JEE Main Solved Paper 2010

Instructions

1. This test consists of 90 questions.

2. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics

having 30 questions in each part of equal weightage. Each question is allotted 4 marks for correct

response.

3. Candidates will be awarded marks as stated above for correct response of each question. 1/4 mark will

be deducted for indicating incorrect response of each question. No deduction from the total score will

be made if no response is indicated for an item in the answer sheet.

4. There is only one correct response for each question. Filling up more than one response in any

question will be treated as wrong response and marks for wrong response will be deducted according

as per instructions.

Physics1. A rectangular loop has a sliding connector PQ of length l and resistance R and it is moving

with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane

ofthepaper.Thethreecurrents I I1 2, and I are

(a) I IB v

RI

B v

R1 2

2= = =

l l, (b) I I

B v

RI

B v

R1 2

3

2

3= = =

l l, (c) I I I

B v

R1 2= = =

l(d) I I

B v

RI

B v

R1 2

6 3= = =

l l,

2. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time

taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time

takenforthechargetoreducetoone-fourthitsinitialvalue.Then,theratiot

t

1

2

willbe

(a) 1 (b)1

2(c)

1

4(d) 2

R

I2I

I1

v R R

Q

P l

Solved Paper 2010

All India Engineering Entrance Examination

Directions Q. Nos. 3-4 contain

Statement I and Statement II. Of the four

choices given after the statdements, choose

the one that best describes the two

statements.

(a) Statement I is true, Statement II is true;Statement II is the correct explanation ofStatementI.

(b) Statement I is true, Statement II is true;Statement II is not the correct explanationofStatementI.

(c) StatementIisfalse,StatementIIistrue.

(d) StatementIistrue,StatementIIisfalse.

3. Statement I Two particles moving in thesame direction do not lose all their energyinacompletelyinelasticcollision.

Statement II Principle of conservation ofmomentum holds true for all kinds ofcollisions.

4. Statement I When ultraviolet light isincident on a photocell, its stoppingpotential is V0 and the maximum kineticenergy of the photoelectrons is Kmax .When the ultraviolet light is replaced byX-rays,both V0 and Kmax increase.

Statement II Photoelectrons are emittedwith speeds ranging from zero to amaximum value because of the range offrequenciespresentintheincidentlight.

5. A ball is made of a material of density where oil water< < with oil and waterrepresenting the densities of oil andwater, respectively. The oil and water areimmiscible. If the above ball is inequilibrium in a mixture of this oil andwater, which of the following picturesrepresentsitsequilibriumposition?

6. A particle is moving with velocityv i j= +k y x( $ $), where k is a constant. The

generalequationforitspathis

(a) y x= +2 constant

(b) y x2 = + constant

(c) xy = constant

(d) y x2 2= + constant

7. Two long parallel wires are at a distance2d apart. They carry steady equal currentflowing out of the plane of the paper asshown. The variation of the magnetic fieldalongtheline XX isgivenby

8. In the circuit shown below, the key K isclosed at t = 0. The current through thebatteryis

AIEEE Solved Paper 2010 105

dd

B

X X'(a)

water

oil water(a) (b)

water

oil

oil

water(c) (d)

oil

dd

B

X X'(b)

dd

B

X X'(c)

dd

B

X X'(d)

L

R2

R1

V K

(a)VR R

R R

1 2

12

22+

at t = 0 andV

R2at t =

(b)V

R2at t = 0 and

V R R

R R

( )1 2

1 2

+at t =

(c)V

R2at t = 0 and

VR R

R R

1 2

12

22+

at t =

(d)V R R

R R

( )1 2

1 2

+at t = 0 and

V

R2at t =

9. The figure shows the position-time( )x t- graph

of one-dimensional motion of a body of mass

0.4kg.Themagnitudeofeachimpulseis

(a) 0.4Ns(b) 0.8Ns (c) 1.6Ns(d) 0.2Ns

Directions Q. Nos. 10-11 are based on the

following paragraph.

A nucleus of mass M m+ is at rest and decaysinto two daughter nuclei of equal mass

M

2each.

Speed of light is c.

10. The binding energy per nucleon for the parentnucleus is E1 and that for the daughter nucleiis E2 .Then

(a) E E2 12= (b) E E1 2>

(c) E E2 1> (d) E E1 22=

11. Thespeedofdaughternucleiis

(a) cm

M m

+

(b) cm

M

2

(c) cm

M

(d) c

m

M m

+

12. A radioactive nucleus (initial mass number Aand atomic number Z) emits 3 -particles and2 positrons. The ratio of number of neutrons tothatofprotonsinthefinalnucleuswillbe

(a)A Z

Z

8

4(b)

A Z

Z

4

8

(c)A Z

Z

12

4(d)

A Z

Z

4

2

13. A thin semi-circular ring of radius rhas a positive charge q distributeduniformly over it. The net field E atthecentre O is

(a)q

r4 2 02pi

$j (b) q

r4 2 02pi

$j

(c) q

r2 2 02pi

$j (d)q

r2 2 02pi

$j

14. The combination of gates shownbelowyields

(a) ORgate (b) NOTgate

(c) XORgate (d) NANDgate

15. A diatomic ideal gas is used in a carengine as the working substance. Ifduring the adiabatic expansion partof the cycle, volume of the gasincreases from V to 32 V, theefficiencyoftheengineis

(a) 0.5 (b) 0.75

(c) 0.99 (d) 0.25

16. If a source of power 4 kW produces1020 photons/second, the radiationbelong to a part of the spectrumcalled

(a) X-rays (b) ultravioletrays

(c) microwaves (d) -rays

17. The respective number of significantfigures for the numbers 23.023,0.0003and 2.1 10 3 are

(a) 5,1,2 (b) 5,1,5

(c) 5,5,2 (d) 4,4,2

106 JEE Main Solved Papers

Oi

j

0 2 4 6 8 10 12 14 16

2

x (m)

t (s)

X

A

B

18. In a series L-C-R circuit, R = 200 andthe voltage and the frequency of the mainsupply is 220 V and 50 Hz respectively.On taking out the capacitance from thecircuit the current lags behind the voltageby 30. On taking out the inductor fromthe circuit, the current leads the voltageby 30. The power dissipated in the L-C-Rcircuitis

(a) 305W (b) 210W

(c) zero (d) 242W

19. Let three be a spherically symmetriccharge distribution with charge density

varying as ( )r rR

=

0

5

4upto r R= ,

and ( )r = 0 for r R> , where r is thedistance from the origin. The electric fieldat a distance r r R( )< from the origin isgivenby

(a)4

3

5

30

0

pi

r r

R

(b)0

04

5

3

r r

R

(c)

4

3

5

40

0

r r

R

(d)

