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
AIEEE Solved Paper 2010 107
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
108 JEE Main Solved Papers
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
AIEEE Solved Paper 2010 109
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
110 JEE Main Solved Papers
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=
AIEEE Solved Paper 2010 111
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
114 JEE Main Solved Papers
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.
116 JEE Main Solved Papers
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
AIEEE Solved Paper 2010 117
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.
118 JEE Main Solved Papers
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
AIEEE Solved Paper 2010 119
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.
120 JEE Main Solved Papers
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
122 JEE Main Solved Papers
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( )
AIEEE Solved Paper 2010 123
Urn A Urn B
3 distinct
red balls
9 distinct
blueballs
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