aieee 2002 pcm solution

8/1/2019 aieee 2002 pcm solution

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1.The inductance between A and D is ;

A~D

(a) 3.66 H (b) 9 H(c) 0.66 H Cd)1H2. A ball whose kinetic energy isE, is projected atan angle of 45" to the horizontal. The kinetic.energy of the ball at the highest point of itsflight Willbe :(a) E (b) E/. .J2(c) E/2 Cd)zero

3. From a building two balls A and B are thrownsuch that A is thrown upwards and Bdownwards (both vertically). If vA and VB aretheir respective velocities on reaching theground, then;(a) vB> VA(b) VA "'Vii(c) VA >VB(d) their velocities depend on their masses

4. If a body Ioses ha If of its velocity on penetra~i~3 em in a wooden block, then how much will Itpenetrate more before coming to rest?(a) 1 em (b) 2 em(c) 3 ern (d) 4 er n

5. If suddenly the gravitational force of attractionbetween earth and a satellite revolving aroundit becomes zero, then the satellite will :(a) continue to move in its orbit with sameveiocity(b) move tangentially to the original orbit withthe same velocity(c) become stationary in its orbit(d) move towards the earth

6. If an ammeter is to be used in place of avoltmeter, then we must connect with theammeter a:

Cal low resistance in parallel(b) high resistance in parallel(c) high resistance in series(d) low resistance in series

7. If in a circular coil A of radius R, current i isflowing and in another coil B of radius 2R acurrent 2i is flowing, then the ratio of themagnetic fields, B A and B E produced by themwill be:(a) 1 (b) 2(c) Ilf (d) 48. If two mirrors arc kept at 60to each other, thenthe number of images formed by them is :(a) 5 (b) 6(c) 7 Cd) 8

9. A wire when connected to 220 V mains supplyhas power dissipation Pl' Now the wire is cutinto two equal pieces which are connected inparallel to the same supply. Power dissipationin this case is P2' Then P~ : 1' 1 is :(a) 1 (b) 4(c) 2 (d) 3

10. If 13.6 eV energy is required to ionize thehydrogen atom, then the energy required toremove an electron from n '" '2 is :(a) 10.2 eV (b) 0 eV(e) 3.4 eV (d) 6.8 eV

11. Tube A has both ends open while tube B has oneend dosed, otherwise they are identical. The ratioof fundamental frequency of tubes A and B is :(a) 1 : 2 (b) 1 : 4(c) 2: 1 (d)4: 1

12. A tuning fork arrangement (pair) produces4 beats/s with one fork of frequency 28~ cps. Alittle wax is placed on the unknown fork and itthen produces 2 beats/so The frequency of theunknown fork is :(a) 286 cps(c) 294 cps (b) 292 cpsCd)288 cps


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13. A wave y =< a sin (rot - l eX) on a string meetswith another wave producing a node at x =< O .Then the equation of the unknown wave is:(a) y = a sin (oor+ Ta)(b)y=-asin (rot + 10:)(e) y = G sin (rot - kx)(d) y = -Gsin (rot - 10 : )

14. On moving a charge of 20 Cby 2 em, 2 J of workis done, then the potential difference betweenthe points is :(a) 0.1 V(c) 2 V (b) 8 VCd) 0.5 V

15. If an electron and a proton having samemomenta enter perpendicularly to a magneticfield, then : .(a) curved path of electron and proton will besame (ignoring the sense of revolution)(b) they will move undeflected. (c) curved path of electron is more curved thanthat of proton(d) path of proton is more curved

1,6. Energy required [0move a body or mass m froma n orbit of radius 2R to 3R is:(a) GMm/12R2 (b) GMm/3R2(c) GMm/8R Cd) GMm/6R

17. If a spring has time period T, and is cut into nequal parts, then the time period of each partwill be:(a) T . J I i(c) nT

(b)l- I n(d) T

UI. A charged particle q is placed a t the centre 0 ofcube of length L (ABCDEFGH). Another same.charge q is plated at a distance L ,fromO. Thenthe electric flux through ABeD Is :E

D~~__ +- ~~Foq - - - - - - -- _ .- - - - q

(al q/411:o L(c ) q/ 21lf,oL

/Gro:-_l3_ L.t(b) zero(d) q l 3 1 1 :& o L

19.1f in the drcuit,power dissipation is 150 W,then R is: R

20

15V

(a) 20ec ) 50

(b) 60Cd) 40

20. Wavelength of light used in an opticalinstrument are ; 0 ' 1 =4000 A . and )'2 = 5000 A ,then ratio of their respective resolving powers(corresponding to )"1 and ),'2) is :(a) 16:25 (b) 9:1(c)4:5 (d) 5:4

21. Two ldentlcal particles move towards eachother with velocity 2 1 1 and 1/ respectively. Thevelocity of centre of mass is :C a) II (b) 1 1 / 3(c) vl2 C d) zero

22. If a current is passed through a spring then thespring will :C a l expand (b) compress(c) remain same (d) none of these

23. Heal given to


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31. Infrared radiations are detected by:(a) spectrometer (b) pyrometer(c) nanometer (d) photometer32. If No is the original mass of rbe substance of

half-life period tl/2 '" 5 years, then the amountofsubstance left after 15 years is :(a) ~o (b) ~6ec ) No2 C d) No4

33. By increasing the temperature, the specificresistance of a conductor and a semiconductor:(a) increases for both(b) decreases for both .(c) increases, decreases respectively(d) decreases, increases respectively

34. If the re are 11 capacitors in parallel connected toV volt source, then the energy stored is equal to :(a) C V(c)GV2

35. Which of the following is more dose to a blackbody?(a) Black board paint (b) Green leaves(c) Black holes Cd) Red roses36. Which statement is incorrect?(a) All reversible cycles have same efficiency(b) Reversible cycle has more efficiency than anirreversible one(c) Carnot cycle is a reversible one

(d) Carnor cycle has the maximum efficiency inall cycles37. Length of a string tied to MO rigid supports is _

40 em. Maximum length (wavelength in ern) ofa stationary wave produced on it, is :(a) 20 (b) 80(c) 40 (d) 120

3S. The power factor of an AC circuit havingresistance R and inductance L (connected inseries) and an angular velocity (()is :(a) _ B _ (b} RroL (R2 + .(j}L2Y 1 2(c) roL C d ) R

R - (R2 - ( f :h?F 239. An astronomical telescope has a large apertureto:(a) reduce spherical aberration(b) have high resolution(c) increase span of observation(d) have low dispersion

