AIEEE-2007

AIEEE - 2007Max. Marks :360 No. of Questions : 120

SECTION I - PHYSICS

1. A circular disc of radius R is removed from abigger circular disc of radius 2R such that thecircumferences of the discs coincide. The centreof mass of the new disc is α / R form the centreof the bigger disc. The value of α is(a) 1/4 (b) 1/3 (c) 1/2 (d) 1/6

2. A round uniform body of radius R, mass M andmoment of inertia I rolls down (without slipping)an inclined plane making an angle θ with thehorizontal. Then its acceleration is

(a)2

gsin

1 MR / I

θ−

(b)2

gsin

1 I / MR

θ+

(c)2

g sin

1 MR / I

θ+

(d)2

gsin

1 I / MR

θ−

3. Angular momentum of the particle rotating witha central force is constant due to(a) constant torque(b) constant force(c) constant linear momentum(d) zero torque

4. A 2 kg block slides on a horizontal floor with aspeed of 4m/s. It strikes a uncompressed spring,and compresses it till the block is motionless.The kinetic friction force is 15N and springconstant is 10,000 N/m. The spring compressesby(a) 8.5 cm (b) 5.5 cm(c) 2.5 cm (d) 11.0 cm

5. A particle is projected at 60o to the horizontalwith a kinetic energy K. The kinetic energy atthe highest point is(a) K/2 (b) K (c) Zero (d) K/4

6. In a Young’s double slit experiment the intensity

at a point where the path difference is 6λ (λ being

the wavelength of light used) is I. If I0 denotes

the maximum intensity, 0

II

is equal to

(a) 34

(b)1

2(c) 3

2(d) 1

2

7. Two springs, of force constants k1 and k2 areconnected to a mass m as shown. The frequencyof oscillation of the mass is f. If both k1 and k2are made four times their original values, thefrequency of oscillation becomes

k2k1m

(a) 2f (b) f/2 (c) f/4 (d) 4f8. When a system is taken from state i to state f

along the path iaf, it is found that Q =50 cal andW = 20 cal. Along the path ibf Q = 36 cal. Walong the path ibf is

f

b

a

i(a) 14 cal (b) 6 cal(c) 16 cal (d) 66 cal

9. A particle of mass m executes simple harmonicmotion with amplitude a and frequency ν. Theaverage kinetic energy during its motion fromthe position of equilibrium to the end is

(a) 2 2 22 maπ ν (b) 2 2 2maπ ν

(c) 2 21ma

4ν (d) 2 2 24 maπ ν

10. The displacement of an object attached to aspring and executing simple harmonic motion isgiven by x = 2 × 10–2 cos πt metre.The time atwhich the maximum speed first occurs is(a) 0.25 s (b) 0.5 s(c) 0.75 s (d) 0.125 s

11. In an a.c. circuit the voltage applied isE = E0 sin ωt. The resulting current in the circuit

is 0I I sin t2π

ω = − . The power consumption

in the circuit is given by

(a) 0 0P 2E I= (b) 0 0E IP

2=

(c) P = zero (d) 0 0E IP

2=

Mathrbhumi EAIEEE-2007 SOLVED PAPER2007-2

12. An electric charge 10–3 µ C is placed at the origin(0, 0) of X – Y co-ordinate system. Two points

A and B are situated at ( )2, 2 and (2, 0)

respectively. The potential difference betweenthe points A and B will be(a) 4.5 volts (b) 9 volts(c) Zero (d) 2 volt

13. A battery is used to charge a parallel platecapacitor till the potential difference between theplates becomes equal to the electromotive forceof the battery. The ratio of the energy stored inthe capacitor and the work done by the batterywill be(a) 1/2 (b) 1 (c) 2 (d) 1/4

14. An ideal coil of 10H is connected in series witha resistance of 5Ω and a battery of 5V. 2secondafter the connection is made, the current flowingin ampere in the circuit is(a) (1 – e–1) (b) (1 – e)(c) e (d) e–1

15. A long straight wire of radius a carries a steadycurrent i. The current is uniformly distributedacross its cross section. The ratio of the magneticfield at a/2 and 2a is(a) 1/2 (b) 1/4 (c) 4 (d) 1

16. A current I flows along the length of an infinitelylong, straight, thin walled pipe. Then(a) the magnetic field at all points inside the

pipe is the same, but not zero(b) the magnetic field is zero only on the axis

of the pipe(c) the magnetic field is different at different

points inside the pipe(d) the magnetic field at any point inside the

pipe is zero

17. If MO is the mass of an oxygen isotope 178 O ,MP

and MN are the masses of a proton and a neutronrespectively, the nuclear binding energy of theisotope is(a) (MO –17MN)c2

(b) (MO – 8MP)c2

(c) (MO– 8MP –9MN)c2

(d) MOc2

18. In gamma ray emission from a nucleus(a) only the proton number changes(b) both the neutron number and the proton

number change(c) there is no change in the proton number and

the neutron number(d) only the neutron number changes

19. If in a p-n junction diode, a square input signalof 10 V is applied as shown

RL

5V

−5VThen the output signal across RL will be

(a)+5V

(b)

10 V

(c)−10 V

(d)−5V

20. Photon of frequency ν has a momentumassociated with it. If c is the velocity of light,the momentum is(a) hν / c (b) ν /c(c) h ν c (d) hν / c2

21. The velocity of a particle is v = v0 + gt + ft2. Ifits position is x = 0 at t = 0, then its displacementafter unit time (t = 1) is(a) v0 + g/2 + f (b) v0 + 2g + 3f(c) v0 + g/2 + f/3 (d) v0 + g + f

22. For the given uniform square lamina ABCD,whose centre is O,

O

A B

CD

E

F

(a) AC EFI 2 I= (b) AC EF2I I=(c) AD EFI 3I= (d) AC EFI I=

23. A point mass oscillates along the x-axisaccording to the law x = x0 cos( t / 4)ω π− . Ifthe acceleration of the particle is written asa = A cos( t )ω δ+ ,then

(a) 20A x ,ω= 3 / 4δ π=

(b) A = x0, / 4δ π= −(c) 2

0A x ,ω= / 4δ π=(d) 2

0A x ,ω= / 4δ π= −

AIEEE-2007 SOLVED PAPER 2007-3

24. Charges are placed on the vertices of a square

as shown. Let Eur

be the electric field and V thepotential at the centre. If the charges on A and Bare interchanged with those on D and Crespectively, then

A B

CD

q

−q

q

−q

(a) Eur

changes, V remains unchanged

(b) Eur

remains unchanged, V changes

(c) both Eur

and V change

(d) Eur

and V remain unchanged25. The half-life period of a radio-active element X

is same as the mean life time of another radio-active element Y. Initially they have the samenumber of atoms. Then(a) X and Y decay at same rate always(b) X will decay faster than Y(c) Y will decay faster than X(d) X and Y have same decay rate initially

26. A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If thework done on the system is 10 J, the amount ofenergy absorbed from the reservoir at lowertemperature is(a) 100 J (b) 99 J(c) 90 J (d) 1 J

27. Carbon, silicon and germanium have fourvalence electrons each. At room temperaturewhich one of the following statements is mostappropriate ?(a) The number of free electrons for

conduction is significant only in Si and Gebut small in C.

(b) The number of free conduction electronsis significant in C but small in Si and Ge.

(c) The number of free conduction electronsis negligibly small in all the three.

(d) The number of free electrons forconduction is significant in all the three.

28. A charged particle with charge q enters a regionof constant, uniform and mutually orthogonal

fields Eur

and Bur

with a velocity vr

perpendicular

to both Eur

and Bur

, and comes out without anychange in magnitude or direction of v

r . Then

(a) 2v B E / E= ×r ur ur

(b) 2v E B / B= ×r ur ur

(c) 2v B E / B= ×r ur ur

(d) 2v E B / E= ×r ur ur

29. The potential at a point x (measured in µ m) dueto some charges situated on the x-axis is givenby V(x) = 20/(x2 – 4) voltThe electric field E at x = 4 µ m is given by(a) (10/9) volt/ µ m and in the +ve x direction(b) (5/3) volt/ µ m and in the –ve x direction(c) (5/3) volt/ µ m and in the +ve x direction(d) (10/9) volt/ µ m and in the –ve x direction

30. Which of the following transitions in hydrogenatoms emit photons of highest frequency?(a) n = 1 to n = 2 (b) n = 2 to n = 6(c) n = 6 to n = 2 (d) n = 2 to n = 1

31. A block of mass m is connected to another blockof mass M by a spring (massless) of springconstant k. The block are kept on a smoothhorizontal plane. Initially the blocks are at restand the spring is unstretched. Then a constantforce F starts acting on the block of mass M topull it. Find the force of the block of mass m.