0

03

5

4

r r

R

20. Thepotentialenergyfunctionforthe

forcebetweentwoatomsinadiatomicmoleculeisapproximatelygivenby

U xa

x

b

x( ) =

12 6,where a and b are

constantsand x isthedistancebetweentheatoms.IfthedissociationenergyofthemoleculeisD U x U D= = [ ( ) ],at equilibrium is

(a)b

a

2

2(b)

b

a

2

12

(c)b

a

2

4(d)

b

a

2

6

21. Two identical charged spheres aresuspended by strings of equal lengths.The strings make an angle of 30 witheach other. When suspended in a liquid ofdensity 0.8 g cm 3 , the angle remains the

same. If density of the material of thesphere is 16 g cm 3 , the dielectric

constantoftheliquidis

(a) 4 (b) 3

(c) 2 (d) 1

22. Two conductors have the same resistanceat 0C but their temperature coefficientsof resistance are 1 and 2 . The respectivetemperature coefficients of their seriesandparallelcombinationsarenearly

(a)

1 2 1 22

++,

(b)

1 21 2

2+

+,

(c)

1 2

1 2

1 2

++

,

(d) 1 2 1 2

2 2

+ +,

23. A point P moves in counter-clockwisedirection on a circular path as shown inthe figure. The movement of P is such thatit sweeps out a length s t= +3 5, where s isin metre and t is in second. The radius ofthe path is 20 m. The acceleration of Pwhen t = 2 s isnearly

(a) 13 ms 2 (b) 12 ms 2

(c) 7.2 ms 2 (d) 14 ms 2

24. Two fixed frictionless inclined planemaking an angle 30 and 60 with thevertical are shown in the figure. Twoblock A and B are placed on the twoplanes. What is the relative verticalaccelerationof A withrespectto B?

(a) 4.9 ms 2 inhorizontaldirection

(b) 9.8 ms 2 inverticaldirection

(c) Zero

(d) 4.9 ms 2 inverticaldirection


Y

B

O AX

P x y( , )

20m

60

A

30

B

25. For a particle in uniform circular motionthe acceleration a at a point P R( , ) on thecircle of radius R is (here is measuredfromthe X-axis)

(a) +v

R

v

R

2 2

cos $ sin $ i j

(b) +v

R

v

R

2 2

sin $ cos $ i j

(c) v

R

v

R

2 2

cos $ sin $ i j

(d)v

R

v

R

2 2$ $i j+

Directions Q. Nos. 26-28 are based on

the following paragraph.

An initially parallel cylindrical beam travels in

a medium of refractive index ( )I I= +0 2 ,where 0 and2 are positive constants and I isthe intensity of the light beam. The intensity

of the beam is decreasing with increasing

radius.

26. Asthebeamentersthemedium,itwill

(a) diverge

(b) converge

(c) diverge near the axis and converge near the

periphery

(d) travelasacylindricalbeam

27. The initial shape of the wavefront of the

beamis

(a) convex

(b) concave

(c) convex near the axis and concave near the

periphery

(d) planar

28. Thespeedoflightinthemediumis

(a) minimumontheaxisofthebeam

(b) thesameeverywhereinthebeam

(c) directlyproportionaltotheintensity I

(d) maximumontheaxisofthebeam

29. A small particle of mass m is projected atan angle with the X-axis with an intialvelocity v0 in the xy-plane as shown in the

figure. At a time tv

g< 0

sin,

the angular

momentumoftheparticleis

(a) mgv t02 cos $ j (b) mgv t0 cos $ k

(c) 1

20

2mgv t cos $ k (d) 12

02mgv t cos $ i

30. The equation of a wave on a string of

linear mass density 0.04 kg m 1 is given

by

y 0.02 (m) sin(s) (m)

=

2

0 04 0 50pi

t

.

x

. .

Thetensioninthestringis

(a) 4.0N (b) 12.5N

(c) 0.5N (d) 6.25N


x

y

v0

Chemistry31. The standard enthalpy of formation of

NH3 is 46.0 kJ mol 1 If the enthalpy of

formation of H2 from its atoms is436 kJmol1 and that of N2 is

712 1kJ mol ,the average bond enthalpy

ofNHbondin NH3 is

(a) 964kJmol1 (b) +352kJmol1

(c) +1056kJmol1 (d) 1102kJmol1

32. The time for half-life period of acertain reaction, A products is 1 h.When the initial concentration of thereactant ,A is 2.0 mol L1, how muchtime does it take for its concentration tocome from 0.50 to 0.25 mol L1, if it is azeroorderreaction?

(a) 4h (b) 0.5h

(c) 0.25h (d) 1h

33. A solution containing 2.675g ofCoCl 6NH3 3 (molar mass = 267.5 gmol1) is passed through a cationexchanger. The chloride ions obtained insolution were treated with excess ofAgNO3 to give 4.78 g of AgCl (molar mass

143.5 g mol )1= . The formula of the

complexis

(Atmassof Ag )= 108 u

(a) [Co(NH ) ]Cl3 6 3 (b) [CoCl (NH ) ]Cl2 3 4

(c) [CoCl (NH ) ]3 3 3 (d) [CoCl(NH ) ]Cl3 5 2

34. Considerthereaction,

Cl H S2 2( ) ( )aq aq+ S 2H 2Cl( ) ( ) ( )s aq aq+ ++

The rate equation for this reaction is,rate [Cl ] [H S]2 2= k

Which of these mechanisms is/areconsistentwiththisrateequation?

I. Cl +H S H +Cl +Cl +HS2 2+ +

(slow)

Cl +HS H +Cl +S+ + (fast)

II.H S H + HS2+

(fast, equilibrium)

Cl +HS 2Cl + H +S2 +

(slow)

(a) (II)only

(b) Both(I)and(II)

(c) Neither(I)nor(II)

(d) (I)only

35. If 10 4 3 dm of water is introduced into a

1.0 dm3 flask at 300 K, how many moles of

water are in the vapour phase whenequilibriumisestablished?(Given, vapour pressure ofH O2 at 300K is3170Pa; R = 8.314 JK mol )1 1

(a) 5.56 10 mol3 (b) 1.53 10 mol2

(c) 4.46 10 mol2 (d) 1.27 10 mol3

36. One mole of a symmetrical alkene onozonolysis gives two moles of an aldehydehaving a molecular mass of 44 u. Thealkeneis

(a) propene (b) 1-butene

(c) 2-butene (d) ethene

37. If sodium sulphate is considered to becompletely dissociated into cations andanions in aqueous solution, the change infreezing point of water ( ),Tf when0.01 mole of sodium sulphate is dissolvedin 1 kg of water is( 1.86 K kg mol )1Kf =

