sol.: Given determinant is 0

1

1

1

2

2

2

z

z

z

By 3211 CCCC

0

1

12

2

zz

zz

z

0

10

102

22

2

z

z

z

0]3[ 222 zz 02 zz 0z

So number of solution is 1.

Ans. 1

8/8/2019 Jee 2010 BT

18/61



18

*52. Let ,100...,,2,1, kSk denote the sum of the infinite geometric series whose first term is!

1

k

kand the common ratio is

k

1.

Then the value of

100

1

22

13!100

100

k

kSkk is

Sol.:

kR

k

kA

k

k

1

!

1

!1

1

kSk for 1k

The value of

100

1

22

13!100

100

k

kSkk

k

k

SkkSS

100

3

221

2

131!100

100

100

3

2

!1!2

110

!100

100

kk

k

k

k

!99

100

!98

99........

!3

4

!2

3

!2

3

!1

210

!100

100 2 3

!99

100210

!100

100 2

Ans. 3

53. The number of all possible values of , where 0 , for which the system of equations 3sin)(3cos)( xyzzy ;

zyx

3sin23cos23sin ; 3sin3cos)2(3sin)( yzyxyz have a solution ),,( 000 zyx with 000 zy , is

Sol.: 03cos1

3cos1

3sin zy

x

03sin21

3cos21

3sin zy

x

03sin3cos1

3cos21

3sin zy

x

The system of equations has solution, so

0

3sin3cos3cos23sin

3sin23cos23sin

3cos3cos3sin

0

3sin3cos00

3sin23cos21

3cos3cos1

3sin

03cos3cos3sin3sin

03cos,03sin , 13tan

But

possiblenotiswhich0

1,03sinfor03siny

03cos ( for 01

,03cos z

, not possible)

13tan ( for 3 solutions4

3,

12

5,

12

)

So number of solution = 3.

Ans. 3

8/8/2019 Jee 2010 BT

19/61



19

54. Letfbe a real-valued differentiable function on R (the set of all real numbers) such that 1)1( f . If the y-intercept of the

tangent at any point ),( yxP on the curve )(xfy is equal to the cube of the abscissa ofP, then the value of )3(f is equal

to

Sol.: According to given condition: 3xdx

dyxy

xdxx

ydxxdy

2

dxx

x

yd c

x

x

y

2

2

But 11 f 2

3c The curve is

2

3

2

3xx

y 93 f

Ans. 9

*55. The number of values of in the interval

2,

2such that

5

nfor ,0n 2,1 and 5cottan as well as

4cos2sin is

Sol.: 5cottan 06cos

2

6

n (i)

3,36

2,

2as

22,

2,

2,

2,

22,

236

Also 4cos2sin

2

2cos4cos Zmm

,2

224

2

26

m or2

22

m (ii)

From (i) and (ii) The number of common solution = 3

Ans. 3

Alternative:

4cos2sin (i)

012sin2sin2 2 012sin212sin 12sin or2

12sin

2,

2 ,2

4

or

12,

12

5

All the three values are satisfying the equation 5cottan

Ans. 3

*56. The maximum value of the expression 22 cos5cossin3sin

1is

Sol.: Maximum value of the expression 22 cos5cossin3sin

1

2

2

53

1

2

2cos42sin33

1

12cos22sin2

31

1

minmin

Ans. 2

8/8/2019 Jee 2010 BT

20/61



20

SOLUTIONS TO IIT-JEE 2010

PHYSICS: Paper-I (Code: 8)

PARTIII

SECTION I


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

57. 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)

IBL(C)

2IBL

(D)4

IBL

Sol.: dRILBdRT 2 2ILBT

Correct choice: (C)

58. An AC voltage source of variable angular frequency and fixed amplitude 0V is connected in series with a capacitance C

and an electric bulb of resistance R (inductance zero). When 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

Sol.:2

2

22 1|| RC

RXZ C

When angular frequency of the source increases the impedance of the circuit decreases, therefore the current in the circuitincreases, hence the power through the resistance increases.

Correct choice: (B)

59. To verify Ohms law, a student is provided with a test resistorTR , a high resistance 1R , a small resistance 2R , two identical

galvanometers1G and 2G , and a variable voltage source V. The correct circuit to carry out the experiment is

G1

G2

V

RT R1

R2

(A)

G1

G2

V

RT R2

R1

(B)

G1

V

RT(C)

R1

G2

R2

G1

V

RT(D)

R2

G2

R1

Sol.: A galvanometer with a high resistance in series behaves as a voltmeter and a low resistance connected in parallel to the

galvanometer behaves as an ammeter.