40. The kinetic energy needed to project a body ofmass m from the earth's surface (radius R ) toinfinity is :(a) 0gR (b) 2mgR

2(c) mgR (d) mgR4

41, Cooking gas containers are kept in a lorry. moving wi th uniform speed. The temperature ofthe gas molecules inside will :(a) increase(b) decrease(c) remain same(d) decrease for some, while increase for others

42. When temperature increases, the frequency of atuning fork:(a) increases(b1 decreases(c) remains same(d) increases or decreases depending on thematerial

43. If mass-energy equivalence is taken intoaCCOUnt , when water is cooled to form ice, themass of water should :(a) increase(b) remain unchanged(c) decrease(d) first increase then decrease

44. The energy band gap is maximum in :(a) metals (b) superconductors(c) insulators (d) semiconductors

45. The part of a transistor which is most heavilydoped to produce large number of majorityca rri eIS is :(a) emitter(b) base(c) collector(d) can be any of the above three

46. In a simple harmonic oscillator, at the meanposition .:(a) kinetic energy isminimum, potential energy

is maximum(b) both kinetic and potential energies aremaximum(c) kinetic energy is maximum, potentialenergy is minimum(d) both kinetic and potential energies areminimum

47. Initial angular velocity of a circular disc of massM is (OJ . Then two small spheres of mass mareattached gently to two diametrically oppositepoints on the edge of the disc. What is the finalangular velocity of the disc?


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(M + m )a).~ @t (M+ m )b) -n-l - .;1 .(e ) ( M . ~ 4n D (j)l C d ) ( M ~ 2 m ) ff\48. The minimum velocity (in ms'") with which acar driver must-traverse a flat curve of radius150 m and coefficient of friction 0.6 to avoidskidding is:(a) 60(e) 15 (b) 30C d ) 25

49. A cylinder of height 20 m is completely filledwith water. The velocity of efflux of water[in ms-1 ) through a small hole on the side wallof the cylinder near its bottom, is ;(a) 10 (b) 20(c) 25.5 (d) 5

50. A spring of force constant 800 N/m has allextension of 5 cm. The work done in extending .it from 5 em to 15 cm is:(a)16J (b)8J(c) 32 J Cd) 24 J

51. A child Swinging on a swing in sitting position,stands up, then the time period ofthe swing will :(a) increase(b) decrease(c) remain same(d) increase if the child is long and decrease ifthe child is short

52. A lift is moving down with acceleration a. Aman in the lift drops a ball inside the lift. Theacceleration of the ball as observed by the manin the lift and a man standing stationary on theground a re respectively;(a) g, g (h) g - a, g - a(c) g - a, g (d) .c, g

53. The mass ofa product liberated on . anode in anelectrochemical cell depends on :(a) (It )1/2 (b) It(c) f/t Cd) J 2 t(where t is the um e period for which the currentis passed) .

54. At what temperature is the rms velocity of ahydrogen molecule equal to that of an oxygenmolecule at 4rC ?(a) 80 K(c) 3 K (b) -73 KC d ) 20 K

55. The time period of a charged particleundergoing a circular motion in a uniformmagnetic field is independent of its;(a) speed (b) mass(c) charge (d) magnetic induction

56. A solid sphere, a hollow sphere and. a ring arereleased from top of an inclined plane(frictionless) so that they slide down the plane.Then maximum acceleration down the plane isfor (no rolling) :(a) solid sphere(c) ring (b) hollow sphere(d) all same

57. In a transformer, numberoftums in the primaryare 140 and that in the secondary are. 280. Ifcurrent in primary is 4 A, then that in thesecondaryis :(a) 4 A(c) 6 A (b) 2ACd) 10 A

58. Even Carnot engine cannot give 1000/0efficiency because we cannot:(a) prevent radiation(b) find ideal sources(c) reach absolute zero temperature(d) eliminate friction

59. Moment of inertia of a circular wire of mass Mand radius R about its diameter is:(a)MR72 Cb) MR2ec ) 2MR2 '(d) MR'l4

60, When forces F1, F2, Faareacring on a particle ofmass m such that F 2 and F 3 are mutuallyperpendicular, then the particle remainsstationary. If the force Fj is now removed thenthe acceleraticnof the particle is :(a) F1/m (b) F2Fj/mFI(c) (F2 - F3)!m (d) F2/m61. Two forces are such that the sum of theirmagnitudes is 18 N and. their r e su lt ant wh i ch hasmagnitude 12 N, isperpendicular to the smaller. force. Then the magnitudes of the forces are:(a) 12N,.6N (b) 13N,5N(c) 10N,8 N Cd) 16 N, 2 N

62 ..Speeds of twO identical cars are ,U and 4u at aspecific instant. The ratio of the respectivedistances at which the two cars are stoppedfrom that instant is :(a) 1 :1(c) 1:8

(b) 1:4Cd) 1 :16

63. 1 mole of a gas with Y"' ' ' 7/5 is mixed with1 mole of a gas with 'Y = 5/3, then the value of yfor th e resulting mixture is :(a) 7/5 (b) 2/5ec ) 24/16 (d) 12/7

64. If a charge q is placed at the centre of the linejoining two equal charges Q such that thesystem is in equllibrium then the value of q is :.(a) Ql2 (b) - Q/2ec ) Ql4 Cd) - Q/4


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65. Capacitance (in F) of a spherical conductorhaving radius 1 m , is ;(a) 1.1 x 10-10 (b) 10-6(c) 9 x 10-9 (d) 10-3

66. A light string passing over a smooth light pulleyconnects MO blocks of masses m l and m2(vertically). Jfthe acceleration of the system isg /8, then the ratio of the masses is ;(a) 8 .;1 (b) 9 ; 7(c) 4: 3 (d) 5 : 3

67.Two spheres of the same material have radii1 m and 4 m and temperatures 4000 K and2000 K respectively. The ratio of the energyradiated per second by the first sphere to that byth e second is:(a) 1 ; I(e) 4: 1 (b) 16 : I(d) 1 : 9

68. Three identical blocks of masses m '" 2 kg are.drawn by a force F ; : 10.2N with anacceleration of 0.6 ms-2 on a frictionlesssurface, then what is the tension (in N) In thestring between the blocks Band C ?