(a) MF(m M)+

(b)mFM

(c) (M m)Fm+ (d) mF

(m M)+32. Two lenses of power –15 D and +5 D are in

contact with each other. The focal length of thecombination is(a) + 10 cm (b) – 20 cm(c) – 10 cm (d) + 20 cm

33. One end of a thermally insulated rod is kept at atemperature T1 and the other at l2. The rod iscomposed of two sections of length l1 and l2 andthermal conductivities K1 and K2 respectively.The temperature at the interface of the twosection is

T1 l1 l2 T2

K1 K2

(a) 1 1 1 2 2 2

1 1 2 2

(K T K T )(K K )

++

l ll l

(b) 2 2 1 1 1 2

1 1 2 2

(K T K T )(K K )

++

l ll l

(c) 2 1 1 1 2 2

2 1 1 2

(K T K T )(K K )

++

l ll l

(d) 1 2 1 2 1 2

1 2 2 1

(K T K T )(K K )

++

l ll l

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AIEEE-2007 SOLVED PAPER2007-4

34. A sound absorber attenuates the sound level by20 dB. The intensity decreases by a factor of(a) 100 (b) 1000(c) 10000 (d) 10

35. If CP and CV denote the specific heats of nitrogenper unit mass at constant pressure and constantvolume respectively, then(a) CP – CV = 28R (b) CP – CV = R/28(c) CP – CV = R/14 (d) CP – CV = R

36. A charged particle moves through a magneticfield perpendicular to its direction. Then(a) kinetic energy changes but the momentum

is constant(b) the momentum changes but the kinetic

energy is constant(c) both momentum and kinetic energy of the

particle are not constant(d) both momentum and kinetic energy of the

particle are constant37. Two identical conducting wires AOB and COD

are placed at right angles to each other. The wireAOB carries an electric current I1 and CODcarries a current I2. The magnetic field on a pointlying at a distance d from O, in a directionperpendicular to the plane of the wires AOB andCOD, will be given by

(a) 2 201 2(I I )

2 dµ

+π

(b)

120 1 2I I

2 dµ +

π

(c) ( )1

2 20 21 2I I

2 dµ

+π

(d) ( )01 2I I

2 dµ

+π

38. The resistance of a wire is 5 ohm at 50°C and 6ohm at 100°C. The resistance of the wire at 0°Cwill be(a) 3 ohm (b) 2 ohm(c) 1 ohm (d) 4 ohm

39. A parallel plate condenser with a dielectric ofdielectric constant K between the plates has acapacity C and is charged to a potential V volt.The dielectric slab is slowly removed frombetween the plates and then reinserted. The network done by the system in this process is

(a) zero (b) 21(K 1) CV

2−

(c)2CV (K 1)

K− (d) 2(K 1) CV−

40. If gE and gM are the accelerations due to gravityon the surfaces of the earth and the moonrespectively and if Millikan’s oil dropexperiment could be performed on the twosurfaces, one will find the ratio

electronic charge on the moonto be

electronic charge on the earth

(a) M Eg / g (b) 1

(c) 0 (d) E Mg / g

SECTION II - CHEMISTRY

41. The equivalent conductances of two strongelectrolytes at infinite dilution in H2O (whereions move freely through a solution) at 25°C aregiven below :

32

CH COONa 91.0 S cm / equiv.Λ =o

2HCl 426.2 S cm / equiv.Λ =o

What additional information/ quantity one needsto calculate A° of an aqueous solution of aceticacid?

(a) Λo of chloroacetic acid (ClCH2COOH)

(b) Λo of NaCl

(c) Λo of CH3COOK(d) the limiting equivalent coductance of

HH ( )+

+ λ° .

42. Which one of the following is the strongest basein aqueous solution ?(a) Methylamine (b) Trimethylamine(c) Aniline (d) Dimethylamine.

43. The compound formed as a result of oxidationof ethyl benzene by KMnO4 is(a) benzyl alcohol (b) benzophenone(c) acetophenone (d) benzoic acid.

44. The IUPAC name of is

(a) 3-ethyl-4-4-dimethylheptane(b) 1, 1-diethyl-2,2-dimethylpentane(c) 4, 4-dimethyl-5,5-diethylpentane(d) 5, 5-diethyl-4,4-dimethylpentane.



45. Which of the following species exhibits thediamagnetic behaviour ?(a) NO (b) O2

2– (c) O2+ (d) O2.

46. The stability of dihalides of Si, Ge, Sn and Pbincreases steadily in the sequence

(a) 2 2 2 2PbX SnX GeX SiX<< << <<(b) GeX2 << SiX2 << SnX2 << PbX2(c) SiX2 << GeX2 << PbX2 << SnX2(d) SiX2 << GeX2 << SnX2 << PbX2.

47. Identify the incorrect statement among thefollowing.(a) Br2 reacts with hot and strong NaOH

solution to give NaBr and H2O.(b) Ozone reacts with SO2 to give SO3.(c) Silicon reacts with NaOH(aq) in the

presence of air to give Na2SiO3 and H2O.(d) Cl2 reacts with excess of NH3 to give N2

and HCl.48. The charge/size ratio of a cation determines its

polarizing power. Which one of the followingsequences represents the increasing order of thepolarizing power of the cationic species, K+,Ca2+, Mg2+, Be2+?(a) Ca2+ < Mg2+ < Be+ < K+

(b) Mg2+ < Be2+ < K+ < Ca2+

(c) Be2+ < K+ < Ca2+ < Mg2+

(d) K+ < Ca2+ < Mg2+ < Be2+.49. The density (in g mL–1) of a 3.60 M sulphuric

acid solution that is 29% H2SO4 (molar mass= 98 g mol–1) by mass will be(a) 1.45 (b) 1.64 (c) 1.88 (d) 1.22

50. The first and second dissociation constants ofan acid H2A are 1.0 × 10–5 and 5.0 × 10–10

respectively. The overall dissociation constantof the acid will be(a) 0.2 × 105 (b) 5.0 × 10–5

(c) 5.0 × 1015 (d) 5.0 × 10–15.51. A mixtuve of ethyl alcohol and propyl alcohol

has a vapour pressure of 290 mm at 300 K. Thevapour pressure of propyl alcohol is 200 mm. Ifthe mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the sametemperature will be(a) 360 (b) 350 (c) 300 (d) 700

52. In conversion of lime-stone to lime,

3(s) (s) 2(g)CaCO CaO CO→ +

the values of H∆ ° and S∆ ° are + 179.1 kJ mol-1and 160.2 J/K respectively at 298 K and 1 bar.

Assuming that H∆ ° and S∆ ° do not change withtemperature, temperature above whichconversion of limestone to lime will bespontaneous is(a) 1118 K (b) 1008 K(c) 1200 K (d) 845 K.

53. The energies of activation for forward andreverse reactions for A2 + B2 2AB are 180kJ mol–1 and 200 kJ mol–1 respectively. Thepresence of a catalyst lowers the activationenergy of both (forward and reverse) reactionsby 100 kJ mol–1. The enthalpy change of thereaction (A2 + B2 → 2AB) in the presence of acatalyst will be (in kJ mol–1)(a) 20 (b) 300 (c) 120 (d) 280

54. The cell,2 2

cellZn | Zn (1 M) || Cu (1 M) | Cu (E 1.10 V)+ + ° =was allowed to be completely discharged at 298K. The relative concentration of Zn2+ to Cu2+

2

2[Zn ]

[Cu ]

+

+

is

(a) 9.65 × 104 (b) antilog (24.08)(c) 37.3 (d) 1037.3.

55. The pKa of a weak acid (HA) is 4.5. The pOHof an aqueous buffer solution of HA in which50% of the acid is ionized is(a) 7.0 (b) 4.5 (c) 2.5 (d) 9.5

56. Consider the reaction, 2A + B → products.When concentration of B alone was doubled, thehalf-life did not change. When the concentrationof A alone was doubled, the rate increased bytwo times. The unit of rate constant for thisreaction is(a) s–1 (b) L mol–1 s–1

(c) no unit (d) mol L–1 s–1.57. Identify the incorrect statement among the

following:(a) 4f and 5f orbitals are equally shielded.(b) d-Block elements show irregular and erratic

chemical properties among themselves.(c) La and Lu have partially filled d-orbitals

and no other partially filled orbitals.(d) The chemistry of various lanthanoids isvery similar.

58. Which of the following has a square planargeometry?(a) [PtCl4]2– (b) [CoCl4]2–

(c) [FeCl4]2– (d) [NiCl4]2–

(At. nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78)



59. Which of the following molecules is expectedto rotate the plane of plane-polarised light?

(a)

COOH

H N2 H

H

(b)

CHO

HO

CH OH2

H

(c)

SH

(d)

NH2

PhPh

H N2

H H

60. The secondary structure of a protein refers to(a) fixed configuration of the polypeptide

backbone(b) α − helical backbone(c) hydrophobic interactions(d) sequence of α − amino acids.

61. Which of the following reactions will yield2, 2-dibromopropane?(a) CH3 – CH = CH2 + HBr →(b) CH3 – C ≡ CH + 2HBr →(c) CH3CH = CHBr + HBr →(d) CH ≡ CH + 2HBr →

62. In the chemical reaction,CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B)+ 3H2O, the compounds (A) and (B) arerespectively(a) C2H5NC and 3KCl(b) C2H5CN and 3KCl(c) CH3CH2CONH2 and 3KCl(d) C2H5NC and K2CO3.

63. The reaction of toluene with Cl2 in presence ofFeCl3 gives predominantly(a) m-chlorobenzene(b) benzoyl chloride(c) benzyl chloride(d) o- and p-chlorotoluene.

64. Presence of a nitro group in a benzene ring(a) deactivates the ring towards electrophilic

substitution(b) activates the ring towards electrophilic

substitution(c) renders the ring basic(d) deactivates the ring towards nucleophilic

substitution.65. In which of the following ionization processes,

the bond order has increased and the magneticbehaviour has changed?

(a) 2 2N N +→ (b) 2 2C C +→

(c) NO NO+→ (d) 2 2O O +→ .

66. The actinoids exhibit more number of oxidationstates in general than the lanthanoids. This isbecause(a) the 5f orbitals extend further from the

nucleus than the 4f orbitals(b) the 5f orbitals are more buried than the 4f

orbitals(c) there is a similarity between 4f and 5f

orbitals in their angular part of the wavefunction

(d) the actinoids are more reactive than thelanthanoids.

67. Equal masses of methane and oxygen are mixedin an empty container at 25°C. The fraction ofthe total pressure exerted by oxygen is(a) 1/2 (b) 2/3

(c) 1 2733 298

× (d) 1/3.