Correct choice: (C)

8/8/2019 Jee 2010 BT

21/61



21

60. 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 resistances R100,R60,R40 respectively, the

relation between these resistances is

(A)604010 0

111

RRR (B)

604010 0RRR (C) 406010 0 RRR (D)

406010 0

111

RRR

Sol.:P

VR

2

, whereR is the resistance at working temperature and not at room temperature

406010 0

111

RRR

Correct choice: (D)

*61. 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

Sol.: Real gases behave like an ideal gas at very low densities.

As

0M

TR

P T

P

Correct choice: (D)

62. Consider a thin square sheet of side L and thickness t, made of a material

of resistivity . The resistance between two opposite faces, shown by theshaded areas in the figure is

(A) directly proportional toL (B) directly proportional to t

(C) independent ofL (D) independent oft

L

t

Sol.: ttL

LR

Correct choice: (C)

*63. A thin uniform annular disc (see figure) of mass Mhas outer radius 4R

and inner radius 3R. The work required to take a unit mass from point P

on its axis to infinity is

(A) 5247

2

R

GM(B) 524

7

2

R

GM

(C)R

GM

4(D) 12

5

2

R

GM

3R4R

4R

P

Sol.: Potential due to disc on its axis =

xxRG 222

(R is radius of disc,x is the distance from the centre of the disc on its axis)

By superposition principle potential due to annular disc at point P,

222

2212 xRxRGVP

PVVW

As RRRR 3,4 21 , Rx 4 and 22 34 RR

M

Work done 5247

2

R

GM 0V

Correct choice: (A)

8/8/2019 Jee 2010 BT

22/61



22

*64. A block of mass m is on an inclined plane of angle . The coefficient of friction between the block andthe plane is and tan . 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

)cos(sin1 mgP to )cos(sin2 mgP , the frictional forcefversus P graph will look like

P

(A)P2

P1 P

f

(B)

P2P1 P

f

(C)

P2

P1

P

f

(D)P2P1

P

f

Sol.: (i) when sinmgP

sinmgfP

Pmgf sin

P

mgsin

f

tan >

(ii) when P = mg sin ,f= 0

(iii) when sinmgP

Pmgf sin

P2

P1 P

f

Correct choice: (A)

P

mgsin f

tan >

SECTIONIIMultiple Correct Choice Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE

OR MORE may be correct.

*65. A student uses a simple pendulum of exactly 1 m 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 Tin measuring T, the time period, is 0.05 seconds

(B) Error Tin measuring T, the time period, is 1 second(C) Percentage error in the determination ofg is 5%

(D) Percentage error in the determination ofg is 2.5%

8/8/2019 Jee 2010 BT

23/61



23

Sol.: t= nT;n

tT

n

tT

05.0

20

1T sec.

g

l

n

t 2

2

224

t

lng

; 100

2100

t

t

g

g= 5%

Correct choices: (A), (C)

66. A ray OP of monochromatic light is incident on the face AB of prism ABCD

near vertexB 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 120 900 750

1350

600

B

C

DA

P

O

Sol.: 3sin

60sin 0

rr= 300

Let ic be the critical angle, then,3

1sin ci

300 < ic < 450

Also, the angle between incident and emergent ray is 900

Correct choices: (A), (B), (C)750

1350

600

B

C

DA

300

600

450

450

300

600

67. 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 ofQ1 the electric field is zero

(D) at a finite distance to the right ofQ2 the electric field is zero

Q1 Q2

Sol.: Number of field lines magnitude of charge and2

r

KQE

Correct choices: (A), (D)

*68. One mole of an ideal gas in initial stateA undergoes a cyclic processABCA, as

shown in the figure. Its pressure atA is P0. Choose the correct option(s) from thefollowing

(A) Internal energies atA andB are the same

(B) Work done by the gas in processAB is 400 nVP

(C) Pressure at Cis4

0P

(D) Temperature at Cis4

0T

A

T

V

V0

4V0 B

C

T0

Sol.: ProcessAB is isothermal, so WAB = P0V0 ln 4 and UAB = 0

ProcessBC is unknown (as its not given whether the lineBCis directed towards origin or not)

Correct choices: (A), (B)

8/8/2019 Jee 2010 BT

24/61



24

*69. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its

direction and moves with a speed of 2 ms1. Which of the following statement(s) is (are) correct for the system of these two

masses?