(a) 0.0372K (b) 0.0558K

(c) 0.0744K (d) 0.0186K

38. From amongst the following alcohols, theone that would react fastest withconc.HClandanhydrous ZnCl ,2 is

(a) 2-butanol (b) 2-methylpropan-2-ol

(c) 2-methylpropanol (d) 1-butanol

39. Inthechemicalreactions,

the compounds A and B respectivelyare

(a) nitrobenzeneandfluorobenzene

(b) phenolandbenzene

(c) benzene diazonium chloride and

fluorobenzene

(d) nitrobenzeneandchlorobenzene

40. 29.5 mg of an organic compoundcontaining nitrogen was digestedaccording to Kjeldahls method and theevolved ammonia was absorbed in 20mLof 0.1 M HCl solution. The excess of theacid required 15 mL of 0.1 M NaOHsolution for complete neutralisation. Thepercentage of nitrogen in the compoundis

(a) 59.0 (b) 47.4

(c) 23.7 (d) 29.5

41. The energy required to break one mole ofClCl bonds in Cl2 is 242 kJ mol

1. Thelongest wavelength of light capable ofbreakingasingleClClbondis(c t= 3 108 ms1and

NA = 6.02 1023 mol 1)

(a) 594nm (b) 640nm

(c) 700nm (d) 494nm


NH2

NaNO2

HCl, 278 KA

HBF4B

42. Ionisation energy of He+ is19.6 10 J atom .18 1 The energy of

the first stationary state ( )n = 1 of Li2+

is

(a) 4.41 10 J atom16 1

(b) 4.41 10 J atom17 1

(c) 2.2 10 J atom15 1

(d) 8.82 10 J atom17 1

43. Considerthefollowingbromides

Thecorrectorderof S 1N reactivityis

(a) (II) (III) (I)> >(b) (II) (I) (III)> >(c) (III) (II) (I)> >(d) (I) (II) (III)> >

44. Which one of the following has anoptical isomer?(en ethylenediamine)=

(a) [Zn(en)(NH ) ]3 22+

(b) [Co(en) ]33+

(c) [Co(H O) (en)]2 43+

(d) [Zn(en) ]22+

45. On mixing, heptane and octane forman ideal solution. At 373 K, thevapour pressures of the two liquidcomponents (heptane and octane)are 105 kPa and 45 kPa respectively.Vapour pressure of the solutionobtained by mixing 25 g of heptaneand 35 g of octane will be (molarmass of heptane = 100 1g mol and of

octane = 114 1g mol ).

(a) 72.0kPa (b) 36.1kPa

(c) 96.2kPa (d) 144.5kPa

46. The main product of the following reaction is

C H CH CH(OH)CH(CH ) ?6 5 2 3 22 4Conc. H SO

(a)

H C

H

C==C

H

CH(CH )

5 6

3 2

(b)

C H CH

H

C==C

CH

CH

6 5 2 3

3

(c)

H C

H

C==C

CH(CH )

H

5 6 3 2

(d)

H C CH CH

H C

C==CH5 6 2 2

3

2

47. Three reactions involving H PO2 4 are given

below

I. H PO + H O H O + H PO3 4 2 3+

2 4

II. H PO + H O HPO + H O2 4

2 42

3+

III. H PO + OH H PO + O2 4

3 42

In which of the above does H PO2 4 act as an

acid?

(a) (II)only (b) (I)and(II)

(c) (III)only (d) (I)only

48. In aqueous solution, the ionisation constantsfor carbonic acid are

K1710= 4.2 and K2

1110= 4.8

Select the correct statement for a saturated0.034Msolutionofthecarbonicacid.

(a) The concentration of CO32 is 0.034 M

(b) The concentration of CO32 is greater than that of

HCO3

(c) The concentration of H+ and HCO3 are

approximatelyequal

(d) Theconcentrationof H+ isdoublethatof CO32

49. The edge length of a face centered cubic cell ofan ionic substance is 508 pm. If the radius ofthecationis110pm,theradiusoftheanionis

(a) 288pm (b) 398pm

(c) 618pm (d) 144pm


Me Br

Me

Br

(II)

(I)

MeMe

Br

(III)

50. The correct order of increasing basicity of thegivenconjugatebases ( )R = CH3 is

(a) R RCOO < HC C < O3+ 2+ + 2

(b) Na > Mg > Al > O > F+ 2+ 3+ 2

(c) Na > F > Mg > O > Al+ 2+ 2 3+

(d) O > F > Na > Mg > Al2 + 2+ 3+

52. Solubility product of silver bromide is5.0 10 13 . The quantity of potassium

bromide (molar mass taken as120 1g mol ) to

be added to 1 L of 0.05 M solution of silvernitratetostarttheprecipitationofAgBris

(a) 1.2 10 g10 (b) 1.2 10 g9

(c) 6.2 10 g5 (d) 5.0 10 g8

53. The Gibbs energy for the decomposition of

Al O2 3 at500Cisasfollows

2

3Al O

4

3Al+O ,2 3 2 rG 966 kJ mol

1= +

The potential difference needed forelectrolytic reduction of Al O2 3 at 500C isatleast

(a) 4.5V (b) 3.0V

(c) 2.5V (d) 5.0V

54. At 25C, the solubility product of Mg(OH) 2 is

1.0 10 11. At which pH, will Mg2+ ions start

precipitating in the form of Mg(OH) 2 from asolutionof0.001M Mg2+ ions?

(a) 9 (b) 10

(c) 11 (d) 8

55. Percentage of free space in cubic closepacked structure and in body centredpackedstructurearerespectively

(a) 30%and26%

(b) 26%and32%

(c) 32%and48%

(d) 48%and26%

56. Out of the following, the alkene thatexhibitsopticalisomerismis

(a) 3-methyl-2-pentene

(b) 4-methyl-1-pentene

(c) 3-methyl-1-pentene

(d)2-methyl-2-pentene

57. Biurettestisnotgivenby

(a) carbohydrates

(b) polypeptides

(c) urea

(d) proteins

58. The correct order of EM M2 +

/values

with negative sign for the foursuccessive elements Cr, Mn, Fe andCois

(a) Mn>Cr>Fe>Co

(b) Cr>Fe>Mn>Co

(c) Fe>Mn>Cr>Co

(d) Cr>Mn>Fe>Co

59. The polymer containing strongintermolecular forces e.g., hydrogenbonding,is

(a) teflon

(b) nylon-66

(c) polystyrene

(d) naturalrubber

60. For a particular reversible reaction attemperature T, H and S were foundto be both +ve. If Te is the temperatureat equilibrium, the reaction would bespontaneouswhen

(a) T Te > (b) T Te>(c) Te is5times T (d) T Te=


Mathematics61. Considerthefollowingrelations

R x y x y= {( , )| , are real numbers and

x wy= forsomerationalnumber w};

Sm

n

p

qm n p=

, , , and q are integers

suchthat n, q 0 and qm pn= } . Then,(a) R is an equivalence relation but S is not an

equivalencerelation

(b) Neither R nor S isanequivalencerelation

(c) S is an equivalence relation but R is not an

equivalencerelation

(d) R and S bothareequivalencerelations

62. The number of complex numbers z suchthat| | | | | |z z z i = + = 1 1 equals

(a) 0 (b) 1 (c) 2 (d)