68. A 5.25% solution of a substance is isotonic with a1.5% solution of urea (molar mass = 60 g mol–1)in the same solvent. If the densities of both thesolutions are assumed to be equal to 1.0 g cm–3,molar mass of the substance will be(a) 210.0 g mol–1 (b) 90.0 g mol–1

(c) 115.0 g mol–1 (d) 105.0 g mol–1.69. Assuming that water vapour is an ideal gas, the

internal energy change ( U)∆ when 1 mol ofwater is vapourised at 1 bar pressure and 100°C,(given : molar enthalpy of vapourisation of waterat 1 bar and 373 K = 41 kJ mol–1 and R = 8.3 Jmol–1 K–1) will be(a) 41.00 kJ mol–1 (b) 4.100 kJ mol–1

(c) 3.7904 kJ mol–1 (d) 37.904 kJ mol–1



70. In a saturated solution of the sparingly solublestrong electrolyte AgIO3 (molecular mass = 283)

the equilibrium which sets in is AgIO3 (s)

(aq) 3 (aq)Ag IO+ −+ . If the solubility product

constant Ksp of AgIO3 at a given temperature is1.0 × 10–8, what is the mass of AgIO3 containedin 100 ml of its saturated saolution?(a) 1.0 × 10– 4 g (b) 28.3 × 10–2 g(c) 2.83 × 10–3 g (d) 1.0 × 10–7 g.

71. A radioactive element gets spilled over the floorof a room. Its half-life period is 30 days. If theinitial velocity is ten times the permissible value,after how many days will it be safe to enter theroom?(a) 100 days (b) 1000 days(c) 300 days (d) 10 days.

72. Which one of the following conformations ofcyclohexane is chiral?(a) Boat (b) Twist boat(c) Rigid (d) Chair.

73. Which of the following is the correct order ofdecreasing SN2 reactivity?(a) R2CH X > R3C X > RCH2 X(b) RCH X > R3C X > R2CH X(c) RCH2 X > R2CH X > R3C X(d) R3C X > R2CH X > RCH2 X.

(X is a halogen)74. In the following sequence of reactions,

2P I Mg HCHO3 2 ether

CH CH OH A B+→ → →

2H OC D→the compound D is(a) propanal (b) butanal(c) n-butyl alcohol (d) n-propyl alcohol.

75. Which of the following sets of quantum numbersrepresents the highest energy of an atom?(a) n = 3, l = 0, m = 0, s = +1/2(b) n = 3, l = 1, m = 1, s = +1/2(c) n = 3, l = 2, m = 1, s = +1/2(d) n = 4, l = 0, m = 0, s = +1/2.

76. Which of the following hydrogen bonds is thestrongest?(a) O – H - - - F (b) O – H - - - H(c) F – H - - - F (d) O – H - - - O.

77. In the reaction,3

(s) (aq) (aq) (aq) 2(g)2A 6HC 2A 6C 3H+ −+ → + +l l l l(a) 11.2 L H2(g) at STP is produced for every

mole HCl(aq) consumed(b) 6 L HCl(aq) is consumed for every 3 L H2(g)

produced(c) 33.6 L H2(g) is produced regardless of

temperature and pressure for every moleAl that reacts

(d) 67.2 H2(g) at STP is produced for everymole Al that reacts.

78. Regular use of the following fertilizers increasesthe acidity of soil?(a) Ammonium sulphate(b) Potassium nitrate(c) Urea(d) Superphosphate of lime.

79. Identify the correct statement regarding aspontaneous process:(a) Lowering of energy in the process is the

only criterion for spontaneity.(b) For a spontaneous process in an isolated

system, the change in entropy is positive.(c) Endothermic processes are never

spontaneous.(d) Exothermic processes are always

spontaneous.80. Which of the following nuclear reactions will

generate an isotope?(a) β - particle emission(b) Neutron praticle emission(c) Positron emission(d) α − particle emission.

SECTION III - MATHEMATICS

81. The resultant of two forces Pn and 3n is a forceof 7n. If the direction of 3n force were reversed,

the resultant would be 19 n. The value of P is(a) 3 n (b) 4 n(c) 5 n (d) 6 n.

82. Two aeroplanes I and II bomb a target insuccession. The probabilities of I and II scoringa hit correctly are 0.3 and 0.2, respectively. Thesecond plane will bomb only if the first missesthe target. The probability that the target is hitby the second plane is(a) 0.2 (b) 0.7(c) 0.06 (d) 0.14.



83. If D =1 1 11 1 x 11 1 1 y

++

for x 0, y 0≠ ≠ , then D

is(a) divisible by x but not y(b) divisible by y but not x(c) divisible by neither x nor y(d) divisible by both x and y

84. For the Hyperbola 2 2

2 2x y

1cos sin

− =α α

, which of

the following remains constant when α varies =?(a) abscissae of vertices(b) abscissae of foci(c) eccentricity(d) directrix.

85. If a line makes an angle of / 4π with the positivedirections of each of x- axis and y- axis, then theangle that the line makes with the positivedirection of the z-axis is

(a)4π (b)

2π (c)

6π (d) 3

π

86. A value of c for which conclusion of Mean ValueTheorem holds for the function f (x) = loge x onthe interval [1, 3] is(a) log3e (b) loge3

(c) 2 log3e (d)12

log3e

87. The function f (x) = tan–1(sin x + cos x) is anincreasing function in

(a) 0,2π

(b) ,

2 2π π −

(c) ,4 2π π

(d) ,

2 4π π −

88. Let A = 5 50 50 0 5

α αα α . If 2A 25= , then α

equals(a) 1/5 (b) 5 (c) 52 (d) 1

89. The sum of series1 1 12! 3! 4!

− + − ....... upto

infinity is

(a) 12e

− (b) 12e

+ (c) e–2 (d) e–1

90. If u and v are unit vectors and θ is the acute

angle between them, then 2 u ×3 v is a unitvector for(a) no value of θ(b) exactly one value of θ(c) exactly two values of θ(d) more than two values of θ

91. A particle just clears a wall of height b at adistance a and strikes the ground at a distance cfrom the point of projection. The angle ofprojection is

(a) 1 bctan

a(c a)−

−(b) 1 bc

tana

−

(c) 1 btan

ac− (d) 45°.

92. The average marks of boys in class is 52 andthat of girls is 42. The average marks of boysand girls combined is 50. The percentage of boysin the class is(a) 80 (b) 60(c) 40 (d) 20.

93. The equation of a tangent to the parabolay2 = 8x is y = x + 2. The point on this line fromwhich the other tangent to the parabola isperpendicular to the given tangent is(a) (2, 4) (b) (–2, 0)(c) (–1, 1) (d) (0, 2)

94. If (2, 3, 5) is one end of a diameter of the spherex2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then thecooordinates of the other end of the diameterare(a) (4, 3, 5) (b) (4, 3, – 3)(c) (4, 9, – 3) (d) (4, –3, 3).

95. Let ˆ ˆ ˆ ˆ ˆ ˆa i j k, b i j 2k= + + = − +rr and ˆ ˆ ˆc xi (x 2) j k= + − −

r .

If the vectors cr

lies in the plane of ar

and br

,then x equals(a) – 4 (b) – 2(c) 0 (d) 1.

96. Let A (h, k), B(1, 1) and C (2, 1) be the verticesof a right angled triangle with AC as itshypotenuse. If the area of the triangle is 1squareunit, then the set of values which 'k' can take isgiven by(a) –1, 3 (b) –3, –2(c) 1, 3 (d) 0, 2



97. Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) bethree point. The equation of the bisector of theangle PQR is

(a)3

x y 02

+ = (b) x 3y 0+ =

(c) 3x y 0+ = (d)3

x y 02

+ = .

98. If one of the lines of my2 + (1– m2) xy – mx2= 0is a bisector of the angle between the lines xy =0, then m is(a) 1 (b) 2(c) –1/2 (d) –2.

99. Let F(x) = f (x) + f 1x

,where

x

l

log tf (x) dt,

1 t=

+∫ Then F(e) equals

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

100. Let f : R R→ be a function defined by

f (x) = min x 1, x 1+ + ,Then which of the

following is true ?(a) f (x) is differentiable everywhere(b) f (x) is not differentiable at x = 0(c) f (x) ≥ 1 for all x R∈(d) f (x) is not differentiable at x = 1

101. The function f : R /0 R→ given by

2x1 2

f (x)x e 1

= −−

can be made continuous at x = 0 by defining f(0) as(a) 0 (b) 1(c) 2 (d) – 1

102. The solution for x of the equation

x

22

dt2t t 1

π=

−∫ is

(a) 32

(b) 2 2

(c) 2 (d) π .

103.dx

cos x 3 sin x+∫ equals

(a) log tan xC

2 12π + +

(b) log tan xC

2 12π − +

(c)12

log tan xC

2 12π + +

(d)12

log tan x

C2 12

π − +

104. The area enclosed between the curves y2 = xand y = | x | is(a) 1/6 (b) 1/3(c) 2/3 (d) 1.

105. If the difference between the roots of the equation

x2 + ax + 1 = 0 is less than 5 , then the set ofpossible values of a is

(a) (3, )∞ (b) ( , 3)−∞ −

(c) (– 3, 3) (d) ( 3, )− ∞ .106. In a geometric progression consisting of positive

terms, each term equals the sum of the next twoterms. Then the common ratio of its progressionis equals

(a) 5 (b) ( )15 1

2−

(c) ( )11 5

2− (d) 1

52

.

107. If 1 1x 5sin cosec

5 4 2− − π + =

, then the values

of x is(a) 4 (b) 5(c) 1 (d) 3.

108. In the binomial expansion of (a – b)n, n ≥ 5, thesum of 5th and 6th terms is zero, then a/b equals

(a) n 56− (b)

n 46−

(c) 5n 4−

(d)6

n 5−.



109. The set S : = 1, 2, 3, ......., 12 is to be partitionedinto three sets A, B, C of equal size. Thus A ∪B ∪ C = S, A B B C A C .∩ = ∩ = ∩ = φ Thenumber of ways to partition S is

(a)3

12!

(4!)(b)

412!

(4!)

(c) 312!

3!(4!)(d) 4

12!

3!(4!)