(A) Total momentum of the system is 3 kg ms1

(B) Momentum of 5 kg mass after collision is 4 kg ms1

(C) Kinetic energy of the centre of mass is 0.75 J

(D) Total kinetic energy of the system is 4 J

Sol.: Applying conservation of linear momentum

2521051 VV (i)

As collision is elastic,

VV 22 (ii)

112 ms3,ms1 VV

Total momentum of the system = 3 kg m/s

m/s2

1

51

01cm

VV

J75.05121

E.K 2cmcm V

Correct choices: (A), (C)

1kg V 5kg 1kg V25kg2m/s

Before collision After collision


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 beanswered. Each of these questions has four choices (A), (B), (C) and (D) out


Paragraph for Question Nos. 70 to 72

When a particle of mass m moves on the x-axis in a potential of the form ,)( 2kxxV it performs simple harmonic motion. The corresponding time period is proportional to

,k

m as can be seen easily using dimensional analysis. However, the motion of the

particle can be periodic even when its potential energy increases on both sides ofx = 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 thex-axis. Its potential energy is

)0()( 4 xxV for | x | near the origin and becomes a constant equal to V0 for

0|| Xx (see figure).

V(x)

V0

X0x

*70. If the total energy of the particle isE, it will perform periodic motion only if

(A)E< 0 (B)E> 0 (C) V0 > E> 0 (D)E> V0Sol.: 0,0 EVE

Correct choice: (C)

*71. For periodic motion of small amplitudeA, the time period Tof this particle is proportional to

(A)m

A (B)m

A

1(C)

mA

(D)

mA

1

Sol.: Using dimensional analysis

Correct choice: (B)

*72. The acceleration of this particle for 0|| Xx is

(A) proportional to V0 (B) proportional to0

0

mX

V(C) proportional to

0

0

mX

V(D) zero

Sol.: 0VU andF = 0dx

dU

Correct choice: (D)

8/8/2019 Jee 2010 BT

25/61



25

Paragraph for Question Nos. 73 to 74

Electrical resistance of certain materials, known as superconductors,

changes abruptly from a nonzero 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 temperatureTC(B) is a function of the magnetic field strength B. The dependence of

TC(B) onB is shown in the figure. B

TC(B)

O

TC(0)

73. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different

magnetic fieldB1 (solid line) andB2 (dashed line). If B2 is larger than B1, which of the following graphs shows the correctvariation ofR with Tin these fields?

T

R

O

(A)

B2B1

T

R

O

(B)

B2

B1

T

R

O

(C)

B1B2

T

R

O

(D)

B1

B2

Sol.: Higher the magnetic field, lower is the temperature at which resistance abruptly becomes zero.

Correct choice: (A)

74. A superconductor has TC (0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its TC decreases to 75 K. 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) = 70 K

Sol.: As the magnetic field decreases, TCincreases

Correct choice: (B)

SECTION IVInteger 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. When two identical batteries of internal resistance 1 each are connected in series across a resistor R, the rate of heat

produced inR isJ1. When the same batteries are connected in parallel acrossR, the rate isJ2. IfJ1 = 2.25J2 then the value of

R in is

8/8/2019 Jee 2010 BT

26/61



26

Sol.: In first case

I rR 2

2

Rate of heat produced inR, RrR

RIJ

2

21

2

2

R

r r

I

In second case

2

'r

R

I

Rate of heat produced inR, Rr

R

J

2

2

2

As 21 25.2 JJ

rR 4 = 4Answer is 4

r

R

r

'I

*76. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum

intensity in the emission spectrum ofA is at 500 nm and in that ofB is at 1500 nm. Considering them to be blackbodies, what

will be the ratio of the rate of total energy radiated byA to that ofB?

Sol.: For a black body,energyradiated per unit time = 4AT (whereR is radius of the body)

42

B

A

B

A

B

A

T

T

R

R

E

E (i)

According to Weins displacement law, Tm constant

42

500

1500

18

6

B

A

E

E

= 9

Answer is 9

*77. When two progressive waves

262sin3and)62sin(4 21 txytxy are superimposed, the amplitude of the

resultant wave is

Sol.: cos2 2122

21 AAAAA = 525

2cos34234 22

Answer is 5

*78. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional areas is

4.9 107 m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic

motion of angular frequency 140 rad s1. If the Youngs modulus of the material of the wire is n 109 Nm2, the value ofn is