63. If and are the roots of the equationx x2 1 0 + = ,then 2009 2009+ is equal to(a) 2 (b) 1 (c) 1 (d) 2

64. Considerthesystemoflinearequations

x 2x x 31 2 3+ + =

2x 3x x 31 2 3+ + =

3x 5x 2x 11 2 3+ + =

Thesystemhas

(a) infinitenumberofsolutions

(b) exactly3solutions

(c) auniquesolution

(d) nosolution

65. There are two urns. Urn A has 3 distinctred balls and urn B has 9 distinct blueballs. From each urn, two balls are takenout at random and then transferred to theother. The number of ways in which thiscanbedone,is

(a) 3 (b) 36 (c) 66 (d) 108

66. If f R: ( , ) 1 1 is a differentiable function

with f( )0 1= and =f ( ) .0 1 Let

g x f f x( ) [ ( ( ) )] .= +2 2 2 Then, g ( )0 is equal to

(a) 4 (b) 4

(c) 0 (d) 2

67. Let f R R: be a positive increasing

function with lim( )

( )x

f x

f x =

31. Then,

lim( )

( )x

f x

f x

2isequalto

(a) 1 (b)2

3(c)

3

2(d) 3

68. Let p x( ) be a function defined on R such

that lim( )

( ),

x

f x

f x =

31 = p x p x( ) ( ),1 for all

x [ , ]0 1, p( )0 1= and p( ) .1 41= Then,

p x dx( )0

1 equals(a) 41 (b) 21

(c) 41 (d) 42

69. A person is to count 4500 currency notes.Let an denotes the number of notes hecounts in the nth minute. Ifa a a1 2 10 150= = = =... and a a10 11, ,...are inAP with common difference 2, then thetimetakenbyhimtocountallnotes,is

(a) 24min (b) 34min

(c) 125min (d) 135min

70. The equation of the tangent to the curve

y xx

= +42

,that is parallel to the X-axis, is

(a) y = 0 (b) y = 1

(c) y = 2 (d) y = 3

71. The area bounded by the curves y x= cos

and y x= sin between the ordinates x = 0

and x =3

2

piis

(a) ( )4 2 2 squnits

(b) ( )4 2 2+ squnits(c) ( )4 2 1 squnits

(d) ( )4 2 1+ squnits

72. Solutionofthedifferentialequation

cos (sin ) ,x dy y x y dx x= <

73. Let a j k= $ $ and c i j k= $ $ $ . Then, the

vector b satisfyinga b c + = 0 anda b = 3,

is

(a) + $ $ $i j k2 (b) 2 2$ $ $i j k +

(c) $ $ $i j k 2 (d) $ $ $i j k+ 2

74. Ifthevectors a i j k= +$ $ $2 ,

b i j k= + +2 4$ $ $ and c i j k= + + $ $ $ aremutually orthogonal, then( , ) is equal to(a) (3,2) (b) (2,3)

(c) (2,3) (d) (3,2)

75. If two tangents drawn from a point P to the

parabola y x2 4= are at right angles, then

thelocusof P is

(a) x = 1 (b) 2 1 0x + =(c) x = 1 (d) 2 1 0x =

76. The line L given byx y

b51+ = passes

through the point (13, 32). The line K is

parallel to L and has the equationx

c

y+ =

31. Then, the distance between L

and K is

(a)23

15(b) 17

(c)17

15(d)

23

17

77. A line AB in three-dimensional spacemakes angles 45 and 120 with thepositive X-axis and the positive Y-axis,respectively. If AB makes an acute angle withthepositive Z-axis,then equals(a) 30 (b) 45 (c) 60 (d) 75

78. Let S be a non-empty subset of R. Considerthefollowingstatement

P : There is a rational number x S suchthat x > 0 .

Which of the following statements is thenegationofthestatement P?

(a) There is a rational number x S such that x 0(b) There is no rational number x S such that

x 0(c) Everyrationalnumber x S satisfies x 0(d) x S and x x 0 isnotrational

79. Let cos( ) + = 45

and sin( ) , = 513

where 04

pi, . Then, tan 2 is equalto

(a)25

16(b)

56

33

(c)19

12(d)

20

7

80. The circle x y x y2 2 4 8 5+ = + +

intersects the line 3 4x y m = at twodistinctpoints,if

(a) < < 85 35m(b) <

84. The number of 3 3 non-singularmatrices, with four entries as 1 andallotherentriesas0,is

(a) lessthan4 (b) 5

(c) 6 (d) atleast7

85. Let f R R: bedefinedby

f xk x x

x x( )

,

,=

+ >

2 1

2 3 1

if

if.

If f has a local minimum at x = 1,thenapossiblevalueof k is

(a) 1 (b) 0 (c) 1

2(d) 1

Directions This section contains 5

multiple choice questions numbered 86

to 90. Each question contains Statement

I (Assertion) and Statement II (Reason).

Each question has 4 choices (a), (b), (c)

and (d) out of which only one is correct.

(a) Statement I is true, Statement II istrue;Statement II is the correct explanationofStatementI

(b) Statement I is true, Statement II istrue;Statement II is not the correctexplanationofStatementI

(c) Statement I is true, Statement II isfalse

(d) Statement I is false, Statement II istrue

86. Four numbers are chosen at random(without replacement) from the set{1,2,3,...,20}.

Statement I The probability thatthe chosen numbers when arranged

insomeorderwillformanAP,is1

85.

Statement II If the four chosen numbersfrom an AP, then the set of all possible valuesof common difference is{ , , , , } . 1 2 3 4 5

87. Let S j j C jj

110

1

10

1= =

( ) ,

S j C jj

210

1

10

=

=

and S j C jj

32 10

1

10

=

=

StatementI S3

955 2=

StatementII

S1890 2= and S2

810 2=

88. Statement I The point A(3, 1, 6) is the mirrorimage of the point B(1, 3, 4) in the planex y z + = 5 .

Statement II The plane x y z + = 5 bisectsthe line segment joining A(3,1, 6) andB(1,3,4).

89. Let f R R: be a continuous function defined

by f xe ex x

( ) =+

1

2.

StatementI f c ,( ) =1

3forsome c R .

StatementII 01

2 2< f x ,( ) forall x R .

90. Let A be a 2 2 matrix with non-zero entriesand A I2 = , where I is 2 2 identitymatrix.

Define Tr( )A = sum of diagonal elements ofA and| |A = determinantofmatrix A.