110. The largest interval lying in ,2 2

−π π

for which

the function,

2x 1 xf (x) 4 cos 1 log (cos x)

2− − = + − +

, is

defined, is

(a) ,4 2π π −

(b) 0,2π

(c) [0, ]π (d) ,2 2π π −

111. A body weighing 13 kg is suspended by two

strings 5m and 12m long, their other ends beingfastened to the extremities of a rod 13m long. Ifthe rod be so held that the body hangsimmediately below the middle point, thentensions in the strings are(a) 5 kg and 12 kg (b) 5 kg and 13 kg(c) 12 kg and 13 kg (d) 5 kg and 5 kg

112. A pair of fair dice is thrown independently threetimes. The probability of getting a score ofexactly 9 twice is(a) 8/729 (b) 8/243(c) 1/729 (d) 8/9.

113. Consider a family of circles which are passingthrough the point (– 1, 1) and are tangent to x-axis. If (h, k) are the coordinate of the centre ofthe circles, then the set of values of k is given bythe interval

(a) 1 1k

2 2− ≤ ≤ (b) 1

k2

≤

(c) 10 k

2≤ ≤ (d)

1k

2≥

114. Let L be the line of intersection of the planes2x + 3y + z = 1 and x + 3y + 2z = 2. If L makesan angle α with the positive x-axis, then cos αequals

(a) 1 (b)1

2(c)

1

3(d) 1

2.

115. The differential equation of all circles passingthrough the origin and having their centres onthe x-axis is

(a) 2 2 dyy x 2xy

dx= +

(b) 2 2 dyy x 2xy

dx= −

(c) 2 2 dyx y xy

dx= +

(d) 2 2 dyx y 3xy

dx= + .

116. If p and q are positive real numbers such that p2

+ q2 = 1, then the maximum value of (p + q) is

(a)12

(b)1

2(c) 2 (d) 2.

117. A tower stands at the centre of a circular park. Aand B are two points on the boundary of the parksuch that AB (= a) subtends an angle of 60° atthe foot of the tower, and the angle of elevationof the top of the tower from A or B is 30°. Theheight of the tower is

(a) a / 3 (b) a 3

(c) 2a / 3 (d) 2a 3 .118. The sum of the series

20 20 20 200 1 2 3C C C C .....− + − +

2010..... C− + is

(a) 0 (b) 2010C

(c) 2010C− (d) 20

101

C2

119. The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin istwice the abscissa of P, then the curve is a(a) circle (b) hyperbola(c) ellipse (d) parabola.

120. If | z + 4 | ≤ 3, then the maximum value of| z + 1 | is(a) 6 (b) 0 (c) 4 (d) 10



SECTION I - PHYSICS

1. (b) Let O be the centre of mass of the dischaving radius 2R. O' is the new C.M.

RO

2R

xm

'O'M

Let m = mass of disc of radius RM' = mass of disc when the disc of radius Ris removed.M = mass of disc of radius 2R

Now, 2m ( R ). ,= π σ

where 2 2

M M

(2R) 4 Rσ = =

π π = the mass

per unit area2 2M ' [ (2R) R ].= π − π σ

= 23 Rπ σ

M = 2 2(2R) . 4 Rπ σ = π σ

We have, M '.x m.R0

M ' m+

=+

( ∵ C.M. of the full disc is at the centre O)or, M'.x + m.R = 0or, M'x = – mR

⇒ x = m

RM'

−

2

2R

.R3 R

π σ= −

π σ =

1R

3 −

But x = Rα

∴Rα

= 13

− .R

There appears misprint in this question.

There must be α R instead of Rα

. Then

1 1R R

3 3 α = − ⇒ α = −

∴1

| |3

α =

2. (b) The acceleration of a solid sphere of massM, radius R and moment of inertia I rollingdown (without slipping) an inclined planemaking an angle θ with the horizontal isgiven by

a = 2

2

gsin

KI

R

θ

+, where, I = MK2

3. (d) Central forces always act along the axis ofrotation. Therefore, the torque is zero. Andif there is no external torque acting on arotating body then its angular momentumis constant.

4. (b) Let the spring be compressed by x.Clearly, Initial K.E. of block = Potentialenergy of spring + workdown againstfriction

or,12

mv2 = 12

kx2 + fx

or,12

× 2 × (4)2 = 2110000 x

2 × ×

+ 15x

or, 16 = 5000 x2 + 15xor, 5000 x2 + 15x – 16 = 0

∴ x = 215 (15) 4 5000 ( 16)2 5000

− ± − × × −×

15 565.8810000

− ±= = 0.055 m

(Ignoring –ve value)∴ x = 5.5 cm.

5. (d) Let K' be the K.E. at the highest point. Then

K' = 12

mvx2 ( ∵ vy = 0 at highest point)

HINTS & SOLUTIONS



u vx

θ

21m(u cos )

2= θ

2 2 21mu cos K.cos

2= θ = θ

21K mu

2 = ∵

or, K' = K.cos260° ( )60θ = °∵21 K

K.2 4

= =

6. (a) In young's double slit experiment, theintensity at a point is given by

I = I0cos22φ

where, I0 = maximum intensityφ = phase difference

Also, φ = 2πλ

× path difference

= 2πλ

× 6 3λ π

=

∴ I = I0 cos2

6π

or,0

II

= cos2 30° = 2

3 32 4

=

7. (a) The two springs are in parallel.∴ Effective spring constant,K = K1 + K2.Now, frequency of oscillation is given by

1 Kf

2 m=

π

or, f = 1 2K K12 m

+π

....(i)

When both K1 and K2 are made four timestheir original values, the new frequency isgiven by

1 24K 4K1f '

2 m+

=π

1 2 1 24(K 4K ) K K1 12

2 m 2 m

+ += = π π

= 2f; from (i)8. (b) For path iaf,

Q = 50 calW = 20 cal

f

b

a

iBy first law of thermodynamics,

U Q W∆ = − = 50 – 20 = 30 cal.For path ibfQ = 36 calW = ?By first law of thermodynamics,Q = ∆ U + Wor, W = Q – ∆ USince, the change in internal energy doesnot depend on the path, therefore∆ U = 30 cal∴ W = Q – ∆ U = 36 – 30 = 6 cal.

9. (b) The kinetic energy of a particle executingS.H.M. is given by

K = 12

ma2 ω2 sin2ωt

where, m = mass of particlea = amplitudeω = angular frequencyt = timeNow, average K.E. = < K >

= <12

mω2 a2 sin2 ωt >

= 12

mω2a2 <sin2 ωt >

= 12

mω2a2 12

2 1sin

2 < θ > = ∵



2 21m a

4= ω

= 14

ma2 2(2 )πν ( w 2 )= πν∵

or, < K > 2 2 2ma= π ν10. (b) Here, x = 2 × 10–2 cos π t

Speed is given by

dxv

dt= = 2 × 10–2 π sin π t

For the first time, the speed to be maximum,sin π t = 1

or, sin π t = sin2π

⇒ t2π

π = or, t = 12

= 0.5 sec.

11. (c) We know that power consumed in a.c.

circuit is given by, rms rmsP E .I cos= φHere, E = E0 sinωt

I = I0 sin t2π ω −

which implies that the phase difference,

2π

φ =

∴ rms rmsP E .I .cos2π

= = 0

cos 02π =

∵

12. (c) The distance of point ( )A 2, 2

Y

Q X(0,0) (2,0)B

A( 2, 2)√ √

r2

r1→

→

from the origin,

OA = 2 21| r | ( 2) ( 2)= +ur

= 4 = 2 units.

The distance of point B(2, 0) from the origin,

OB = 2 22| r | (2) (0)= +ur

= 2 units.

Now, potential at A, VA = 0

1 Q.

4 (OA)π ∈

Potential at B, VB = 0

1 Q.

4 (OB)π ∈

∴ Potential difference between thepoints A and B is given by

VA – VB = 0 0

1 Q 1 Q. .

4 OA 4 OB−

π ∈ π ∈

0

Q 1 14 OA OB

= − π ∈ 0

Q 1 14 2 2

= − π ∈

0

Q0

4= ×

π ∈ = 0.

13. (a) Required Ratio

= Energy stored in capacitorWork done by the battery

2

2

1CV

2Ce

= , where C = Capacitance of

capacitor V = Potential difference, e = emf of battery

2

2

1Ce

2Ce

= .( ∵ V = e)

12

=

14. (a) We have, I = R

tL

oI 1 e−

−

(When current is in growth in LR circuit)

Rt

LE1 e

R

− = −

52

1051 e

5

− × = −

= (1 – e–1)



15. (d) Here, current is uniformly distributed acrossthe cross-section of the wire, therefore,current enclosed in the amperean path

formed at a distance 1a

r2

=

a/2

P1 P2

a

212

rI

a

π= ×

π , where I is total current

∴ Magnetic field at

01 1

current enclosedP (B )

Pathµ ×

=

21

0 2

1

rI

a

2 r

πµ × ×

π =

π 0 1

2I r

2 a

µ ×=

π

Now, magnetic field at point P2,

(B2) 0 I

.2 (2a)µ

=π

0I4 aµ

=π

.

∴ Required Ratio = 0 112

2 0

IrB 4 aB I2 a

µ π= ×

µπ

1

a22 r 2

a a

×= = = 1.

16. (d) There is no current inside the pipe andhence Ampere's law can not be applied.

17. (c) Binding energy= [ZMP + (A – Z)MN – M]c2

= [8MP + (17 – 8)MN – M]c2

= [8MP + 9MN – M]c2

= [8MP + 9MN – M0]c2

But the option (c) is negative of this.

18. (c) There is no change in the proton numberand the neutron number as the γ-emissiontakes place as a result of excitation or de-excitation of nuclei.

19. (a) The current will flow through RL when thediode is forward biased.

20. (a) Energy of a photon of frequency ν is givenby E h= ν .

Also, E = pc, where p is the momentum ofphoton

∴ h pcν = ⇒ p = hcν

.

21. (c) We know that

dxv

dt= ⇒ dx = v dt

Integrating, x t

0 0

dx v dt=∫ ∫

orx

20

0

x (v gt ft )dt= + +∫

t2 3

00

gt ftv t

2 3

= + +

or,2 3

0gt ft

x v t c2 3

= + + +

where c is the constant of integration.By question,x = 0 at t = 0.