Sol.:m

K

Lm

YA

1.01

109.410140

792

n

Lm

YA

n = 4

Answer is 4

8/8/2019 Jee 2010 BT

27/61



27

*79. A binary star consists of two starsA (mass 2.2MS) and B (mass 11MS), whereMS is the mass of the sun. They are separated

by distance dand 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

Sol.:CM

2.2MS 11MS

5d/6 d/6

11 IL and 22 IL

66/11

6/52.26/112

22

2

1

2

1

dM

dMdM

I

I

L

L

S

SS

Answers is 6

*80. Gravitational acceleration on the surface of a planet is ,11

6g where g is the gravitational acceleration on the surface of the

earth. The average mass density of the planet is3

2times 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 kms1

will be

Sol.:2

2

EE

PP

P

E

E

P

eE

eP

R

R

R

R

M

M

V

V

(i)

As2

3

2 3

4

R

RG

R

GMg

gR

P

E

E

P

E

P

g

g

R

R

(ii)

22

E

P

P

E

E

P

eE

eP

g

g

V

V2

11

6

2

3

VeP = 3 km/s

Answer is 3

*81. A piece of ice (heat capacity = 2100 J kg1 C1 and latent heat = 3.36 105 J kg1) of mass m grams is at 5C 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 ofm is

Ans.: Heat required for melting ice = 3361036.3100

1 5 J

Heat used for raising temperature of ice from (50C to 00C) = 336420Q = 84 J

TmsQ 5210084 m

m = 0.008 kg = 8 gm

Answer is 8

*82. 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% off0. 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 ms1.

8/8/2019 Jee 2010 BT

28/61



28

Sol.: Apparent frequency received by car is

V

VVff 001 (i)

The frequency of the reflected wave is

0

11 'VV

Vff =

0

00

VV

VVf (ii)

Differentiating '1f with respect to V0

3302

330

660

330

330330' 02

0

002

0

00

0

1 f

V

ff

V

VV

dV

df

0

10

'165

f

dfdV 7 km/h

Answer is 7

83. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm,

the magnification of its image changes from m25 to m50. The ratio50

25

mm is

Sol.:20

1

25

11

1

V

cm1001 V

Similarly

100

3

100

25

50

1

20

11

2

V

3

1002 V cm

503

100

25

100

50

25

m

m= 6

Answer is 6

84. An -particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de Broglie

wavelengths are and p respectively. The ratio

p

, to the nearest integer, is

Sol.: The momentum of a charged particle accelerated by potential difference V, qmVP 2

de-Broglie wavelength,P

h

3222

422

Vme

Vmep

Answer is 3

8/8/2019 Jee 2010 BT

29/61



29

SOLUTIONS TO IIT-JEE 2010Paper-II (Code: 8)

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

A. General:1. This Question Paper contains 24 pages having 57 questions. The question paper CODE is

printed on the right hand top corner of this sheet and also on the back page of thisbooklet..

2. No additional sheets will be provided for rough work.3. Blank papers, clipboards, log tables, slide rules, calculators, cellular phones, pagers and

electronic gadgets in any form are not allowed.4. Log and Antilog tables are given in the paper.

5. The answer sheet, a machine-gradable Objective Response Sheet (ORS), is providedseparately.

6. Write your Registration No., Name, Name of the centre and sign with pen in theappropriate boxes. Do not write these anywhere else.

B. Question paper format and Marking scheme:7. The question paper consists of 3 parts (Chemistry, Mathematics and Physics), and each

Part consists offour Sections.

8. For each question in Section I,you will be awarded 5 marks if you have darkened onlythe bubble corresponding to the correct answer and zero mark if no bubbles aredarkened. In all other cases, minus two (2) markwill be awarded.

9. For each question in Section II, you will be awarded 3 marks if you darken only thebubble corresponding to the correct answer and zero mark if no bubble is darkened. Nonegative marks will be awarded for incorrect answers in this Section.

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

11. For each question in Section IV,you will be awarded 2 marks for each row in which youhave darkened the bubble(s) corresponding to the correct answer. Thus, each question inthis section carries a maximum of8 marks.

8/8/2019 Jee 2010 BT

30/61



30

SOLUTIONS TO IIT-JEE 2010CHEMISTRY: Paper-II (Code: 8)

PARTI

SECTION I


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

Note: Questions with (*) mark are from syllabus of class XI.

1. In the reaction H3C C

O

NH2

(1) NaOH/Br2

(2) C

O

Cl

T

the structure of the Product T is

(A)H3C C

O

OC

O(B) NH

C

O

CH3

(C) H3C NH

O

C

(D)H3C C

O

NHC

O

Sol.: H3C C

O

NH2

(1) NaOH/Br2Me NH2

HCl

C

O

ClMe NHC

O

Correct choice: (C)

*2. Assuming that Hunds rule is violated, the bond order and magnetic nature of the diatomic molecule B 2 is

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

Sol.: Molecular orbital configuration of B2 as per the condition will be

1s2, *1s2, 2s2, *2s2, 2yP2

Bond order of B2 =2

46= 1.