StatementI Tr( )A = 0

StatementII | |A = 1


Answers1. (b) 2. (c) 3. (a) 4. (d) 5. (b) 6. (d) 7. (a) 8. (b) 9. (b) 10. (c)

11. (b) 12. (b) 13. (c) 14. (a) 15. (b) 16. (a) 17. (a) 18. (d) 19. (b) 20. (c)

21. (c) 22. (d) 23. (d) 24. (d) 25. (c) 26. (b) 27. (d) 28. (a) 29. (c) 30. (d)

31. (b) 32. (c) 33. (a) 34. (d) 35. (d) 36. (c) 37. (b) 38. (b) 39. (c) 40. (c)

41. (d) 42. (b) 43. (a) 44. (b) 45. (a) 46. (a) 47. (a) 48. (c) 49. (d) 50. (d)

51. (d) 52. (b) 53. (c) 54. (b) 55. (b) 56. (c) 57. (a) 58. (a) 59. (b) 60. (b)

61. (c) 62. (b) 63. (c) 64. (d) 65. (d) 66. (b) 67. (a) 68. (b) 69. (b) 70. (d)

71. (a) 72. (a) 73. (a) 74. (a) 75. (c) 76. (d) 77. (c) 78. (c) 79. (b) 80. (b)

81. (b) 82. (b) 83. (c) 84. (d) 85. (c) 86. (c) 87. (a) 88. (a) 89. (a) 90. (c)

Solutions

Physics

1. A moving conductor is equivalent to a battery ofemf

= vBl (motionalemf)Equivalentcircuit I I I= +1 2

Applying Kirchhoffslaw

I R IR vB1 0+ =l ...(i)

I R IR vB2 0+ =l ...(ii)AddingEqs.(i)and(ii),weget

2 2IR IR vB+ = l

IvB

R=

2

3

l

I IvB

R1 2

3= =

l

2. Energy stored in capacitor

Uq

C Cq e

q

Cet z t= = =

1

2

1

2 2

2

002

22( / ) /

(where, = CR)

Now, U U eit

=2 /

1

2

2 1U U ei it

= /

1

2

2 1=

e t / t12

2=

ln

Now, q q e t= 0/

1

40 0

2q q e t= /

t 2 4 2 2= = ln ln

t

t1

2

1

4=

3. If it is a completely inelastic collision, thenm v m v m v m v1 1 2 2 1 2+ = +

vm v m v

m m=

+

+1 1 2 2

1 2

KE = +p

m

p

m12

1

22

22 2

As p1 and p2 both simultaneously cannot be zero,therefore total KE cannot be lost.

4. Since, the frequency of ultraviolet light is less thanthe frequency of X-rays, the energy of eachincident photon will be more for X-rays

KEphotoelectron = hStopping potential is to just stop the fastestphotoelectron

Vh

e e0 =

So, KEmax and V0 bothincreases.

But KE ranges from zero to KEmax because of lossof energy due to subsequent collisions before theelectron to be ejected and not due to range offrequencies of the incident light.

5. Given, oil water< , Bin between = ($ )j

Towards x, net magnetic field will add up and

direction will be ( $ ) j .

Towards x, net magnetic field will add up and

direction will be ($ )j .

R

I

I

I2

I2

I1

I1

RR

m1 m2

v1 v2

8. At t = 0, inductor behaves like an infinite resistanceSo, at t = 0, i

V

R=

2

and at t = , inductor behaves like a conductingwire i.e., resistance less wire.

Then, iV

R

V R R

R R= =

+

eq

( )1 2

1 2

(QR1 and R2 are in parallel)

9. From the graph, it is a straight line, so uniformmotion. Because of impulse direction of velocitychanges as can be seen from the slope of thegraph.

(Qimpulse ( )I f t= I m a t=

=

= m v v

tt mv mv

( )2 12 1)

Initialvelocity, v1 = =2

21ms 1

Finalvelocity, v2 = = 2

21ms 1

pi mv= =1 0 4. N-s

pf mv= = 2 0 4. N-s

J p p= = f i 0 4 0 4. .

= 0 8. N-s (J = impulse)

| | .J = 0 8 N-s

10. After decay, the daughter nuclei will be more stablehence, binding energy per nucleon will be morethan that of their parent nucleus.

11. Conserving the momentum0

2 21 2=

Mv

Mv (Qinitialvelocity = 0)

v v1 2= ...(i)

Now, from energy conservation and mass-energyequivalence

mc M v M v2 12

221

2 2

1

2 2= + ...(ii)

mc M v2 12

2=

2 2

12mc

Mv=

v cm

M1

2=

12. In positive beta decay, a proton is transformed intoa neutron and a positron is emitted.

p n e+ + +0

Number of neutrons initially was A Z

Number of neutrons after decay ( )A Z 3 2(due to alpha particles) + 2 1 (due to positivebeta decay)

The number of protons will reduce by 8.

[as 3 2 (due to alpha particles)

+ 2 (due to positive beta decay)]

Hence, atomic number reduces by 8.

So, the ratio of number of neutrons to that of

protons =

A Z

Z

4

8

13. Linear charge density, pi

=

q

r

Qpi

pi= = =

q

l

q

rl r, ( )for semi-circle

Smallcharge, dq =q

rrd

pi

E dEK dq

r= =

sin ( $ ) sin ( $ ) j j2(Qcos componentwillbecancelledout)E

K

r

qr

rd= 2 pi sin ( $ )j

= Kr

qd

2 0pi

pisin ( $ )j

= q

r2 2 02pi

( $ )j

14. Truth table for given combination isA B X

0 0 0

0 1 1

1 0 1

1 1 1

ThiscomesouttobetruthtableofORgate.

Gates P and Q willworkasNOTgate.


X

Y

ddEcos

dE

rd

r

dEsin

X

A

B

P

Q

R

15. The efficiency of cycle is = 1 21

T

T

Foradiabaticprocess,

T V =1 constant

Fordiatomicgas, = 75

T V T V1 11

2 2

1 =

T TV

V1 2

2

1

1

=

T T1 2

7

51

32=

( ) = T25 2 52( ) / = T2 4

T T1 24=

= = =11

4

3

4075.

16. As, power of source = = 4 10 103 20 h

QP

E=

number of photons =

4 10

10 6 023 10

3

20 34.

= .6 64 1016 Hz

TheobtainedfrequencyliesinthebandofX-rays.

17. From the rules of significant figures, it is obviousthat, 23.023 has significant digits. 0.0003 has onlyone non-zero after decimal, so has 1 significantdigit and 2.1 10 3 have only two significant digits.