∴ 0g f

0 v 0 0 0 c2 3

= × + × + × +

⇒ c = 0.

∴2 3

0gt ft

x v t2 3

= + +

At t = 1,

0g f

x v2 3

= + + .



22. (d) By the theorem of perpendicular axes,Iz = Ix + Iy or, Iz = 2 Iy(∴ Ix = Iy by symmetry of the figure)

A BE

D CIz Iy

Ix

F

∴ zEF

II

2=

... (i)Again, by the same theoremIz = IAC + IBD = 2 IAC(∴ IAC = IBD by symmetry of the figure)

∴ zAC

II

2=

... (ii)From (i) and (ii), we getIEF = IAC.

23. (a) Here,x = x0 cos (ωt – / 4π )∴ Velocity,

0dx

v x sin tdt 4

π = = − ω ω −

Acceleration,

20

dva x cos t

dt 4π = = − ω ω −

20x cos t

4 π = ω π + ω −

= x0 ω2

cos 3t

4π ω +

But by question,Acceleration, a = A cos (ωt + δ )Comparing the two accelerations, we get

A = x0ω2 and 34π

δ = .

24. (a) As shown in the figure, the resultant electricfields before and after interchanging thecharges will have the same magnitude butopposite directions.Also, the potential will be same in bothcases as it is a scalar quantity.

A B

CD

q q

−q−q

→E

A B

CDq q

−q −q

→E

25. (b) By question,Half life of X, T1/2 = avτ , average life of Y

⇒x

n 2 1

γ=

λ λl or,

x Y( n2).λ = λl

⇒ x Y(0.693).λ = λ

∴ x Y .λ < λNow, the rate of decay is given by

R = t0R e−λ

For X, xtx 0R R e−λ=

For Y, yty 0R R e

−λ=

Hence, x yR R> .

Thus, X will decay faster than Y.26. (c) The efficiency ( η ) of a Carnot engine and

the coefficient of performance ( β ) of arefrigerator are related as

1− ηβ =

η



Here, 110

η =

∴

11

10 9.1

10

−β = =

Also, Coefficient of performance ( β ) is

given by 2Qw

β = , where Q2 is the energy

absorbed from the reservoir.

or, 2Q9

10=

∴ Q2 = 90 J.27. (a) Si and Ge are semiconductors but C is an

insulator. Also, the conductivity of Si andGe is more than C because the valenceelectrons of Si, Ge and C lie in third, fouthand second orbit repsectively.

28. (b) Here, Er

and Br

are perpendicular to eachother and the velocity v

r does not change;

therefore

qE = qvB ⇒ Ev

B=

Also, 2 2

E B E Bsin

B B

× θ=

r r

2E B sin 90 E

| v | vBB

°= = = =

r

∴ Option (b) is correct.

29. (a) Here, V(x) = 220

voltx 4−

We know that E = dvdx

− 2

d 20dx x 4

= − −

or, E = 2 240x

(x 4)+

−

At x = 4 mµ ,

E = 2 240 4

(4 4)

×+

− 160 10

volt / m.144 9

= + = + µ

Positive sign indicates that Er

is in +ve x-direction.

30. (d) We have to find the frequency of emittedphotons. For emission of photons thetransition must take place from a higherenergy level to a lower energy level whichare given only in options (c) and (d).Frequency is given by

2 21 2

1 1h 13.6

n n

ν = − −

For transition from n = 6 to n = 2,

1 2 213.6 1 1 2 13.6h 9 h6 2

− ν = − = ×

For transition from n = 2 to n = 1,

2 2 213.6 1 1 3 13.6h 4 h2 1

− ν = − = ×

.

∴ 1 2ν > νHence option (d) is the correct answer.

31. (d) Writing free body-diagrams for m & M,

m

mg

m

M

M

Mg

F

F

K

T Ta

N N

we getT = ma and F – T = Mawhere T is force due to spring⇒ F – ma = Maor, F = Ma + ma

∴F

aM m

=+

.

Now, force acting on the block of mass m is

ma = F

mM m

+

mFm M

=+

.

32. (c) Power of combination is given byP = P1 + P2 = (– 15 + 5) D = – 10 D.

Now, P = 1f

⇒1 1

fP 10

= =−

metre

∴ 1f 100 cm 10 cm.

10 = − × = −



33. (d) Let T be the temperature of the interface.As the two sections are in series, the rateof flow of heat in them will be equal.

1T 2T

2K1K

1l 2l

∴ 1 1 2 2

1 2

K A(T T) K A(T T )− −=l l

,

where A is the area of cross-section.or, 1 1 2 2 2 1K A(T T) K A(T T )− = −l lor, 1 1 2 1 2 2 1 2 2 1K T K T K T K T− = −l l l lor, 2 1 1 2 1 1 2 2 2 1(K K )T K T K T+ = +l l l l

∴ 1 1 2 2 2 1

2 1 1 2

K T K TT

K K+

=+

l ll l

1 2 1 2 1 2

1 2 2 1

K T K T.

K K+

=+

l ll l

34. (a) We have, 11

0

IL 10log

I

=

22

0

IL 10log

I

=

∴ L1 – L2 = 10 log 1 2

0 0

I I10log

I I

−

or, 1 2

0 0

I IL 10log

I I

∆ = ×

or, 1

2

IL 10log

I

∆ =

or, 1

2

I20 10log

I

=

or, 1

2

I2 log

I

=

or, 21

2

I10

I=

or, 12

II .

100=

⇒ Intensity decreases by a factor 100.

35. (b) We have,Molar heat capacity = Molar mass × Specificheat

capacity per unit mass∴ Cp = 28 Cp (for nitrogen)and Cv = 28 CvNow, Cp – Cv = Ror, 28 Cp – 28 Cv = R

⇒ Cp – Cv = R28

.

36. (b) When a charged particle enters a magneticfield at a direction perpendicular to thedirection of motion, the path of the motionis circular. In circular motion the directionof velocity changes at every point (themagnitude remains constant). Therefore,the momentum will change at every point.But kinetic energy will remain constant as

it is given by 21mv

2 and v2 is the square

of the magnitude of velocity which doesnot change.

37. (c) Clearly, the magnetic fields at a point P,equidistant from AOB and COD will havedirections perpendicular to each other, asthey are placed normal to each other.

C

B

O D

P

A

d

2I

1I

B

1B

2B

∴ Resultant field, 2 21 2B B B= +

But 0 1 0 21 2

I IB and B

2 d 2 dµ µ

= =π π

∴ ( )2

2 201 2B I I

2 dµ = + π

or, ( )1/ 22 201 2B I I

2 dµ

= +π



38. (d) We know thatRt = R0 (1 + tα ),⇒ R50 = R0 (1 + 50 α ) ... (i)R100 = R0 (1 + 100 α ) ... (ii)

From (i), R50 – R0 = 50 0Rα ... (iii)

From (ii), R100 – R0 = 100 0Rα ... (iv)

Dividing (iii) by (iv), we get

50 0

100 0

R R 1R R 2

−=

−

Here, R50 = 5Ω and R100 = 6Ω

∴ 0

0

5 R 16 R 2

−=

−

or, 6 – R0 = 10 – 2 R0or, R0 = 4Ω .

39. (a) The potential energy of a charged capacitor

is given by 2Q

U .2C

=

If a dielectric slab is inserted between the

plates, the energy is given by 2Q

2KC, where

K is the dielectric constant.Again, when the dielectric slab is removedslowly its energy increases to initialpotential energy. Thus, work done is zero.

40. (b) Electronic charge does not depend onacceleration due to gravity as it is auniversal constant.So, electronic charge on earth= electronic charge on moon∴ Required ratio = 1.

SECTION II - CHEMISTRY

41. (b) According to Kohlrausch’s law, molarconductivity of weak electrolyte acetic acid(CH3COOH) is given as follows:

3 3CH COOH CH COONa HCl NaClΛ = Λ + Λ − Λo o o o

∴ Value of NaClΛo should also be

known for calculating value of 3CH COOHΛo .

42. (d) Aromatic amines are less basic than aliphaticamines. Among aliphatic amines the order ofbasicity is 2° > 1° > 3° (∵ of decreasedelectron density due to crowding in 3°amines)∴ dimethylamine (2° aliphatic amine) isstrongest base among given choices.

43. (d) When alkyl benzene are oxidised with alkalineKMnO4, the entire alkyl group is oxidised to–COOH group regardless of length of sidechain.

CH CH2 3

Ethyl benzene

4KMnO→

COOH

Benzoic aicd

44. (a)

3

3 2 2 2 3

3 2

33 ethyl 4,4 dimethyl heptane

CH7 6 5 4 | 3 2 1CH CH CH C CH CH CH

| |CH CH

|CH

− − −

− − − − − −

45. (b) Diamagnetic species have no unpairedelectrons

22O − ⇒ σ1s2, σ*1s2, σ∗s2, 2

z2pσ , π2px2,

π2py2, π*2px

2, π*2py2

46. (c) Reluctance of valence shell electrons toparticipate in bonding is called inert paireffect. The stability of lower oxidation state(+2 for group 14 element) increases on goingdown the group. So the correct order isSiX2 < GeX2 < PbX2 < SnX2

47. (d) Chlorine reacts with excess of ammonia toproduce ammonium chloride and nitrogen.