(a) 9.2ee ) 4

F

(b) 7.8Cd) 9.8

69. One end of massless rope, which passes over amassless andfrictionlesspulley P is tiedto a hook Cwhile the otherend is free.Maximumtension thatthe rope can ~~~9rnm~~m~ii.liif%mlbear is 360 N.With what value of maximum safe acceleration(in ms-2) can a man of 60 kg climb on the rope?(a) 16(e) 4

70. A particle: ofmass m movesalong line PCwith velocity vas shown.What is theangularmomentum ofthe particleabout O ?

(b) 6(d) 8

(a) m vL(c) mvr

(b) m vlCd) Zero

71. Wires 1 and 2 carryingcurrents (1 and i2respectively are i,inclined at an angle eto each other. What isthe force on a smallelement dl of wire 2 ata distance r from wire1as shown in figure)due to the magneticfield of wire 1 ?(a) i : r iji Z dl tan e (h) i : r ii dl sin 0(c) 2 J! 0 _ 1 11 2 dt cos 0 (d) 4 1 1 0 i1i2 dl sin 8nr - 1 1 : r

2

c

72. At a specific instant emission of radioactivecompound is deflected in a magnetic field. Thecompound can emit;(I) electrons (ii) protons(iii) He2+ (iv) neutronsThe emission at the instant can be :(a) i, ii, iii (b) I, ii, iii, iv(c) iv (d) i i, i ii

73. Sodium and copper have work functions2.3 eV and 4.5 eVrespectively. Then the ratio ofthe wavelengths' is nearest to : .(a) 1 : 2 (b) 4 : 1(e) 2: 1 (d) 1:4

74. Formation of covalent hands in compoundsexhibits :.(a) wave nature of electron(b) particle nature of electron(c) both wave and particle nature of electron(d) none of the abovex xxxxxx75. A conducting isquare loop of x x x x x X xside Landresistance R x L x x x x x x vmoves in its plane !with a uniform x XXX)(Xxvelocity v x xxxxxxperpendicular toone of its sides. Amagnetic induction B constantin time and space, pointing perpendicular andinto the plane at the loop exists everywherewith half the loop outside the field, as shown infigure. The induced emf is : .(a) zero (b) RvB(c) vBL Cd) vBLR


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- . .~ ~ - - - - - - - - - - ~ - - - - - - - - - - - - - - - -76, Which of.the following is a redox. reaction ?(a) NaC! + KN03 -----,) NaN03 + KC!(b) CaCp4 + 2HCl -----)0 CaC!z + HaCP4(c) Ca(OH)z + 2NH4Cl ~ CaClz + 2NH3+2HzO

(d ) 2 K[A g C C N)2 J + Zn -----)0 2A g+ Kz[Zn(CN)4]77. For an ideal gas, number of mol per litre interms of its pressure P, temperature T and gasconstant R is :PT(a) If C O ) PRTec ) % T (d) R J

78, Number of P-O bonds in PPIO is :C a l 17 (b) 16(c) 15 (d) 6

79, KO a is used in space and .submarines because it :(a) absorbs CO:2 and increases Oj concentration(b) absorbs moisture(c) absorbsCOz(d) produces ozone80. Which of the following ions has the maximummagnetic moment?

(a) Mn2+(c) Ti2 .,"

(b) Fe2T(d) Cr2-t

81. Acetylene does not react with :(a) Na (b) ammoniacal AgN03ec ) HCl (d) NaOH

82. CompouudA given below is:

~~60HA

(a) antiseptic(c) analgesic. " (b) antibiotic(d) pesticide

83. For the following cell with hydrogen electrodesat two different pressures P : J . and P 2PtCHa) I HT(aq) I Pt(H2)PIemf is given by:

RTI P l(a) -P " og,- p zRT P 2(c)f log, P J1M

RT PI(b) 2F log. P 2(d ) RT I. f J - , 22F og. Pl

84. Acetylene reacts with hypochlorous acid to form :(a) C I2CHCHO (b) C IC H2COOH(c) CH 3COCI (d ) C I CH 2CHO85. On heating benzyl amine with chloroform andethanollc KOH, product obtained is :

(a) benzyl alcohol (b) benzaldehyde(c) benzonitrile (d) benzyl isocyanide86, Which of the following reaction is possible atanode?(a) F2 + 2e- ------) 2F-

(b) 2H+ +~ O2 + u- ,---')0 Hp(c) 2Cr]+ + 7HP ---+ Cip?-+ 14H+ + 6e-(d) Fez+ -----)0 Fe3+ + e:

87. which of th e following concentration factor isaffected by change intemperature?(a) Molarity (b) Molality(c) Mole fraction Cd)We. igh t fraction88. Cyanide process is used for the extraction of :(a) barium (b) silver :

(c) boron (d) zinc89. Following reaction

(CH3hCBr + HzO -----)0 (CH3)3COH + HBris an example of :(a) elimination reaction(b) f ree radlcal substitution(c) nucleophilic substitution(d)electrophilic substitution

90. A metal M forms water soluble MSO 4 and inertMO. MO in aqueous solution forms insolubleM(OH)2 soluble inNaOH. Metal Mis:(3) Be (b) Mgec ) Ca (d) Si

91. Half-life of a substance A foito\ilg first order'kinetics is 5 days . . Starting with 100g of A,amount left after 15 days is :(a) 25 g (bJ 50 g "(c) 12.5 g (d) 6.25 g

92, The most stable ion is:(a) (Fe(OH)s]3- (b) [FeC16]3-(c) [Fe(CN)6]J- Cd) [Fe(Hp)6]3T

93. A substance forms Zwitter ion. It can havefunctional groups:(a) ~NH2' ~COOH (b) ~NH2' ~S03H(c) both (a) and (b) (d) none of these


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94. If FeHand CrHboth are present in group III ofqualitative analysis, then distinction can bemade by:(a) addition of NH40H in presence of NH4CIwhen only Fe(OH)3 is precipitated(b) addition of NH40H in presence of NH4Clwhen Cr(OHh and Fe(OH)3 both areprecipitated and on adding B r2 water andNaOH, Cr(OH)3 dissolves(c) precipitate of Cr(OHh and Fe(OH)3 asobtained in (0) are treated with conc. HClwhen only FeCOH)3dissolves(d) both (b) and (c)

95. In an organic compound of molar mass 108 gmol" C, Hand N atoms are present in 9: 1 : 3.5by weight. Molecular formula can be :(a) C6H a N z (b) C7HlON(c) CSH6N3 (d) C4H1SN3

96. Solubility ofCa(OHh is s mol 1,-1. The solubilityprod uct (Ksp ) under the same condition is :(a) 4 53 (b) 354(c) 4s 2 Cd)s3

97. Heat required to raise the temperature of1mole of a substance by lOis called:(a) specific heat(b) molar heat capacity(e) water equivalent(d) specific gravity