B2 will be diamagnetic.

Correct choice: (A)

3. The compounds P, Q and S

HO

COOH

P H3C

OCH3

Q

CO

O

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

(A)

HO

COOH

H3C

OCH3 CO

O

NO2 NO2

O2N

(B)

HO

COOH

H3C

OCH3 CO

O

NO2NO2

NO2

8/8/2019 Jee 2010 BT

31/61



31

(C)

HO

COOH

H3C

OCH3 CO

O

NO2NO2

NO2

(D)

HO

COOH

H3C

OCH3 C O

O

NO2NO2

NO2

Sol.:

OH

CO2H

(P)

HNO3/H2SO4

OH

CO2H

NO2

CH3

OCH3

(Q)

HNO3/H2SO4NO2

CH3

OCH3

O

(S)

HNO3/H2SO4O

O

O NO2

Correct choice: (C)

*4. The species having pyramidal shape is

(A) SO3 (B) BrF3 (C) 23SiO (D) OSF2

Sol.: OSF2:2

N=

2

26 = 4.

It has 1 lone pair.

S (Shape is trigonal pyramidal)

FO F

..

The shapes of SO3, BrF3 and2

3SiO are triangular planar, T-shaped and triangular planar respectively.

Correct choice: (D)

5. 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

Sol.: Ni(CO)4 and Ni(PPh3)4 have Ni in zero oxidation state, while CO and Ph 3P are strong field ligands. Thus, both these

molecules have sp3 hybridization and are diamagnetic. [Ni(CN)4]2 has electronic configuration of Ni2+ as 3d8 and

hybridization is dsp2. It is diamagnetic. [NiCl4]2 has sp3 hybridization with 2 unpaired electrons giving spin only magnetic

moment as 8 or 2.82 B.M.

Correct choice: (B)

6. 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%

8/8/2019 Jee 2010 BT

32/61



32

Sol.: Packing efficiency =squareofArea

circleseffectivebyoccupiedArea

= 100L

r22

2

= 100)r22(

r2

2

2

= 1004

= 78.54%.

Correct choice: (D)

SECTIONIIInteger Type

This section contains a group of 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 no. in the ORS is to be bubbled.

*7. 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

graphs below. If the work done along the solid line path is ws and that along the dotted line path is wd, then the integer closest

to the ratio wd/ws is

0.5

1.0

1.5

2.0

25

3.0

3.5

4.0

4.5

0.00.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

P

(atm)

V (lit.)

b

a

Sol.: wd =

2

5

3

211

2

34 =

3

516

wd =3

26 L atm

ws =2.303 RT log2/1

5.5=2.303 PV log 11

ws =4.606 1.04 =4.8 L atm

s

d

w

w=

8.4

326

= 1.80 ~2.0

The answer is 2.

*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

Sol.: Fluorine generally shows O and1 oxidation states while sodium shows 0 and +1 oxidation state.

The answer is 2.

9. Silver (atomic weight = 108 g mol1

) has a density of 10.5 g cm3

. The number of silver atoms on a surface of area 1012

m2

can be expressed in scientific notation as y 10x. The value of x is

Sol.: Silver has cubic close packed structure. It has fcc unit cell of edge length a cm.

10.5 =323 )a(10023.6

1084

a =3/1

23 5.1010023.6

1084

= 4.087 108 cm.

(i) Assuming that surface area given is the area of any extended face of a FCC unit cell.

Area of square face = (4.087 108)2 cm2

Number of Ag atoms per unit area =2

a

2

Surface area = 1012 m2 = 108 cm2.

8/8/2019 Jee 2010 BT

33/61



33

Number of Ag atoms in the given surface area =28

8

)10087.4(

102

= 1.197 107.

x = 7.

(ii) Assuming that surface area given is the area corresponding to closest packed layer.

2 a = 4r, r =4

a2 =4

210087.4

8

= 1.44 108

In a closest packed layer, the arrangement appears like

2)r2(4

36 =

4

36(2 1.44 108)2 =

161007.244

36

= 21.5 1016 cm2

Number of silver atoms on surface =16

8

105.21

103

= 0.139 108 = 1.39 107.

x = 7.

The answer is 7.

10. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)] is

Sol.: Rh

Cl PPh3

CO NH3

Rh

Cl PPh3

NH3 CO

Rh

Cl CO

NH3 PPh3

The number of geometrical isomers is 3.The answer is 3.

*11. The total number of diprotic acids among the following is

H3PO4 H2SO4 H3PO3 H2CO3 H2S2O7

H3BO3 H3PO2 H2CrO4 H2SO3

Sol.: Diprotic acids are H2SO4, H3PO3, H2CO3, H2S2O7, H2CrO4 and H2SO3.

The answer is 6.


This Section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered.

Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for questions 12 to 14

Two aliphatic aldehydes P and Q react in the presence of aqueous K 2CO3 to give compound R, which upon treatment with

HCN provides compound S. On acidification and heating, S gives the product shown below.

O O

H3C

H3C OH

8/8/2019 Jee 2010 BT

34/61



34

12. The compounds P and Q respectively are

(A)

H3C

CH

CH3

CH

andH3C

CH

OO

(B)

H3C

CH

CH3

CH

andH

CH

OO

(C)H3C CH2

CH3

CH

andH3C

CH

OO

CH (D)H3C CH2

CH3

CH

andH

CH

OO

CH

Sol.: CH3CHCHO + OH CH3C

CHO + H2O

CH3 CH3

CH3CCHO + CH2 = O CH3CCHO CH3CCHO

H2O

OH

CH3 CH3 CH3

CH2O CH2OH

CH3CCH=O + HCN CH3CCHCN CH3CCHCOOHH+

/H2O

CH3 H3C OH OH

CH2OHCH2OH CH2OH

(S)

O OH3C

H3C OH

Intramolecular esterificationH3C

(P)

(Q)

(R)

Correct choice: (B)

13. The compound R is

(A)

H3C

CH3C

C

CH2OH

O

H (B)

H3C

CH3C

C

CHOH

O

H

H3C

(C)CH

CHC

CH2OH

O

HH3C

CH3

(D)

CH

CH

H3C

C

CHOH

O

HH3C

CH3

Sol.: Correct choice: (A)

14. The compound S is

(A)CH

CH

C

CH2

CN

O

HH3C

CH3

(B)H3C

CH3C

C

CH2

CN

O

H

(C)CH

CHCH

CH2OH

CN

OHH3C

CH3

(D)H3C

CH3C

CH

CH2OH

CN

OH

Sol.: Correct choice: (D)

8/8/2019 Jee 2010 BT

35/61



35

Paragraph for questions 15 to 17

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

Sol.: The spherically symmetric state S1 of Li2+ with one radial node is 2s. Upon absorbing light, the ion gets excited to state S2,

which also has one radial node. The energy of electron in S2 is same as that of H-atom in its ground state

12

2

n En

ZE where E1 is the energy of H-atom in the ground state

2

12

n

E)3( for Li2+

3nEE 1n

State S2 of Li2+ having one radial node is 3p.

Orbital angular momentum quantum number of 3p is 1.

Energy of state 112

2

1 E25.2E)2(

)3(S

Correct choice: (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

Sol.: Correct choice: (C)

*17. The orbital angular momentum quantum number of the state S2 is

(A) 0 (B) 1 (C) 2 (D) 3

Sol.: Correct choice: (B)

SECTION IV(Matrix Type)

This section contains 2 questions. Each question contains has four statements (A, B, C and D) given in Column I and fivestatements (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.

18. Match the reactions is Column I with appropriate options in Column II.

Column I Column II

(A) N2Cl + OHNaOH/H2O

0C N=N OH (p) Racemic mixture

(B) H3C

C

CH3

O

CH3CCCCH3

OHOH

H3C CH3

H2SO

4

CH3

CH3

(q) Addition reaction

(C) C CH

O

CH3

1. LiAlH4

2. H3O+

OH

CH3

(r) Substitution reaction

(D) HS ClBase S (s) Coupling reaction

(t) Carbocation intermediate

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

8/8/2019 Jee 2010 BT

36/61



36

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

Sol.: (A)(p), (s)

(CH3)2SiCl2 + 2H2O HCl2 (CH3)2Si(OH)2 .Polymn

[(CH3)2SiO]n.