8NH3 + 3Cl2 → N2 + NH4Cl48. (d) Smaller the size and higher the charge more

will be polarising power of cation. So thecorrect order of polarising power isK+ < Ca2+ < Mg2+ < Be2+

49. (d) Mass of 3.6 moles of H2SO4= Moles × Molecular mass= 3.6 × 98 g = 352.8 g∴ 1000 ml solution has 352.8 g of H2SO4



Given that 29 g of H2SO4 is present in = 100g of solution∴ 352.8 g of H2SO4 is present in

= 100

352.829

× g of solution

= 1216 g of solution

Density = Mass

Volume =

12161000

= 1.216 g/ml

= 1.22 g/ml

50. (d) 2H A H HA+ −+

∴ K1 = 1.0 × 10–5 = 2

[H ][HA ][H A]

+ −

HA H A− + −→ +

∴ 102

[H ][A ]K 5.0 10

[HA ]

+ −−

−= × =

2 2

1 22

[H ] [A ]K K K

[H A]

−+= = ×

= (1.0 × 10–5) × (5 × 10–10) = 5 × 10–15

51. (b) Given 0Ap ?= , 0

Bp 200mm= , xA = 0.6,

xB = 1 – 0.6 = 0.4, P = 290

P = pA + pB = 0 0A A B Bp x p x+

⇒ 290 = 0Ap × 0.6 + 200 × 0.4

∴ 0Ap = 350 mm

52. (a) ∆G° = ∆H° – T∆S°For a spontaneous reaction ∆G° < 0

or ∆H° – T∆S° < 0 ⇒H

TS

∆ °>

∆ °

⇒ T > 3179.3 10

1117.9K160.2

×> ≈ 1118K

53. (a) R f bH E E∆ = − = 180 – 200 = – 20 kJ/molThe nearest correct answer given in choicesmay be obtained by neglecting sign.

54. (d) Ecell = 0; when cell is completely discharged.

Ecell = E°cell

2

2

Zn0.059log

2 Cu

+

+

−

or 0 = 1.1

2

2

Zn0.059log

2 Cu

+

+

−

2

2

Zn 2 1.1log 37.3

0.059Cu

+

+

× = =

∴

237.3

2

Zn10

Cu

+

+

=

55. (d) For acidic buffer pH = pKa + logAHA

−

Given pKa = 4.5 and acid is 50% ionised.[HA] = [A–] (when acid is 50% ionised)∴ pH = pKa + log 1∴ pH = pKa = 4.5

pOH = 14 – pH = 14 – 4.5 = 9.556. (b) From the given data we can say that order of

reaction with respect to B = 1 because changein concentration of B does not change halflife. Order of reaction with respect to A = 1because rate of reaction doubles whenconcentration of A is doubled keepingconcentration of A constant.∴ Order of reaction = 1 + 0 = 1 and unitsof first order reaction are L mol–1 sec–1.

57. (a) 4f orbital is nearer to nucleus as compared to5f orbital therefore, shielding of 4f is morethan 5f.

58. (a) Complexes with dsp2 hybridisation aresquare planar. So [PtCl4]2– is square planarin shape.

59. (b) The organic compounds which have chiralcarbon atom and do not have plane ofsymmetry rotate plane polarised light.

2

CHO|*HO C H

|CH OH

− − (* is asymmetric carbon)

60. (b) Proteins have two types of secondarystructures α-helix and β-plated sheet.



61. (b) The reaction follows Markownikoff rulewhich states that when unsymmetricalreagent adds across unsymmetrical doubleor triple bond the negative part adds tocarbon atom having lesser number ofhydrogen atoms.

3CH C CH HBr− ≡ + → 3 2CH C CH|Br

− =

HBr→ 3 3

Br|

CH C CH|

Br

− −

2, 2-dibromo-propane62. (a) This is carbylamine reaction.

CH3CH2NH2 + CHCl3 + 3KOH

→ C2H5NC + 3KCl+ 3H2O

63. (d) FeCl3 is Lewis acid. In presence of FeCl3 sidechain hydrogen atoms of toluene aresubstituted.

Toluene

CH3

+ Cl2 3FeCl→ o-chloro toluene

CH3 Cl

+

p-chloro toluene

CH3

Cl

64. (a) Nitro is electron withdrawing group, so itdeactivates the ring towards electrophilicsubstitution.

65. (c)(a) N2 : bond order 3, paramagnetic

N2– : bond order, 2.5 paramagnetic

(b) C2 : bond order 2, diamagneticC2

+ : bond order 1.5, paramagnetic(c) NO : bond order 2.5, paramagnetic

NO+ : bond order 3, diamagnetic(d) O2 : bond order 2, paramagnetic

O2+ : bond order 2.5, paramagnetic

∴ (c) is correct answer66. (a) More the distance between nucleus and outer

orbitals, lesser will be force of attraction onthem. Distance between nucleus and 5forbitals is more as compared to distancebetween 4f orbital and nucleus. So actinoidsexhibit more number of oxidation states ingeneral than the lanthanoids.

67. (d) Let the mass of methane and oxygen = m gm.Mole fraction of O2

= 2

2 4

Moles of OMoles of O + Moles of CH

= m / 32

m / 32 m /16+ =

m / 323m / 32

= 13

Partial pressure of O2 = Total pressure × mole

fraction if O2 = P × 13

= 1

P3

68. (a) Osmotic pressure of isotonic solutions (π)are equal. For solution of unknownsubstance (π = CRT)

15.25 / M

CV

=

For solution of urea, C2 (concentration)

= 1.5 / 60

VGiven π1 = π2

CRTπ =∵∴ C1RT = C2RT or C1 = C2

or 5.25 / M 1.8 / 60V V

=

∴ M = 210 g/mol69. (d) Given ∆H = 41 kJ mol–1 = 41000 J mol–1

T = 100°C = 273 + 100 = 373 Kn = 1∆U = ∆H – ∆nRT = 41000 – (2 × 8.314 × 373)= 37898.88 J mol–1 ; 37.9 kJmol–1

70. (c) Let x = solubility

3 3AgIO Ag IO −+ +Ksp = [Ag+] [IO3

–] = x × x = x2

Given Ksp = 1 × 10–8

∴ spx K= = 81 10−× = 1.0 × 104

mol/lit= 1.0 × 10–4 × 283 g/lit

= 41.0 10 283 100

gm/100ml1000

−× × ×

= 2.83 × 10–3 gm/ 100 ml



71. (a) Let activity of safe working = AGiven A0 = 10A

1/ 2

0.693 0.693t 30

λ = =

t½ = 0A2.303log

Aλ =

2.303 10Alog

0.693 / 30 A

= 2.303 30

log100.693

×× = 100 days.

72. (b) Chiral conformation will not have plane ofsymmetry. Since twisted boat does not haveplane of symmetry it is chiral.

73. (c) In SN2 mechanism transition state is

pentavelent. For bulky alkyl group it will havesterical hinderance and smaller alkyl groupwill favour the SN

2 mechanism. So thedecreasing order of reactivity of alkyl halidesisRCH2X > R2CHX > R3CX

74. (d) 2P I3 2 3 2

ACH CH OH CH CH I+→

MgEther

→ 3 2(B)

CH CH MgI HCHO→

3 2CH CH|

H C OMgI|

H

− − 2H O→ 3 2

n propylalcohol

CH CH|

H C OH|

H(D)

−

− −

75. (c)

(a) n = 3, 0=l means 3s-orbital

(b) n = 3, 1=l means 3p-orbital

(c) n = 3, 2=l means 3d-orbital

(d) n = 4, 0=l means 4s-orbitalIncreasing order of energy among theseorbitals is3s < 3p < 4s < 3d∴ 3d has highest energy.

76. (c) Greater the difference betweenelectronegativity of bonded atoms, strongerwill be bond.∴ F – H ...... F is the strongest bond.

77. (c) 2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl–(aq) + 3H2(g)

∴ 6 moles of HCl produces = 3 moles of H2= 3 × 22.4 L of H2

∴ 1 mole of HCl produces

= 3 22.4

6×

= 11.2 L of H2

∵ 2 moles of Al produces 3 moles of H2 = 3 × 22.4 L of H2∴ 1 mole of Al produces

= 3 22.4

33.62

×= L of H2

78. (a) (NH4)2SO4 + 2H2O → 2H2SO4 + NH4OHH2SO4 is strong acid and increases theacidity of soil.

79. (b) Spontaneity of reaction depends ontendency to acquire minimum energy stateand maximum randomness. For aspontaneous process in an isolated systemthe change in entropy is positive.

80. (b) Isotopes are atoms of same element havingsame atomic number but different atomicmasses. Neutron has atomic number 0 andatomic mass 1. So loss of neutron willgenerate isotope.

SECTION III - MATHEMATICS

81. (c) Given : Force P = Pn, Q = 3n, resultant R = 7n

& P' = Pn, Q' = (–3)n, R' = 19

P 7

3–3

√19

α

We know thatR2 = P2 + Q2 + 2PQ cos α

⇒ (7)2 = P2 + (3)2 + 2 × P × 3 cos α⇒ 49 = P2 + 9 + 6P cos α .....(i)⇒ 40 = P2 + 6P cos α

and ( )219 = P2 + (–3)2 + 2P × –3 cos α

⇒ 19 = P2 + 9 – 6P cos α⇒ 10 = P2 – 6P cos α .....(ii)Adding (i) and (ii)

50 = 2P2

⇒ P2 = 25 ⇒ P = 5n.



82. (d) Given : Probabilities of aeroplane I, i.e.,P(I) = 0.3 probabilities of scoring a targetcorrectly by aeroplane II, i.e. P(II) = 0.2

∴ P( I ) = 1 – 0.3 = 0.7 and P( II ) = 1 – 0.2 = 0.8∴ The required probability= P ( I II) P( I ).P(II)∩ = = 0.7 × 0.2 = 0.14

83. (d) Given, D = 1 1 11 1 x 11 1 1 y

++

Apply R2 → R2 – R1 and R → R3 – R1

∴ D = 1 1 10 x 00 0 y

= xy

Hence, D is divisible by both x and y84. (b) Given, equation of hyperbola

2 2

2 2x y

1cos sin

− =α α

We know that the equation of hyperbola is2 2

2 2x y

1a b

− =

Here, 2 2a cos= α and 2 2b sin= α

We know that, 2 2 2b a (e 1)= −

⇒ 2 2 2sin cos (e 1)α = α −

⇒ 2 2 2 2sin cos cos .eα + α = α

⇒ 2 2 2e 1 tan sec= + α = α ⇒ e sec= α

∴ ae = cos α . 11

cos=

αCo-ordinates of foci are ( ae,0)± i.e. ( ± 1, 0)Hence, abscissae of foci remain constantwhen α varies.