(Refer to Rules)

18. The given circuit is under resonance as X XL C= .Hence,powerdissipatedinthecircuitis

PV

R= =

2

242 W

19. Apply shell theorem, the total charge upto distancer can be calculated as follows

dq r dr= 4 2pi

=

4

5

4

20pi r dr

r

R

( )Qdq d v=

=

45

40

23

pi r dr rR

dr

dq q r drr

Rdr

r = = 4

5

40

23

0pi

=

45

4 3

1

40

3 4

pi rR

r

Astheelectricfield, Ekq

r=

2

=

1

4

14

5

4 3 402 0

3 4

pipi

r

r r

R

Er r

R=

0

04

5

3

20. The potential energy for diatomic molecule isU x

a

x

b

x( ) =

12 6

U x( )= = 0

As, FdU

dx

a

x

b

x= = +

12 613 7

Atequilibrium, F = 0

xa

b

6 2=

Ua

a

b

b

a

b

b

aat equilibrium =

=

2 2 42

2

D U x Ub

a= = =[ ( ) ]at equilibrium

2

4

21. From FBD of sphere, using Lamis theorem

F

mg= tan ...(i)

When suspended in liquid, as remainsunchanged,

F

mgd

=

1

tan ...(ii)

Using Eqs. (i) and (ii), we get

F

mg

F

mgd

=

1

,where FF

K =

F

mg

F

mg Kd

=

1

or K

d

.

.

=

=

=

1

1

1

10 8

16

2


mg

F

T

22. Let R0 be the initial resistance of both conductors. Attemperature their resistances willbe,

R R1 0 11= +( ) and R R2 0 21= +( ) For series combination, R R Rs = +1 2

R R Rs s0 0 1 0 21 1 1( ) ( ) ( )+ = + + + where, R R R Rs 0 0 0 02= + =

2 1 20 0 0 1 2R R Rs( ) ( )+ = + + or

s =

+1 22

Forparallelcombination,

RR R

R R =

+1 2

1 2

RR R

R Rp p0

0 1 0 2

0 1 0 2

11 1

1 1( )

( ) ( )

( ) ( )+ =

+ +

+ + +

where, RR R

R R

Rp 0

0 0

0 0

0

2=

+=

R R

Rp

0 02

1 2 1 22

0 1 221

1

2( )

( )

( )+ =

+ + +

+ +

as 1 and 2 aresmallquantities.

1 2 isnegligible.

or

p=

+

+ +1 2

1 22 ( )

=+

+

1 2 1 22

12

As ( ) 1 22+ isnegligible

p =+1 22

23. Given that, s t= +3 5 Speed, v

ds

dtt= = 3 2

andrateofchangeofspeed, at = =dv

dtt6

Tangentialaccelerationat t = 2 s,

at = =6 2 12 ms 2

andat t v= = = 2 3 2 122 1s ms, ( )

Centripetalacceleration, av

Rc =

2

=

144

20

2ms

Netacceleration = +a at c2 2

14 2ms

24. For the motion of block along inclined planemg masin =

a g= sinwhere, a is along the inclined plane.

The vertical component of acceleration is gsin2 .Therefore, the relative vertical acceleration of A

with respect to B is

gg

(sin sin )2 2 260 302

= =4.9 ms

(inverticaldirection)

25. For a particle in uniform circular motion,

a =v

R

2

towards centre ofcircle

a i j= v

R

2

( cos $ sin $ )

or a i j= v

R

v

R

2 2

cos $ sin $

26. As intensity is maximum at axis, therefore will bemaximum and speed will be minimum on the axisof the beam. So, beam will converge.

27. For a parallel cylindrical beam, wavefront will beplanar.

28. The speed of light at the surface or interface of themedium is maximum while on the axis of themedium is minimum.

29. The angular momentum of the projectile is givenby

L r v= m( )

L i j= + m v t v t gt0 0

21

2cos $ ( sin $

+ [ cos $ ( sin ) $]v v gt0 0 i j

=

mv t gt0

1

2cos $ k

= 1

20

2mgv t cos $ k

30. The tension in the stringOncomparingwithstandardequation,

y a t kx= sin ( )

T vk

= = 22

2= 0 04

2 0 004

2 0 50

2

2.

( / . )

( / . )

pi

pi= 6.25 N

where, = linearmassdensity,and k = waveconstant.


ac

acP R,( )

X

Y

Chemistry

31. Given, 12

3

2N H NH2 2 3( ) ( ) ( )g g g+

H .f = 46 0 kJ mol 1 (i)

2H H2( ) ( )g g ; Hf = 436 kJ mol 1 (ii)

2N( ) ( )g g N2 ; Hf = 712 kJ mol 1 (iii)

Now,onmultiplyingEq.(i)by2andEq.(ii)by3,weget

N H NH2 32 3 2( ) ( ) ( );g g g+

H f = 92 0. kJmol1

6 3H H2( ) ( )g g ; H f = 1308 kJmol1

2N N2( ) ( )g g ; H f = 712 kJmol1

OnaddingEqs.(i),(ii)and(iii),weget

2 6 2 3N H NH( ) ( ) ( )g g g+ ;

H f = 2112 1kJ mol

Thus,theenergyrequiredtobreakthe NH3molecule

= =2112

21056 kJmol1

Average bond enthalpy of NH bond in NH3

molecule = =1056

3352 kJ mol1

AlternateSolution

NH N H2 231

2

3

2( ) ( ) ( )g g g +

H H f = ( )NH3 = ( )46 = 46 kJmol1

Also, H H =

3 N H + 1

2HN N +

3

2HH H

46 3=

HN H + 1

2712( )+

3

2436( )

HN H =1

31056[ ]= +352 kJmol1

32. Given that, [ ]A 0 12= mol Lt1 2 1/ = h

Forzeroorderreaction,

tA

k1 2

0

02/

[ ]= or, k

A

t0

0

1 22=

[ ]

/

Onputtingthevalue,weget

k02

2 11=

=

[ ]

So, k0 1=

and kX

t0 =

or t =

=0.50 0.25

10.25 h

33. Mole of CoCl 6NH 2.675267.5

= 0.013 3 =

AgNO Cl AgCl (white)3( ) ( )aq aq+ Molesof AgCl

4.78

143.5= 0.03=

0.01mol CoCl 6NH3 3 gives0.03molAgCl.

1 mol CoCl 6NH3 3 ionises to give 3 mol Cl.

Hence, the formula of compound is[Co(NH ) ]Cl3 6 3.

34. Slowest step is the rate determining step. Thus, incase (I), rate law is given as

rate = k[Cl ][H S]2 2

While for the reaction given in case (II), rate law isgiven as

rate [Cl ][HS ]2=k

Hence, only mechanism (I) is consistent with thegiven rate law.

35. The volume occupied by water molecules invapour phase is ( )1 10 4 dm3, that is

approximately ( )1 10 3 3 m .

p V n RTvap H O2=

3170(Pa) 1 10 (m ) n3 3 H O2 = (mol)

8 314. (JK1 mol1) 300(K)

n OH23170 10

8 314 3001 27 10

33

=

=

.. mol

36. / (i) The general formula of aldehyde compoundis C H O2n n . First calculate the value of n.

(ii) Alkene is symmetrical, therefore, only singletype of aldehyde is produced as a product.

C H O 442n n =

C Hn n2 44 16 28= =

n = 2

Since,thealkeneissymmetrical,thenthestructureis CH CH CH CH3 3 ==

Thus, CH CH==CH CH3 32-butene (ii) Zn/H O

(i) O

2

3

2CH CH ==O3Acetaldehyde

37. Na SO 2Na + SO2 4 + 42vant Hoff factor of dissociation for Na SO = 32 4

T i K mf f=

= . .3 1 86 0 01 Q m.