(B)(p), (q), (r), (t)

XeF4 + H2O Xe + XeO3 + H2F2 + O2

SiO2 + 4HF SiF4 + 2H2Oglass

SiF4 + 2HF H2[SiF6]Soluble hexathorosilicic(IV) acid

(C)(p), (q)

Cl2 + H2O HCl + HOCl

(D)(p), (q)

VCl5 + 7H2O [V(H2O)6]3+ + 3Cl+ HCl + HOCl

8/8/2019 Jee 2010 BT

37/61



37

SOLUTIONS TO IIT-JEE 2010MATHEMATICS: Paper-II (Code: 8)

PARTII

SECTION I


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

20. Letf be a real-valued function defined on the interval (1, 1) such that x

x txfe

0

4 12)( dt, for all ),1,1(x and

let 1f be the inverse function of f . Then )2()'( 1f is equal to

(A) 1 (B)3

1(C)

2

1(D)

e

1

Sol.: dttxfe

x

x 0

4 12)( (i)

Putting 0x

2)0( f 0)2(1 f xxff ))(( 1 1)()'(.))((' 11 xfxff )0('

1)2()'( 1

ff

Differentiate equation (i) with respect tox

1)()(' 4 xxfexfe xx

Putting 0x

12)0(' f 3)0(' f 3

1)2()'( 1 f

Correct choice: (B)

21. A signal which can be green or red with probability5

4and

5

1respectively, is received by station A and then transmitted to

stationB. The probability of each station receiving the signal correctly is4

3. If the signal received at station B is green, then

the probability that the original signal was green is

(A)5

3(B)

7

6(C)

23

20(D)

20

9

Sol.: GR Received signal is green

GO Original signal was green

If the eventXYZdenotes original single sent toA wasX, received atA was Yand received atB isZ

)(

)(

G

GG

G

G

RP

ROP

R

OP

)()()()(

)()(

RGGPRRGPGGGPGRGP

GGGPGRGP

4

3

4

1

5

1

4

1

4

3

5

1

4

3

4

3

5

4

4

1

4

1

5

4

4

3

4

3

5

4

4

1

4

1

5

4

23

20

80

4680

40

Correct choice: (C)

8/8/2019 Jee 2010 BT

38/61



38

22. If the distance of the point )1,2,1( P from the plane zyx 22 , where 0 , is 5, then the foot of the perpendicular

from P to the plane is

(A)

3

7,

3

4,

3

8(B)

3

1,

3

4,

3

4(C)

3

10,

3

2,

3

1(D)

2

5,

3

1,

3

2

Sol.: 5441

241

155 ( )0

10

Plane is 01022 zyx

Let the foot of perpendicular be F, equation of line PFis

2

1

2

2

1

1

zyx

So the point Fcan be taken as

F

P

)21,22,1(

It will satisfy equation of plane.

01042441 3

5

Point Fis

3

7,

3

4,

3

8

Correct choice: (A)

*23. Let }4,3,2,1{S . The total number of unordered pairs of disjoint subsets ofS is equal to

(A) 25 (B) 34 (C) 42 (D) 41

Sol.: Given set 4,3,2,1S

For disjoint subsetsA andB

Sa

Ba

Aa

Ba

Aa

Ba

Aa

Then total number of disjoint subsets will be 12

134

= 41

Correct choice: (D)

*24. For 10,....1,0r let rr BA , and rC denote, respectively, the coefficient ofr

x in the expansions of ,)1(10

x 20)1( x and

30)1( x . Then

10

1

1010 )(

r

rrr ACBBA is equal to

(A) 1010 CB (B) )( 101021010 ACBA (C) 0 (D) 1010 BC

Sol.:

10

1

21010

302010

2010

r

rrr CCCCC =

10

1

10

1

21010

30102010

20

r r

rrr CCCCC

= 11 1020103010301020 CCCC = 101010201030 BCCC Correct choice: (D)

25. Two adjacent sides of a parallelogram ABCD are given by kjiAB 11102 and kjiAD 22 . The side AD isrotated by an acute angle in the plane of the parallelogram so that AD becomes 'AD . If 'AD makes a right angle with theside AB , then the cosine of the angle is given by

(A) 9

8

(B) 9

17

(C) 9

1

(D) 9

54

8/8/2019 Jee 2010 BT

39/61



39

Sol.: ||||. ADABADAB )90cos(

31540 sin

sin9

8;

9

17cos

90

D C

BA

'D

Correct choice: (B)

SECTIONII(Integer Type)

This section contains 5 questions. The answer to each question is a singledigit integer, ranging from 0 to 9. The correctdigit below the question no. in the ORS is to be bubbled.

26. Let kbe a positive real number and let

02

2021

120

and

122

212

2212

kk

kk

kk

B

kk

kk

kkk

A .