85. (b) Let the angle of line makes with the positivedirection of z-axis is α direction cosines ofline with the +ve directions of x-axis, y-axis,and z-axis is l, m, n respectively.

∴ l = cos4π , m = cos

4π , n = cos α

as we know that, l2 + m2 + n2 = 1

∴ cos24π

+ cos24π

+ cos2 α = 1

⇒12

+ 12

+ cos2 α = 1

⇒ cos2 α = 0 ⇒ α = 2π

Hence, angle with positive direction of the

z-axis is 2π

86. (c) Using Lagrange's Mean Value TheoremLet f(x) be a function defined on [a, b]

then, f (b) f (a)f '(c)

b a−

=−

....(i)

c ∈[a, b]∴ Given f(x) = logex

∴ f'(x) = 1x

∴ equation (i) become1 f (3) f (1)c 3 1

−=

−

⇒ e elog 3 log 11c 2

−= elog 3

2=

⇒e

2c

log 3= ⇒ c = 2 log3e

87. (d) Given f(x) = tan–1 (sin x + cos x)

f'(x) = 21

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

−+ +

2

1 12. cos x sin x

2 21 (sin x cos x)

− =

+ +

2

cos .cos x sin .sin x4 4

1 (sin x cos x)

π π − =

+ +

∴ f'(x) = 2

2 cos x4

1 (sin x cos x)

π +

+ +if f ’ (x) >O then f(x) is increasing function.

Hence f(x) is increasing, if x2 4 2π π π

− < + <

⇒ 3x

4 4π π

− < <

Hence, f(x) is increasing when n ,2 4π π ∈ −

88. (a) Given A = 5 50 50 0 5

α α α α

and | AA2 | = 25

∴ A2 = 5 5 5 50 5 0 50 0 5 0 0 5

α α α α α α α α

=

2 2

2 225 25 5 5 25 50 5 250 0 25

α + α α + α + α α α + α



∴ | AA2 | = 25 2(25 )α

∴ 25 = 25 2(25 )α ⇒ 1

| |5

α =

89. (d) We know that ex = 1 + x + 2 3x x

........2! 3!

+ + ∞

Put x = – 1

∴ e–1 = 1 1 11 1 ........

2! 3! 4!− + − + ∞

∴ e–1 = 1 1 1 1

........2! 3! 4! 5!

− + − ∞

90. (b) Given ˆ ˆ| 2u 3v | 1× = and θ is acute angle

between u and v , ˆ ˆ| u | 1, | v | 1= =

⇒ ˆ ˆ6 | u | | v | | sin | 1θ =

⇒ 6 | sin | 1θ = ⇒ 1sin

6θ =

Hence, there is exactly one value of θ for

which 2 u × 3 v is a unit vector..91. (a) Let B be the top of the wall whose

coordinates will be (a, b). Range (R) = c

A

B

C

cD

u

a

b

(a,b)

α

B lies on the trajectory

y = x tan α – 12

g 2

2 2x

u cos α

⇒ b = a tan α – 12

g 2

2 2a

u cos α

⇒ b = a tan α2 2

ga1

2u cos tan

−

α α

= a tan α 22

a1

2u sincos .

g cos

− α α

α

= a tan α2

a1

u .2sin cosg

− α α

= a tan α2

a1

u sin 2g

− α

= a tan αa

1R

−

2 2u sinR

g

α=

∵

⇒ b = a tan αa

1c

−

⇒ b = a tan α . c ac−

⇒ tan α = bc

a(c a)−

The angle of projection, α = tan–1 bca(c a)−

92. (a) Let the number of boys be x and that of girlsbe y.⇒ 52x + 42y = 50(x + y)⇒ 52x – 50x = 50y – 42y

⇒ 2x = 8y ⇒ x 4y 1

= and x 4

x y 5=

+

Required % of boys = x

100x y

×+

= 4

1005

× = 80 %

93. (b) Parabola y2 = 8xY

FX

(2,0)'X

'Y

02

x=

+

x8y2 =

Point must be on the directrix of parabola∵ equation of directrix x + 2 = 0 ⇒ x = –2Hence the point is (–2, 0)

94. (c) We know that equation of sphere isx2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0where centre is (–u, –v, –w)given x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0∴ centre ≡ (3, 6, 1)Coordinates of one end of diameter of thesphere are (2, 3, 5). Let the coordinates ofthe other end of diameter are ( , , )α β γ



∴ 23

2α +

= , 36

2β +

= , 51

2γ +

=

⇒ α = 4, β = 9 and γ = –3∴ Coordinate of other end of diameter are(4, 9, –3)

95. (b) Given ˆ ˆ ˆa i j k= + +r

, b i j 2k= − +r $ $ $ and

c xi (x 2) j k= + − −r $ $ $

If cr

lies in the plane of ar

and br

, then

[a b c]r r r

= 0

i.e. 1 1 11 1 2 0x (x 2) 1

− =− −

⇒ 1[1 – 2(x – 2)] – 1[– 1 – 2x] + 1[x – 2 + x] = 0⇒ 1 – 2x + 4 + 1 + 2x + 2x – 2 = 0⇒ 2x = –4 ⇒ x = – 2

96. (a) Given : The vertices of a right angled triangleA(l, k), B(1, 1) and C(2, 1) and Area of ABC∆= 1 square unit

Y

O X

A (1, k)

B (1, 1) C (2, 1)

We know that, area of right angled triangle

= 12

× BC × AB = 1 = 12

(1) | (k – 1)|

⇒ (k 1) 2± − = ⇒ k = – 1, 397. (c) Given : The coordinates of points P, Q, R are

(–1, 0), (0, 0), (3,3 3) respectively..

'X

'Y

Y

(0, 0)P (-1, 0)X

3/π3/2π

M

)33,3(R

Q

Slope of QR = 2 1

2 1

y y 3 3x x 3

−=

−

⇒ tan 3θ =

⇒3π

θ = ⇒ RQX3π

∠ =

∴2

RQC3 3π π

∠ = π − =

∴Slope of the line QM = tan 23π

= – 3

∴Equation of line QM is (y – 0) = – 3 (x – 0)

⇒ y = – 3 x ⇒ 3 x + y = 098. (a) Equation of bisectors of lines, xy = 0 are y = ± x

(0, 0)

y

x

y = xy = -x

∴ Put y = ± x in the given equationmy2 + (1 – m2)xy – mx2 = 0

∴ mx2 + (1 – m2)x2 – mx2 = 0⇒ 1 – m2 = 0 ⇒ m = ± 1

99. (c) Given f(x) = f(x) + f 1x

,wheref(x) = x

1

log tdt

1 t+∫∴ F(e) = f(e) + f 1

e

⇒ F(e) = e 1/ e

1 1

log t log tdt dt

1 t 1 t+

+ +∫ ∫ ....(A)

Now for solving, I = 1/ e

1

log tdt

1 t+∫∴ Put

21 1

z dt dzt t

= ⇒ − = ⇒ dt = – 2

dz

zand limit for t = 1 ⇒ z = 1 and for t = 1/e ⇒z = e

∴ I = e

21

1log

dzz1 z1z

− +∫

e

21

(log1 log z).z dzz 1 z

− = − +∫

e

1

log z dz(z 1) z

= − − +∫ [∴ log1 = 0]

e

1

log zdz

z(z 1)=

+∫∴ I =

e

1

log tdt

t(t 1)+∫



[By property b b

a af (t)dt f (x)dx=∫ ∫ ]

Equation (A) be

F(e) = e e

1 1

log t log tdt dt

1 t t(1 t)+

+ +∫ ∫e

1

t.log t log tdt

t(1 t)+

=+∫

e

1

(log t)(t 1)dt

t(1 t)+

=+∫

⇒ F(e) = e

1

log tdt

t∫Let log t = x

∴ 1dt dx

t=

[for limit t = 1, x = 0 and t = e, x = log e = 1]

∴ F(e) = 1

0x dx∫

F(e) =

12

0

x2

⇒ F(e) = 12

100. (a) f(x) = min x + 1, | x | + 1⇒ f(x) = x + 1 ∨ x ∈ R

'X

'Y

Y

(-1, 0)X

y = - x + 1

(0, 1)

y = x + 1

Hence, f(x) is differentiable everywhere forall x ∈ R.

101. (b) Given, f(x) = 2x1 2x e 1

−−

⇒ f (0) = 2xx 0

1 2lim

x e 1→−

−2x

2xx 0

(e 1) 2xlim

x(e 1)→

− −=

−

0form

0

∴ using, L'Hospital rule

f (0) 2x

2x 2x 2xx 0

4elim

2(xe 2 e .1) e .2→=

+ +

2x

2x 2x 2xx 0

4elim

4xe 2e 2e→=

+ +

0form

0

2x

2x 2xx 0

4elim

4(xe e )→=

+

0

04.e

4(0 e )=

+ = 1

102. (c)x

22

dt2t t 1

π=

−∫

∵ 12

dxsec x

x x 1

−=−

∫

∴ x1

2sec t

2− π =

⇒ sec–1 x – sec–1 2 = 2π

⇒ sec–1x – 4π =

2π

⇒ sec–1x = 2 4π π

+

⇒ sec–1x = 34π

⇒ x = sec 34π

⇒ x = – 2

103. (c) dxI

cos x 3 sin x=

+∫dx

I1 3

2 cos x sin x2 2

=

+

∫

1 dx2 sin cos x cos sin x

6 6

=π π +

∫

1 dx.