.= = 0 01

10 01

= 0.0558 K


38. The reaction of alcohol with conc. HCl andanhydrous ZnCl2 follows S 1N pathway, so greaterthe stability of carbocation formed, faster is thereaction.

2-methylpropan-2-ol gives 3 carbocation. Hence,it reacts rapidly with conc. HCl and anhydrousZnCl2 (Lucas reagent).

CH OH H H C OH3 3 2 C

CH

CH

C

CH

CH

++

+

3

3

3

3

+

H O

3Fast

+ Cl3

2

H C H CC

CH

CH

C

CH

CH

3

3

3

3

Cl

39.

40. Weight of organic compound .= 29 5 mgNH + HCl NH Cl3 4

HCl + NaOH NaC(Remaining) 15 0.1 M

= 1.5 mmol

l +H O2

TotalmillimoleofHCl = 2

Millimoleusedby NH 2 1.5 0.53 = =

Weightof NH 0.5 17 mg 8.5 mg3 = =

Weightofnitrogen14

178.5 mg 7 mg= =

%nitrogen7

29.5100 23.7%= =

AlternateSolution

Massoforganiccompound = 29.5 10 3 g

InitialmolesofHCltaken

= 0.1 = 20 10 2 103 3

MolesofNaOHreactedwithexcessofHCl

= 0.1 15 10 3

MolesofexcessofHCl = 1.5 10 3

MolesofHClthatreactedwithNH3= 2 10 103 31.5

Molesof NH3 absorbed = 0.5 10 3

MolesofNin NH 0.53 = 10 3

MassofNinorganiccompound

= 14 10 30.5 g

%Ninorganiccompound =

14 10

10100

3

3

0.5

29.5

= 23.7%

41. Energy, E N hA= = Nhc

A or = N hc

EA

or =

. .6 626 10 3 10 6 02 10

242 10

34 8 23

3

or = =494 10 m 494 nm9

42. / According to Bohrs atomic model,E

Z

n

2

2

or EKZ

n=

2

2(where, k = constant)

For n = 1,wecanwrite

( )

( )

( )

( )

E

E

Z

Z

1

1

2

2

He

Li

He

Li

+

2+

+

2+

=

E1 for He+

= .19 6 10 18 J atom 1

=

19 6 10 4

9

18

1

.

E( )Li 2+

or E.

12

1819 6 10 9

4( )Li +

=

= 4.41 10 J atom17 1

43. Higher the stability of carbocation, faster is thereaction because S 1N reactions involve theformation of carbocation intermediate.

44. Complex [Co(en) ]3 3+ has no plane of symmetryand centre of symmetry, thats why it is opticallyactive.


NH2 NaNO2HCl, 278 K

N NCl+

HBF4

Balz-Schiemann

reaction

Benzene

diazonium chlorideF

+ N + BF + HCl2 3

>CH2

Me(2 Hs)

+

>Me

+

(6 Hs)

Me

(III)

(I)

Me

+

(2 allylic)

(II)

Me

+

Co

en

en

enCo

en

en

en

3+3+

45. / Vapour pressure of the solution can becalculated by the formulap X p X pT H H O O= + . So, first calculate themole fractions of heptane and octane.

p X p X pT

= + H H O O

X .H =

+=

25

10025

100

35

114

0 45

X O =

+=

35

11425

100

35

114

0.55

p . .T = + =0 45 105 0 55 45 72 kPa

46.

47. Only in reaction (II), H PO2 4 gives H+ to H O2 , thusbehaves as an acid.

48. H CO H + HCO2 3 + 3 ; K1710= 4.2

HCO H + CO3 +

32

; K .2114 8 10=

K K1 2>>

[H ] [HCO ]3+

= ; K2 =+[H ][CO ]

[HCO ]

32

3

So, [ ]CO32

= = K .2114 8 10

AlternateSolution

H CO H HCO H O2 3 2 3( ) ( ) ( ) ( )a Oq l aq aq+ + +

30.034 a a b a b+

HCO H CO H O2 332

3 ++ +( ) ( ) ( ) ( )a Oq l aq aqa b b a b+

TheCO32 concentration cannot be 0.034as H CO2 3

and HCO3 are weak acids. HCO3

dissociates very

little in second step, such that [ ]~HCO3

M and

since a b> , so [ ] [ ]HCO3 3 > CO 2 .

Therefore, concentration of [ ]~H3O+

a

So, [ ] [ ]CO H O3 3+2

Concentration of [ ]H O3+ and [ ]HCO3

are almostequal and its value is aM.

49. For fcc arrangement,2( )r r+ + = edge length

2 110 508( )+ =r

So, r = 144 pm

50. R RC

O

O C ==

O

O

: In carboxylate ion,

the negative charge is present on oxygen, a mostelectronegative element here, thus it is resonancestabilised.

HC C : Carbon is sp hybridised, so itselectronegativity is increased higher relative tonitrogen.

NH :

2 Nitrogen is more electronegative than sp3

hybridised C-atom.

From the above discussion, it is clear that the order

of the stability of conjugated bases is as

R RCOO HC C N H

2 > > >

And higher is the stability of conjugated bases,

lower will be basic character. Hence, the order of

basic character is as

R RCOO HC C NH

2 < < > > >

AlternateSolution

Theorderofacidityofthegivencompoundsis

RCO H HC CH NH2 3> > > R H

Theorderofbasicityoftheirconjugatebasesis

R RCO NH2 2 < <

54. Mg(OH) Mg 2OH2 2+ +

Ksp2+ 2[Mg ][OH ]=

[OH ][Mg ]

10sp

2

4

+

= =

= =

K 1 10

1010

11

3

8 M

pOH 4= and pH 10=

55. Packing fraction of ccp is 74%%freespacein ccp = 26%

Packingfractionof bccis68%

%freespacein bcc 32%=

56. CH CH C CH

H

CH ==CH3 2

3

*2

3-methyl-1-pentene

It has one chiral centre.

Therefore, it shows optical isomerism.

57. / Biuret test is characteristically given by the

compound having C

O

NH

functional

group.

Polypeptides, proteins and urea have C

||NH

O

(peptide) linkage while

carbohydrates have glycosidic linkages. So, testof carbohydrates should be different from that ofother three. Biuret test produces violet colour onaddition of dil. CuSO4 to alkaline solution of acompound containing peptide linkage.

58. Usually across the first transition series, thenegative values for standard electrode potential

decrease except for Mn due to stable

d 5-configuration.

So, correct order : Mn > Cr > Fe > Co

59. Nylon-66, has amide linkage and hydrogen bonds

are formed between C

O

NH

group of

successive chains.