If det 610)adj(det)adj( BA , then [k] is equal to

[Note: adjMdenotes the adjoint of a square matrixMand [k] denotes the largest integer less than or equal to k]

Sol.: 122

12120

2212

det

kk

kk

kkk

A

kkkkk

kkk

k 81212

1122

100

2412

12 2

312 k

0det B [B is skew symmetric matrix of order 3]

66 1012 k 1012 k 29k [k] = 4

Ans. 4

27. Letf be a function defined onR(the set on all real numbers) such that

,)2012()2011()2010)(2009(2010)(' 432 xxxxxf for all Rx . Ifg is a function defined on R with values in theinterval ),0( such that ))(ln()( xgxf , for all Rx , then the number of points inR at which g has a local maximum is

Sol.:

xgxg

xf .1

xgxfxg

As 0xg , So xg will change sign as

2009 2011

++

one point of maxima is 2009x

Ans. 1

*28. Let 11321 ,....,, aaaa be real numbers satisfying 151 a , 0227 2 a and 212 kkk aaa for 11...,,4,3k .

If 9011

... 21122

21

aaa, then the value of

11

... 1121 aaa is equal to

8/8/2019 Jee 2010 BT

40/61



40

Sol.: 151 a

As 211 kkkk aaaa

So ......,,, 321 aaa are in A.P.

Also 0227 2 a

2

312 aa

2

3d

9011

......211

22

21

aaa 90115

11

111

1

2 r

dr

9035150225 2 dd 027307 2 dd 0973 dd 7

9,3

d

But7

9d

2

3d

Now 030302

11030

2

1

11

...... 1121

daaa

Ans. 0

29. Consider a triangle ABCand let ba, and c denote the lengths of the sides opposite to vertices A, B and C respectively.

Suppose 6a , 10b and the area of the triangle is 315 . If ACB is obtuse and ifrdenotes the radius of the incircle of

the triangle, then 2r is equal to

Sol.: 315

315sin1062

1C

2

3sin C

obtuseis

3

2CC

120

10036cos

2c

C

14c

Hence semi perimeter 15s

3r 32 r

Ans. 3

30. Two parallel chords of a circle of radius 2 are at a distance 13 apart. If the chords subtend at the center, angles ofk

and

k

2, where 0k , then the value of [k] is [Note : [k] denotes the largest integer less than or equal to k]

Sol.: 1321 pp

132

cos2cos2

kk

2

13

2coscos

kk

3k

Ans. 3

/2k

/k p1

p22

2

A B

DC

O

8/8/2019 Jee 2010 BT

41/61



41

SECTION III(Paragraph Type)

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered.

Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Consider the polynomial 32 4321)( xxxxf . Let s be the sum of all distinct real roots of )(xf and let || st .

31. The real number s lies in the interval

(A)

0,

4

1(B)

4

3,11 (C)

2

1,

4

3(D)

4

1,0

Sol.: Given function 32 4321)( xxxxf is monotonically increasing

as Rxxxxf 0]136[2)(' 2 ]48/5)4/1[(12 2 x

)(xf will have only one real root

Since 0

2

1.

4

3

ff s lies in

2

1,

4

3

Correct choice: (C)

32. The area bounded by the curve )(xfy and the lines 0x , 0y and tx , lies in the interval

(A)

3,

4

3(B)

16

11,

64

21(C) (9, 10) (D)

64

21,0

Sol.: As || st

4

3,

2

1t

Hence required area will be the shaded region shown in

the figure.

t1/23/4

O

Area t

dxxxxtA

0

32 )4321()( 432 tttt which is increasing

Area

4

3,

2

1AA

256

525,

16

15

Correct choice: (A)

33. The function )(' xf is

(A) increasing in

4

1,t and deceasing in

t,

4

1(B) decreasing in

4

1,t and increasing in

t,

4

1

(C) increasing in ),( tt (D) decreasing in ),( tt

Sol.:

4

124 xxf xf increases

,

4

1x

xf decreases

4

1,x

Correct choice: (B)

8/8/2019 Jee 2010 BT

42/61



42


Tangents are drawn from the point )4,3(P to the ellipse 149

22

yx

touching the ellipse at pointsA andB.

*34. The coordinates ofA andB are

(A) (3, 0) and (0, 2) (B)

15

1612,

5

8and

5

8,

5

9

(C)

15

1612,

5

8and (0, 2) (D) (3, 0) and

5

8,

5

9

Sol.: Let 49 2 mmxy be the tangent passing through point 4,3 to the given ellipse.

4934 22 mm 2

1m and indeterminate.

one tangent is2

5

2

1 xy at

5

8,

5

9

jee 2010 bt

Documents