2 sin x6

=π +

∫

⇒ I = 1

. cosec x dx2 6

π + ∫But we know that

cosec x dx log | (tan x / 2) | C= +∫∴ I = 1

2. log tan x

C2 2

π + +



104. (a) The area enclosed between the curvesy2 = x and y = | x |From the figure, area lies between y2 = x andy = x

Y

X'X

'Y

y = -x

y = x

A(1, 1)

(1, 0)(0, 0) O

xy2 =

1y2y

∴Required area = 1

2 10

(y y )dx−∫1

0( x x)dx= −∫

13/ 2 2

0

x x3 / 2 2

= −

∴Required area 1 13 / 2 20 0

2 1x x

3 2 = −

2 1 13 2 6

= − =

105. (c) Let α and β are roots of the equationx2 + ax + 1 = 0α + β = – a and αβ = 1

given | | 5α − β <

⇒ 2( ) 4 5α + β − αβ <

( )2 2( ) ( ) 4α − β = α + β − αβ∵

⇒ 2a 4 5− < ⇒ a2 – 4 < 5⇒ a2 – 9 < 0 ⇒ a2 < 9 ⇒ – 3 < a < 3⇒ a ∈ (–3, 3)

106. (b) Let the series a, ar, ar2, ..... are in geometricprogression.given, a = ar + ar2

⇒ 1= r + r2

⇒ r2 + r – 1 = 0

⇒ r = 1 1 4 1

2− ± − × −

⇒ r = 1 5

2− ±

(taking +ve value)

⇒ r = 5 12−

107. (d) 1 1x 5sin cosec

5 4 2− − π + =

⇒ 1 1x 5sin cosec

5 2 4− −π = −

⇒ 1 1x 4sin sin

5 2 5− −π = −

1 1[ sin x cos x / 2]− −+ = π∵

⇒ 1 1x 4sin cos

5 5− − =

....(1)

Let 1 4 4cos A cos A

5 5− = ⇒ =

⇒ A = cos–1 (4/5)

⇒ sin A = 35

A B

C

5

4

3

⇒ A = sin–1 35

∴ cos–1(4/5) = sin–1 (3/5)∴ equation (i) become,

1 1x 3sin sin

5 5− −=

⇒ x 35 5

= ⇒ x = 3

108. (c) Tr + 1 = (–1)r. nCr (a)n–r. (b)r is an expansion of(a – b)n∴ 5th term = t5 = t4+1

= (–1)4. nC4 (a)n–4.(b)4 = nC4 . an–4 . b4

6th term = t6 = t5+1 = (–1)5 nC5 (a)n–5 (b)5

Given t5 + t6 = 0∴ nC4 . an–4 . b4 + (– nC5 . an–5 . b5) = 0

⇒ n n 5

44 5

n! a n! a b. .b . 0

4!(n 4)! 5!(n 5)!a a− =

− −

⇒n 4

4n!.a b 1 6

0(n 4) 5.a4!(n 5)!.a

− = −−

or, 1 6

n 4 5a−

− = 0 ⇒ a x 4

b 5−

=

109. (a) Set S = 1, 2, 3, ...... 12A B C S, A B B C A C∪ ∪ = ∩ = ∩ = ∩ = φ∴ The number of ways to partition

= 12C4 × 8C4 × 4C4 12! 8! 4!4!8! 4!4! 4!0!

= × ×

312!

(4!)=



110. (b) f(x) = 2x 1 x

4 cos 1 log(cosx)2

− − + − +

f(x) is defined if – 1 x

1 12

≤ − ≤ and cos x > 0

or x0 2 and x

2 2 2π π

≤ ≤ − < <

or 0 x 4 and x2 2π π

≤ ≤ − < <

∴ x 0,2π ∈

111. (a) Given : A body weighing 13 kg is suspendedby two strings OB = 5m and OA = 12 m.Length of rod AB = 13 M.Let T1 is tension in string OB and T2 istension in string OA.∴ T2 sin θ = T1 cos θ ....(i)and T1 sin θ + T2 cos θ = 13 ....(ii)

'X X

A

B

O

TC

θsinT1θcosT2

13 Kgs

12 m

13 m

5 m

2T

1T

θcosT1θsinT2

θ

θ

θ−π 2/

θ−

π2/

But givenOC = CA = CB∴ AOC OAC∠ = ∠ = θ (let)

and COB OBC∠ = ∠Now in ∆ AOB

5sin sin A

13θ = = and 12

cos13

θ =

Now putting the value of sin θ and cos θ inequation (i) and (ii) we get

2 15 12

T T13 13

= and 1 25 12

T . T . 1313 13

+ =

⇒ 12T1 – 5T2 = 0 ...(iii)⇒ 5T1 + 12T2 = 169 ....(iv)Solving equation (iii) and (iv)60T1 – 25T2 = 0–60T1 ± 144T2 = –169 × 12–––––––––––––––––––––––––––169 T2 = – 169 × 12⇒ T2 = 12 and T1 = 5∴ Tensions in strings are 5kg and 12 kg

112. (b) A pair of fair dice is thrown, the sample spaceS = (1, 1), (1, 2) (1, 3) .... = 36Possibility of getting 9 are (5, 4), (4, 5), (6, 3), (3, 6)∴ Possibility of getting score 9 in a single throw

= 4 1

36 9=

∴ Probability of getting score 9 exactly twice

= 3C2 × 21 1

. 19 9

− 3! 1 1 82! 9 9 9

= × × ×

3.2! 1 1 8 82! 9 9 9 243

= × × × =

113. (d) Equation of circle whose centre is (h, k)i.e (x – h)2 + (y – k)2 = k2

'X X

(h, k)

)1,1(−

(radius of circle = k because circle is tangentto x-axis)Equation of circle passing through (–1, +1)∴ (–1 –h)2 + (1 – k)2 = k2

⇒ 1 + h2 + 2h + 1 + k2 – 2k = k2

⇒ h2 + 2h – 2k + 2 = 0 D ≥ 0∴ (2)2 – 4 × 1.(–2k + 2) ≥ 0⇒ 4 – 4(–2k + 2) ≥ 0 ⇒ 1 + 2k – 2 ≥ 0

⇒ 1

k2

≥

114. (c) Let the direction cosines of line L be l, m, n, then2l + 3m + n = 0 ....(i)and l + 3m + 2n = 0 ....(ii)on solving equation (i) and (ii), we get

m n6 3 1 4 6 3

l= =

− − − ⇒

m n3 3 3l

= =−

Now 2 2 2

2 2 2

m n m n3 3 3 3 ( 3) 3

l l + += = =

− + − +∵ l2 + m2 + n2 = 1

∴ m n 13 3 3 27

l= = =

−

⇒ 3 1 1 1,m , n

27 3 3 3l = = = − =

Line L, makes an angle α with +ve x-axis∴ l = cos α

⇒ cos α = 13



115. (a) General equation of circles passing throughorigin and having their centres on the x-axis isx2 + y2 + 2gx = 0 ...(i)On differentiating w.r.t x, we get

2x + 2y . dydx

+ 2g = 0 ⇒ g = – dyx y

dx +

∴ equation (i) be

x2 + y2 + 2 dy

x y .x 0dx

− + =

⇒ x2 + y2 – 2x2 – 2x dydx

.y = 0

⇒ y2 = x2 + 2xydydx

116. (c) Since, p and q are positive real numbersp2 + q2 = 1 (Given)Using AM ≥ GM

∴ 2

2p q(pq)

2+ ≥

2 2p q 2pq

pq4

+ += ≥

1 2pqpq

4+

≥ 1 + 2pq ≥ 4pq

1 ≥ 2pq or, 2pq ≤ 1

pq ≤ 12

or, pq ≤ 12

Now, (p + q)2 = p2 + q2 + 2pq

⇒ (p + q)2 ≤ 1 + 2 × 12

⇒ p + q 2≤

117. (a) In the ∆ AOB,∠ AOB = 60°,and ∠ OBA =∠ OAB (since OA =OB = AB radius ofsame circle). ∴ ∆AOB is a equilateraltriangle. Let theheight of tower is h

A

B

C

O

h

a

a a

30°

30°

60°

m. Given distancebetween two pointsA & B lie onboundary of circular park, subtends an angleof 60° at the foot of the tower AB i.e. AB = a.A tower OC stands at the centre of a circularpark. Angle of elevation of the top of thetower from A and B is 30°. In ∆ ΟΑΧ

htan 30

a° =

∴ ∠ OBA = ∠ AOB = ∠ OAB = 60°

⇒1 h

a3= ⇒ h =

a

3

118. (d) We know that, (1 + x)20 = 20C0 + 20C1x + 20C2x2 + ...... 20C10 x10 + ..... 20C20 x

20

Put x = –1, (0) = 20C0 – 20C1 + 20C2 – 20C3 +...... + 20C10 – 20C11 .... + 20C20⇒ 0 = 2[20C0 – 20C1 + 20C2 – 20C3

+ ..... – 20C9] + 20C10⇒ 20C10 = 2[20C0 – 20C1 + 20C2 – 20C3

+ ...... – 20C9 + 20C10]⇒ 20C0 – 20C1 + 20C2 – 20C3 + .... + 20C10

= 12

20C10

119. (b,c) Equation of normal at p(x, y) is

Y – y = – dx(X x)

dy−

Coordinate of G at X axis is (X, 0) (let)

∴ 0 – y = – dx(X x)

dy−

⇒ dyy X x

dx= − ⇒ X = x + y dy

dx

∴ Co-ordinate of G dy

x y ,0dx

+ Given distance of G from origin = twice ofthe abscissa of p.

∴ dyx y | 2x |

dx+ =

⇒ x + y dydx

= 2x or x + y dydx

= – 2x

⇒ ydydx

= x or ydydx

= – 3x

⇒ ydy = xdx or ydy = – 3xdxOn Integrating

⇒ 2 2

1y x

c2 2

= + or2 2

2y 3x

c2 2

= − +

⇒ x2 – y2 = –2c1 or 3x2 + y2 = 2c2∴ the curve is a hyperbola and ellipse both

120. (a) z lies on or inside the circle with centre(–4, 0) and radius 3 units.

'XX

)0,4(− )0,1(−)0,7(− Real

Im.

Y

'YFrom the Argand diagram maximum value of| z + 1| = 6Second method : | z + 1 | = | z + 4 – 3 |

≤ | z + 4 | + | –3 | ≤ | 3 | + | – 3|⇒ | z + 1 | = 6


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AIEEE-2007