9S. ~particle is emitted in a radioactive reactionwhen:(a) a proton changes to neutron(b) a neutron changes to proton(cja neutron changes to electron(d) an electron changes to neutron99. In a mixture of A and B, components shownegative deviation when :

(a) A-B interaction is stronger than,A-A andB-B interaction(b) A-B interaction is weaker than A-A andB-B interaction

(c) LHt;nix > 0, f:..Sm ix > 0'; (d ) t.Vm ix = = 0, f:..Sm ix > 0

100. Refining of impure copperwith zinc impurity isto be done by electrolysis using electrodes as :

cathode Anode(a) pure copper(b) pure zincec) pure copperCd)pure.zinc

pure zincpure copperimpure copperimpure zinc

101. Aluminium is extracted by the electrolysis of:(a) alumina(b) bauxite(c) molten cryolite(d) alumina mixed with molten cryolite

102. For an aqueous solution, freezing point is- O.186C. Elevation of the boiling point of thesame solution is (K f '" 1.86 mol' kg andKb '" 0.512 mol:' kg) :(a) 0.] 86 o 0.0512ec) ] .860 (d) 5.120

103. Underlined carbon issl hybridised in :(a) CH3~H =CH2 (b) CH3~H2NH2e c) CH3~ONH2 (d) CH3CH2~N

104. Bond angle of 10928' is found in :(a) NH] (b)H20 (e) CH5 (d) NH4

105. For a reaction A + 2B ---* C, rate is given bydrC]+dt= k[A] [BJ, hence the order of thereaction is :(a) 3(c) 1

(b) 2(d) 0

106. CH3MgIis an organometallic compound due to :(a) Mg-I bond (b) C-I bondCc)C--Mg bond (d) C-H bond107. One of the following species acts as bothBronsted acid and base;

(a) H2P02 (b) HPO~-(c) HPO~- (d) all of theselOS. Hybridisa tion ofthe underline atom changes in :(a) , & n - I3 changes to A I H : ;(b) H2Q changes to H30+

(c) ~3 changes to NH~(d) in all cases

109. Racemic mixture is formed by mixing two : \(a) isomeric compounds(0) chiral compoundsec) meso compounds(d) enantiomers with chiral carbon110. The number of lone pairs on Xe in XcF2 , XeF4

and XeF6 respectively are :(a) 3, 2, 1 (b) 2, 4, 6(c) 1, 2, 3 (d) 6, 4, 2

111. An aqueous solution of 1MNaCI and 1MHC]is:(a) not a buffer but pH < 7(0) not a buffer but pH > 7(c) a buffer with pH < 7(d) a buffer with pH > 7


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125. A square planar complex is formed byhybridisation of the following atomic orbitals;(a) s, p " " Py, P:(b) s, p " " Py' P., d(c) d, 5, p" , v,(d) 5, p " , > Py' P., d, d

129, Type cf isomerism shown by[Cr(NH3)sNOz]Clzis:(a) optical(c) geometrical (b) ionisation(d) linkage

127. One of the following equilibria isnot affected bychange in volume of the flask :(a) PCls( g) ~ PC13(g) + CI2(g)(b) N2(g) + 3H2(g) ---,..> 2NH3{g)e c ) Nz(g) + 02(g)~ 2NO(g)(d) SOzC12(g) ~ 80 2(g) + CI2(g)128. Uncertainty in position of a particle of 25 g inspace is 10-5m Hence, uncertainty in velocity(ms" lis (Planck's constant h ""6.6 X 10~34 Js) :(a) 2.1 x 1O-2~ (b) 2.1x 10-34(c) 0.5 x 10-34 Cd) 5.0 x 10-24

129. Consider the following reactions at 11000 C(I) 2C + O2 ----). 2CO, t.Go=- 460 kJ mol"!(II) 2Zn + 02 ~ 22nO,6 .Go =- 360 k J mol'"Based on these, select correct alternate:(a) zinc can be oxidised by CO(b) zinc oxide can be reduced by carbonec ) both (a) and (b)(d) none is the correct

130. A reaction is non-spontaneous at the freezingpoint of water but is spontaneous at the boilingpoint of water then: .Mf Ll.S

(a) +ve +ve(b) - ve -ve(c) - ve +ve(d) +ve -ve

131. Monomers are converted to polymer by:(a) hydrolysis of monomers .(b) condensation reaction between monomersec ) protonation of monomers(d) none. of the above

132. Increasing order of bond strength ofo, 0; O~- and O~ is :(a) O~


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(a) (CH3hC---~> (CH:3)zCH- > CHjCI-f2-(b) CH3CH2 - > (C1-13)zCH- > (CH3hC-(c) (CH3)2CH - > CH3CH1 - > (CH3)3C--(d ) (CH3 )2C - > CH3CHz ,- > eC A3)3CH-141. PC 1 3 and I?Cls both exist; NC I3 exists but NC lsdoes not exist. It is due to ;C a l lower electronegatlvity ofPthan N,(b) lower tendency of N to form covalent bondee ) availability of vacant d-obital in P but not in

N(d) statement is itselfincorrect

,142. Following types of compounds (as I, II)(I) CHJCH=CHCH3 (II) CH3-CH-OHICll2CH3are studied in terms of isomerism in :(a) chain'isomerism (b) position isomerism(e) conformers (d) stereoisomerism

143. C.onductivity (Seimen's S) is directlyproportional to area of the vessel and the, concentration of the solution in it and isinversely proportional to the length of thevessel, then constant of proportionality isexpressed in :(a) S m marl(c) 5 m2 mol'"

(b) S 2 m2 mor~2(d) 52 m 2 mol

144. A heat engine absorbs heat 'b from a source attemperature T ] and heat q2 from a source attemperature T2' Work done is found to beJ (ill + q 2 \ This is in accordance with :(a) first law of thermodynamics(0) second law of thermodynamics(c) Joules equivalent law(d) none of the above

145. Select correct statement:(a) when a covalent bond is formed, transfer ofelectrons takes place(b) pure H?O does not contain any ion(e) a bond is formed when attractive ,forcesovercome repulsive forces(d) HF is less polar than HBr

146. The metallic sodium dissolves in liquidammonia to fonn a deep blue coloured solution.The deep blue colour is due to formation of:(a) solvated electron, e- (NH3);

(b) solvated atomic. sodium, Na(NH~J ye e ) (Na+ + Na-)C d ) NaNHz + I-f2147. Maximum dehydration takes place that of:

(a) & O H (b)60 HCH3

(d)6H

148. SN 1 reaction is feasible in :(a) ~ Cl+ KOB---l>

C 1eb)N + KOH--'"ec ) @-CI + KOH--+Cd) @-CH2CH zCl + I


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---------------------1.Ifa 7- p and a 2 =Sa - 3, 1 3 2 =5 P - 3, then theequation having a/~ and 1 3 1 a as its roots, is :(a) 3x2.+ 19x + 3= 0 (b) 3x2 -19x + 3= 0(c) 3x2 -19x - 3=0 (d) x2 - 16x + 1 = 0

2.lf Y '" (x + ~1 + x2 t,then. d2y dy(1 + X2) -dx2 + x dx is:

(a) n2y(c) - y

(b) - n2y(d) 2x2y

s.ir i, log3le31-X + 2), log3 (4,3'" -1)areinAP,then x equals :(a) 1 0g3 4ec ) 1 -log4 3 (b) 1 -log3 4Cd) \Og4 34. A problem in mathematics is. given to threestudents A, B, C and their respective frobabilityof sol~ng the problem is ~, " 3 and ~.Probability that the problem is solved, is :(a) 3/4 (b)1/2(c) 2/3 Cd) 1(3

5, The period of sin 2 . e is :(a) rr2 (b) n(c) 2n (d) n/2

6. t m, 11are the pth, qth and rth term of an GP andlog I p 1all positive, then logm '1 1 equals:

logn r ](a) 3 (b) 2ee ) 1 (d) zeroI, $- cos 2 x .7 _ lin r: IS :x-;O ,12x(a) "-ee ) zero (b) ~ 1(d) does not exist

8. A triangle with vertices (4, 0), (-1, -1), (3, 5) is :(a) isosceles and right angled(b) isosceles but not right angled(c) right angled but not isosceles(d) neither right angled nor isosceles

9. In a class of 100 students there are 70 boyswhose average marks in a subject are 75. If theaverage marks of the complete class is 72, thenwhat is the average of t h e girls?(a) 73 (b) 65C e) 68 (d ) 74

chensin x is equal to :C a l ta;12 ( % J(c) tan u

(b) cot2 ( % )(d) cot ( % )

11. The order and degree of the differemial" ( d ) 2 / 3 dJequation 1 1 - 3 g : , ' " 4 dx ; are :

(a) ( J , ~ ) (b) (3, 1)ee ) (3, 3) (d) 0, 2)

12. A plane which passes through the point (3, 2,0). x-4 y-7 z-4.and the line -. -1- = = -5- =-4-IS :(a) x - y + z = 1 (b) x + y + z = = 5(c) x + 2y - z '" 1 (d) 2 x - y + z =5

13. The solution of the equation d2~ = = e.. .> : is :dxe'2x(a)-4I'(c) - e-2x + cx l + d4(

2 )"x+Sx+3.14. lim -z--- IS equal to :X-;oo X + X + 2(a) e1 (b) e2(c) e3 (d) e

IS. The domain of sin -1 [log3 - e x 13)] is :(a) [1,9] (b) [-1.9]ec ) [-9, 1 J (d) [-9, -:1]

16. The value of il4 .41/8,81116 ... cois :(a) 1(c) 3/2 (b) 2(d) 4

17. Fifth term of a GP is 2, then the product of its 9tenus is:(a) 256C e ) 1024

18. f ; O ~ I sin x I dx is:(a) 20C c ) 10

(b) 512(d) none of these

(b) 8Cd) 18


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19. In ~ f01lf4 "dx- J , tan x , then lim n [I" + 1,,+2]Pl---+o!)

equals;(a)!2(e) 00

20. I ol ( x2] dx is;(a) 2-. . /2ee ) J2-1

(b) 1C d ) zero

(b)2+J2(d) - J2 -.J3 + 5

21 I n 2x(l + sin x) dx' . IS -l[ 1+eos2x

112(a) -4ec ) zero

(b) 112C d ) ~2

22. The period of thef(x)= sin" x + cos" x is: function

(a) 11(c) 21t

(b) 2 !_2(d) none of these

23. The domain of definition of the functionfex) = IOg10 ( s x ~ X l ) is:

(a) [1, 4](e) [0, 5)

(b) [1, 0]C d ) (5, 0]

24. If sin y = x sin (a + y), then :t is :(a) . sina (b)Sin2.(a+y)sm2 (a+ y) sin a(c) sin a sin 2 (a + y) (d) sin 2 .(a - y)sin a

25. If xY = e" ' -Y, then Z is ;(a) 1+ x b 1- Jd g x1 + log x ( ) 1 + log x(c) not defined (d) log x(1 + log xi

x3 _ 3xy2 + 2 =0 and6. The two curves3x2y - y3 - 2= 0:(a) cut at Tight angle (b) touch each other(c) cut at an angleiCd)cut at an angle %27. The function f(x) = cot" x + x increases in theinterval:(a) (1,00)(c) (- 00, 00)

(b)(-l, 00)C d ) (0, 00)

28. The greatest value offex) = ex + 1)113 - e x - IF3 On [0, 1] is ;(a) 1 (b) 2ec ) : 3 (d) 1/3

29. Evaluate f O W 2 ~ dx :JSin x + .lcos x(a) 2 ! _ (b) 2 !_4 2(e) zero (d) 1

30. f dx is equal to :x (x" + 1)1 ( X " )a) - log -- + cn x" + 1

(b) _! log ( x t ! + 1 ) + cn x"( X" )ec ) log _- _ .. + cx" + 1

(d) none of these31. The area bounded by the curve y O " 2x - Xl and

the straight line y =- x is given by :()9. b 43 .a " 2 sq urut ( ) (; sq urut(c 3S .e) (; sq unit (d) none of these

32. The differential equation of all non-verticallines in a plane is ;

d2y(a) dx2 .=0dy(c) d X ' = 0

2(b) d ~ = 0dy~dx(d) --.= 0dy

33. Given two vectors are i-and i+ 2) the unitvector coplanar with the two vectors andperpendicular to first is :

1 - - 1 - A(a) J2 (i + j) (b) 15 (2 i+ j)(d) none of these

34. The vector i+ xl + 3 k is rotated through anangle 0 and doubled in magnitude, then itbecomesa] + (4x - 2) 1+ 2k. Thevalueofx are:( a ) { - ~ , 2 } ( b ) H , 2 }( C ) H , o } (d){2,7}


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35. A parallelepiped is formed by planes drawnthrough the points (2, 3, 5) and (5, 9, 7), parallelto the co-ordinate planes. The length of adiagonal of the parallelopiped is :(a) 7 unit (b) ' . / 3 8 unit(e) Ms unit (d) none of these