60. G H T S= . At equilibrium, G = 0For a reaction to be spontaneous, G should benegative, so T should be greater than Te .

Mathematics

61. Since, the relation R is defined asR x y x y= {( , )| , are real numbers and x wy= for

some rational number w}

(i) Reflexive xRx x wx =

w = 1 Rationalnumber

Therelation R isreflexive.

(ii) Symmetric xRy yRx/ as 0 1R 0 1( )but1R0 1 0= w ( )

whichisnottrueforanyrationalnumber.

Therelation R isnotsymmetric

Thus, R isnotequivalencerelation.

Now, for the relation S is defined as

Sm

n

p

qm n p q=

, , , and integers such that

n q, 0 and qm pn= }

(i) Reflexivem

nR

m

nmn mn = (true)

The relation S is reflexive.

(ii) Symmetricm

nR

p

qmq np =

np mqp

qR

m

n=

The relation S is symmetric.

(iii) Transitivem

nR

p

qand

p

qR

r

s

mq np= and ps rq=

mq ps np rq =

ms nr=

m

n

r

s

m

nR

r

s=

The relation S is transitive.

The relation S is equivalence relation.

62. / If z x iy= + is complex number, then anequation of a circle with centre at z0 and radiusr is| |z z r =0 .

We have,| | | | | |z z z i = + = 1 1


n nH N (CH ) NH + HO C (CH ) CO H2 2 6 2 2 2 4 2

(2 1)H On 2

Nylon 66-

NH(CH ) NH C ( CH ) C)2 6 2 4

O

O n(1, 0)(1,0)

O

==

| + |z i

(0, 1)

Clearly, z is the circumcentre of the triangle formedby the vertices (1, 0) and (0, 1) and (1, 0), which isunique.

Hence, the number of complex number z is one.

63. / The quadratic equation ax bx c2 0+ + = hasroots and .Then, = + =b

a

c

a,

Also, if ax bx c2 0+ + =

Then, xb b ac

a=

2 42

We know that 1, , 2 are cube roots of unity.

1 02+ + = ( )Q3 1=

and = +

= 1 3

2

1 3

2

2i i,

Since, and are roots of the equationx x2 1 0 + = .

+ = =1 1, x

i=

1 32

xi

=+1 3

2

or =1 3

2

i

x = or 2

Thus, = 2, then = or = , then = 2 (where, 3 1= )Hence, 2009 2009 2009 2 2009+ = + ( ) ( )

= + [( ) ( ) ] 3 669 2 3 1337

= +[ ] 2

= =( )1 1 ( )Q1 02+ + =

64. The given system of linear equations can be put inthe matrix form.

UsingCramersrule,

D =

1 2 1

2 3 1

3 5 2

=

1 2 1

0 1 1

0 1 1

(

)

by

and

R R R

R R R

2 2 1

3 3 1

2

3

= =

1 2 1

0 1 1

0 0 0

0 (by R R R3 3 13 )

D1

3 2 1

3 1 1

5 0 0

=

= + 3 0 0 2 0 5 1 0 5( ) ( ) ( )

= =10 5 5

D1 0As, D1 0 and D = 0.Hence,thesystemhasnosolution.

AlternateSolution

1 2 1

2 3 1

3 5 2

3

3

1

1

2

3

=

x

x

x

~

1 2 1

0 1 1

0 1 1

3

3

8

1

2

3

=

x

x

x

(by

and

R R R

R R R

2 2 1

3 3 1

2

3

)

~

1 2 1

0 1 1

0 0 0

3

3

5

1

2

3

=

x

x

x

(by R R R3 3 2 )

Clearly, the given system of equations has nosolution.

Alternate Solution (Trick)

Subtracting the addition of first two equations fromthird equation, we get 0 5= , which is an absurdresult.

Hence, the given system of equations has nosolution.

65.

The number of ways in which two balls from urn A

and two balls from urn B can be selected

= 3 29

2C C = =3 36 108

66. We have, f R: ( , ) 1 1f f( ) ( )0 1 0 1= =and

g x f f x( ) [ ( ( ) )]= +2 2 2

= + + g x f f x f f x f x( ) [ ( ( ) )] ( ( ) ) ( )2 2 2 2 2 2

= + + g f f f f f( ) [ ( ( ) )] ( ( ) ) ( )0 2 2 0 2 2 0 2 2 0

= 2 0 0 2 0[ ( )] ( ) ( )f f f

= = 2 1 1 2 1 4( )


Urn A Urn B

3 distinct

red balls

9 distinct

blueballs

67. Since, f x( ) is a positive increasing function. 0 2 3< <

73. We have, a b c + = 0 a a b a c + =( ) 0

( ) ( )a b a a a b a c + = 0

3 2 0a b a c + =

2 3b a a c= +

2 3 3 2b j k i j k= $ $ $ $ $ = + 2 2 4$ $ $i j k

b i j k= + $ $ $2

74. Since, the given vectors are mutually orthogonal,therefore

a b = + =2 4 2 0

a c = + = 1 2 0 ...(i)b c = + + =2 4 0 ...(ii)

OnsolvingEqs.(i)and(ii),weget

= 2 and = 3Hence, ( , ) ( , ) = 3 2

75. We know that, the locus of point P from which twoperpendicular tangents are drawn to the parabola,is the directrix of the parabola.

Hence, the required locus is x = 1.

76. Since, the line L is passing through the point( , )13 32 .

Therefore,13

5

321

32 8

520+ = = =

b bb

The line K is parallel to the line L, then its equationmust be

x ya

5 20 = or

x

a

y

a5 201 =

Oncomparingwithx

c

y+ =

31,weget

20 3 53

4a c a= = = ,

Hence,thedistancebetweenlines

=

+

| |a 1

1

25

1

400

=

=

3

201

17

400

23

17

77. / If a line makes angles with the positive X -axis , and . Then, cos cos cos2 2 2 1 + + = .

Weknowthat, cos cos cos2 2 245 120 1+ + =

1

2

1

412+ + =cos

cos21

4 =

cos = = 12

60 or120

78. P : There is rational number x S such that x > 0.~ :P Every rational number x S satisfies x 0.

79. / Here, use the formula sin cos2 2 1 + = to findthe values of sin( ) + and cos( ) , thentan

sin

cos

= .

cos( ) + = + 45

1stquadrant

and sin( ) = 513

1stquadrant

2 = + + ( ) ( ) tan

tan( ) tan( )

tan( )tan( )2

1

=

+ +

+

=

+

=

3

4

5

12

13

4

5

12

56

33

80. Since, the coordinates of the centre of the circleare ( , )2 4 .

Also, r2 4 16 5 25= + + =

The line will intersect the circle at two distinctpoints, if the distance of ( , )2 4 from 3 4x y m = isless than radius of the circle.

| |6 16

55

0

ee

x

x+ 2 2 2

01

2

1

2 2

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JEE Main Solved Paper 2010