36. The equation of the plane containing the linex - Xl = Y - y\ = Z - Z isI In na(x - Xj)+ b (y - y\)+ C (z - zd= 0, where:(a) ax] + by) + cz] e- (b) 01 + bm + en =0abc(c)-=-=-, I m nC d) !xl + mY l + nZl " " 0

37. A and B playa game where each is asked toselect a number from 1 to 25. If the twOnumbers match, both of them win a prize. Theprobability that they will hot win a prize in asingle trial, is :1(a) 2S

2(c) 25(b) 242S(d) none of these

38. If A and B are two mutually exclusive events,then:(a) PCA)

P C B )(d) none of these

39. The equation of the directrix of the parabolay2 + 4y + 4x + 2"" 0 is ;(a)x=.-l (bjxel(c)x=-3/2 (d)x=3/2

40. Let T" denote the number of triangles which canbe formed using the vertices of a regular polygonofn sides. IfT,.,] - Til = 21, then n equals:(a) 5 (b) 7(e) 6 (d) 4

41. In a triangle ABC,(a) a2 4- b2 _ ,2(c) b 2 - c2 _ a2

. A-B+C. al2ea Sill --2-- ISequ, to:(b) c2 + a2 _ b2C d) c2 - a2 _ b2

42. For x ER, lim (X - 3 ) Xx + 2 is equal to ;(b) e-1C d) eS(a) e(c) e-5

43. The incenrre of the triangle with vertices(1, J3), (0, 0) and (2, 0) is :(a) ( 1 , 1) (b) ( ~, 1)

( 2 . . / 3 )(c) 3' 2'44. If the vectors;i,'; and 7 from the sides BC, C 4 .

and AB respectively of a triangle ABC, then:(a) "1 .ti = ti .1 = = '1.-g = 0(b) "1 x ti = 'Ii x1~1'1ee ) 1'ti ;;B ' 1;;7 ."1 = = 0C d) '1x;i +1" 1 + " 1 x ' 1 ' " a45. If ())is an imaginary cube root of unity, then(1 + ro - (fll equals:(a) 128 ro (b) -128 (Oee ) 128 o . l . C d ) -128 0,)26i -3i 1

46. If 4 3i -1 = x + iy, then:20 3 i

(a) x = = 3, y = = 1 (b) x = = 1, Y = 3ee) x ,,:,0, y ""3 (d) x '= 0, y = 047. SiD 2 e = 4xy 2 is true i f and only if:(x+ y)

(a) x + y 0(c) x = y (b) x '= y, x if: 0, Y '" 0(d) x '" 0, y ...0

48. The radius of the circle passing through the foci2 2of the ellipse ~ 6 + ~ = 1 and having its centreat (0, 3), is :(a) 4 unit'ec ) . J U unit

(b) 3 unit(d) j _ unit

49. The probability of lndra winning a test matchagainst West-Indies is 1/2 assumingindependence from match to match. Theprobability that in a match series India's secondwin occurs at the third test is :1(a) -81ec ) -2

50. If 0 ;0'1) is a cubic root of unity, then1 l+i+fJ/ 0 ) 2

1- i -1 w 2 -1 equals:-i -l+ro-i -1

(a) zero (b) 1(c)i C d) 00

51. A biased coin with probability P, < p < 1, ofheads istossed until a head appears for the firsttime. If the probability that the number of tossesrequired is even, is 2/5, then p equals:


14/38

Cal 1/3(c) 2/5 (b)'2/3C d) 3/552. A fair die is tossed eight times. The probabilitythat a third six is observed on the eight throw, is :7C 55 7C 55Cal 2 x (b) _2_x_67 68

(c) 7 C 2 6X 55 I f(c ) none 0 these653. Let f(2) = = 4 and f '(2) =4. Then

I x f(2) - 2f(x). . bim IS given y:x.,2 X - 2(a) 2(e) -4 (b) - 2(d) 3'

54. Three straight lines 2x + l1y - 5", 0,24x + 7y - 20=0 and 4x - 3y - 2= 0:(a) fonn a triangle(b) are only concurrent(c) are concurrent with one line bisecting theangle between the other two(d) none of the above

55. A straight line through the point (2, 2) intersectsthe lines 1 3x + Y =0 and -J3 x - Y = 0 at thepoints A . and B. The equation to the line An sothat the triangle DAB is equilateral, is :(a) x"- 2", 0(c) x + Y - 4 = 0

(b) y - 2= 0(d) none of these

56. The greatest distance ofthe point P(10, 7)fromthe circle x2 + y2 - 4x - 2y - 20 = 0 is :(a) 10 unit (b) 15 unit(c) 5 unit (d) none of these

57. The equation of the tangent to the circlex2 + y2 + 4x - 4y + 4", 0 which make equalintercepts on the positive co-ordinate axes, is :(a) x + y '" 2 (b) x + y =2Ii(c)x+y=4 (d)x+y=8

58. The equation of the ellipse whose foci are( 2, 0) and eccentricity is 1/2, is :_ x2 y2 x2 y2(a) 12 + 16 = 1 (b) 16 + 12' = = 1

2 2(c) ~6 + Ys =1 (d) none of these59. The equation of the chord joining two points

(xI' YI) and ( x~, Y2) on the rectangularhyperbola ry = c

2is:(a) - - - - .3 . . . _ _ + y = = 1

Xl + X2 Y J + Y2X Y(b)--- + --=1XI - X2 Yl - yz

(c) x + "-Y--==lY l + Y2 Xl + X2

X YCd)--+--=lYI - Y2 Xl - X2--+ ~ ~ ~60. If the vectors a = = X i+ Y j + z k and such that

~ , - t _ and Itform a right handed system, then_~ . -C IS:Ca)zl-xk (b)O(c)y] (d)-zi+xk

61.The centre of the circle given by- - + ~ ~ ~ - - + A A. . . C i+ 2 j + 2k) '" 15 and I r - (j + 2k) I = 4is:(a) (0, I, 2)(c) (-1,3,4) (b) (1, 3, 4)(d) none of these

}- tan2 15062. The value of 2 is :1+ tan '150

(a) 11 3ec) ~-2

(b) " r 3Cd)2

63. If tan (~= - i,then sin 0 is :4 4(a) - - but not-5 . 5

4 4(c) - but not --5 5

4 4(b) -" 5 or 5(d) none of these

64. If sin (a + ( 3 ) '" 1, sin (u - B) =~ ,tan (o, + 211) tan (2a + 0)is equal to :(a) 1 (b)-1(e) zero (d) none of these

65. If y = sin2 0 + cosec'B, B;


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69. If (l is a root of 25eos2 a + 5 cos e -12~0 (a) log ex - 1) (b) log x~ < a < 1t, then sin 20: is equal to : (c) x (d) none of these73. The coefficient of x5 in (1 -I - 2x + 3x2 + ... rJ/224 eb) _ 24(a) 25 is:25 (a) 21 (b) 2S13 (d) _ 13ee) 18 18 ee) 26 (d) none of these

70. tan -1 (~ ) + tan -I (~) is equal to : 74. If I x I < 1, then the coefficient of x'] inexpansion of (1 + x + x2 + x3 + ... )2 i s :(a) ~ cos-1 (~) (b) j sin"! (~) (a) n (b) n-1ee ) 1 1 + 2 (d) n + 1z(e) ~ tan " (~) C d) -: G J 75. The number of real roots of 321


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H I N T S IIO L U T I O N S

1.A

Here, inductors are in parallel

c

111 11= '3+ '3+"3L =1

2. At the highest point of its flight, verticalcomponent of velocity is zero and onlyhorizontal component is left which isUx =tz cos eGiven: e = 45

uu~""u cos 45" = r ; : ; : .. . . , . 2Hence, at the highest point kinetic energyE ' ' ' ' ' ~ m u ; ",~m ( J z f = ~ m ( ~ )

= ~ ( . .'. ! . mu2 = E ). 23.From conservation of energy,potential energy at height h

'" KEat groundTherefore, at height h, PE of ball A

PE"' mAgh1 2KEat ground = 2 " m AY A

$0, 1 2mAgh = = " 2 mAYAVA ",.J2ghVB"' fiihimilarly,Therefore, VA'" VB

Note : In question it is not mentioned thatmagnitude of thrown velocity of both balls aresame which is assumed in solution.

4. Let initial velocity of body at point A is v, AB is3cm.vA ~ ~+-3cm~-x~vl2

y 2 " , u2 - 2 a s( ~ r " , y 2 -2ax3v2a"'g

Let on penetrating 3 em in a wooden block, thebody moves x distance from B to c.So, for B toC

From

vu =2' V = 0,v2S = x, a = 8' (deceleration)

2 tv) 2 v2(0) '" ( 2 . - 28-xx =1 em

Note : Here, it is assumed that retardation isuniform.5. When gravitational force becomes zero, thencentripetal force required can not be provided.So the satellite will move with the velocity as ithas at the instant when gravitational forcebecomes zero, i .e., it moves tangentially to theoriginal orbit,6. A voltmeter is a high resistance device and isalways connected in parallel with the circuit.While an ammeter is a low resistance deviceand is always connected in series with thecircuit. So, to use ammeter in place ofvoltmetera low resistance must be connected in parallelwith the ammeter to make its resistance small.

7. Magnetic Held in circular coil A isBA = Ilo!!i2R

where R isradius and i iscurrent flowing in coiL


17/38

Similarly

360"S. Number of images n= --8 ~ 1where 0 is angle between minors,Thus, 8= 60So, number of images

3600n = = - - " . , - - 1 = = 5600

(given)

9.In 1stcase:V 2Using the formula P = = If ... (1)

where.R is resistance of wire, Vis voltage acrosswire and P is power dissipation in wire andR - e . ! (2-)- A ...

From Eqs, (1) and (2)VZ V2PI "" pI!A ""pI" AV2PI =pr.A ....(3)

In 2nd case:Let Rz is net resistance.R xR RR2 "" R + ]f="2where, R is the resistance of half wire.

. p , ( i )R'2 = = _ _ 2__ = = . E ! _ _A2 4AV2p - ) _ =~4Api ....(4)

Hence, from Eqs, (3) and (4)PI 1Pz = " 4P z 4~ PI "'f

10. Energy required to remove an electron from nth .orbit is ,

Here, n = 2Therefore, E 2 =~ 13~6=-3.4 V2

11. Let the tubes A and ~ have equal length calledas L. Since, tube A is opened at both the ends,"therefore, its fundamental frequency

....(1)Since, tube B is closed at one end, therefore, irsfundamental frequency vl ie'" 41From Rqs. (1) and (2), we get

! _ ! . ! 1 = = v /2/ ""_ :1 : " " 2: 1riB v/41 2

12. The tuning fork of frequency 288' Hz isproducing 4 beats/s with the unknown tuningfork r.e., the frequency difference betweenthem is 4. Therefore, the frequency of unknowntuning fork'" 288 4", 292 Or 284On placing a little wax on unknown tuning fork,its frequency decreases hut now the number ofbeats produced per second is 2 ie., the

frequency difference now decreases. It ispossible only when before placing the wax, thefrequency of unknown fork is greater than th efrequency of given tuning fork. Hence, (hefrequency of unknown tuning fork= 292 Hz.13. Equation of a wave

y '" a sin (rot - kx)Let equations of another wave may be,y '" a sin (M + 10:) .. :(2)Y '" - a sin (e M + le x ) . . . (3)

If Eq. (1) propagates with Eq. (2), then we getY '" 2a co s kx sin ror ... (4)

... (2)

... (1)

If Eq. (1), propagates with Eq. (3), then we get .y '" - 2a sin kx cos W{ ... (5)After putting x = 0 in Eqs, (4) and (5)respectively. we gety '" 2a sln mr and Y '" 0Hence, Eq. (3) is a equation of unknown wave,

14. Potential difference between 1:\\10 points in anelectric field is, WV II- V IJ =-1 1 0where, W is work done by moving charge %from point A to D ,(Here: W '" 2J, 1 1 0 = 20C)2So, V A - V B = 20

= 0.1 V15. Since momenta are same and masses forelectron and proton are different, so they willattain different velocities and hence experienceditferenr forces, But radius of circular path isdepending on momenta so both will be movingon same trajectory (curved path).


18/38

16.Gravitational potential energy of body will beE =_9M 1 ,nr

where, M '" mass of earth,Tn = mass of the body,R '" radius of earth

At r='2R,

At r= 3R, G IWmE2 =- -(3R)Energy required to move a body 0[' mass Tn froman orbit of radius 2R to 3 R is

~E ' =0 f f ~ [ ~ _ . j ] ' " G~~17. As we know that spring 'constant of spring isinversely proportional 'to length of spring, so

new spring constant for each part is given byk' = nk : where, k is the spring constant of wholespring. From [he theory of spring pendulum, weknow that time period or spring pendulum isinversely proportional to square root of spring

. T 1constant Le" o: .. c=.J kT' cc 1- lnk:T' =' _:~ .

-Ju[or any surface

So,18. Electric flux 1$ defined as

,---7 ---t~=j :s .d sED,I""'_____......:::__-l-----r~~F~ c 1o p,,,I/~l1------ --- GAo.._,:_____:~----!:8__'

The flux through ABeD can be calculated, byfirst taking a small elemental surface and then

---7 -}writing the E . d sfor this element, keep ill mindthat electric field at the location of this element.is the resultant of both the charges.It is quite obvlous the flux through ABeD comesout to be non-zero because at every point of the

---7 -)surface, the ang le between E and ds is less tha n90 giving a positive non-zero value for rheentire surface. So, option (b) cannot be theanswer.The options (a), (c) and (d) are dimensionallyincorrect, so they cannot be answer.

Note : In th i s part i cular question IIux throughBCFG would be zero because component ofe le ctri c f ie l d a I0 Ilg the a rea vector is zero at everypoint on this surface.V1

19. P '" --- ..R , , , , ,

-----I"5V150'"' _psi = > II . = 6 nR2

2+ R20. Resolving power of an optical instrument is

inversely proportional to i-i. e., RP:x ~i.

I~esolving power a!_~':L= }'2 =_~~02 . ,. .5' 4Resolving power at )'2 }.! 4000 .21..Let mass of each body is In. Their motion is

represented as shown in figure.mf---.v I % l

From

mx~v-ll1v 1/VC~ I =: - m ' -1:-ni- '2

[The direcnon of motion of first panicle is takenas positive)So, velocity or centre of mass of the system L~~in the direction o r -motion of panicle havinglarger speed.

22. Due to flow of current in same direction in twoadjacent sides, an attracrive magnetic forcewillbe produced due to which spring will getcompressed.

23. A s we know that thermal rapacity of asubstance is defined as the amount of hemrequired to raise lts temperature by 1C.


19/38

24. At absolute zero, the energy band is large andhence, Si acts as insulator or in other words, wecan say that there is no free charge carriers;26.Optical fibres works on the principle of total

internal reflection ..27. Escape velocity '" .J2gR c

So, escape velocity is independent of rn. So, itdepends upon mass as m",29. The dimensions of torque and work are[MT}r2].30. In Seeback (thermoelectric) effect, the

temperature o f hot junction at which thethermo emf is maximum is called the neutraltemperature (011), and the temperature at whichthermo emf changes its sign is called inversiontemperature (0;). If 8, is the temperature of coldjunction then these three are related by theexpression. e n - G o : : fJ i- A "o ' " Ai+_~

II 231. Spectrometer is an instrument which is used toobtain a pure spectrum of white light. Itis used

to determine th e wavelength of differentcolours of white light. (or wavelength ofmonochromatic light source), the refractiveindex of the material of [he prism and thedispersive power of the material of the prism.Pyromerers is infrared sensitive devices, sothese are used to detect infrared radiations.Nanometer istJ1C small unit of distance and isnot a device.Photometer is used to measure luminuousintensity, illuminance and other photometricquantities.

32. No is the initial amount of substance and N isthe amount left after decay.( 1 ) "N =No l 2hus,

f} 11' r 15 3fl~no. ala t lves = - - = -,:-=t 1/2 ~,Therefore, N '" N 0 ( ~ r~o

33. If we increase the temperature, the specificresistance (resistivity) of conductor increaseswhile that of semiconductor decreases.34. Energy stored by any system of capacitors is

1 2=2C""IV

where, V is source voltageThus, 1 1 capacitors are connected in parallel.so, Cnet =nC

E 1 C V 2not = " 2 II36. Efficiency of all reversible cycles depends upontemperature of source and sink whichwill he

different.37. When the string is plucked in the middle, it

vibrates in one loop with nodes at fixed endsand an antinodc in the middle.

N~NSo that,

1 ' 1 ~ 2 x 40 = = 80 em3S. Prom the relation, tan $ '" 7 f 1Power factor, cos (j I = = -f'-===,/ 1 + tan2 (P

1"'T-=1 1 . ( . m L )2V "T. If

R

40. The minimum kinetic energy required to projecta body of mass m from earth's surface to infinityis known as escape energy, Therefore,( GM )....gR = -R.MIl"!KE= _"_ = I'll"RH "

41. Temperature of a gas is determined by Ihe totaltranslational KE measured with respect to thecentre of mass of the gas. Therefore, the motionof centre of mass of the gas does not affect thetempera ture, Hence, the tempera rurc of g(1S willremain same.

43. According to the mass-energy equivalence,mass and energy remain conserved. So, whenwater is cooled to form ice, water loses itsenergy $0, change in energy increases the massof water.

44. Metals have minimum energy gap andinsulators have' maximum energy gap andsemiconductors have energy gap lying betweeninsulators and metals.

46. Kinetic energy of particle o r mass m in SHM at .any point is,

1 ) 2 '1=-nlw(a -)c)2


20/38

and potential energy '" ~ ITwl x2where, a is amplitude of particle and x is thedistance from mean position.So, at mean position, x=:O

K" 1 2 2~\= ' 2 mo) a (maximum)PE "" a (minimum)

47. Conservation of angular momentum gives~MR2(!)1 " " ( ~ M R 2 + 2nlR2) 0)21 2 1 2= = ' > 2 MR ( i)1 ""2 R (M + 4m) (])2

( )2 ' " ( M ~ 4 1 l l ) CO l48. Using the relation

mv2 --=R R:=mg.r e-my2 2-r-'"~llJlg or v = llrg

or y'l =: 0.6 x ISO x 10= = ' > v=30m!s

49. Apply the Bernoulli's theorem just inside andoutside the hole. Take reference line forgravitational potentia) energy at the bottom ofthe vessel. ILet Po isthe atmospheric pressure, p the densityof liquid and vthe velocity at which water iscoming out.

pv2P,nside + pgh + 0", PoulSide +2pv2= = > Po + pgh v = = \ . I 2 , _ r { i i = J2 x 10 x 20 = = 20 m/s50. The work is stored as the PE of [he body and isgiven by, 'U = = I X 2 Fe~terl\.l dxXlIX), 1 2 2orU = = kx dx =- k(x2 - Xl )Xl 2

800 2 2 '=2-rCO.IS) - (0.05) ][k '" 800 (given))= 400[0.2 x 0.1] = = 8 J

51. As the child 'stands up, the effective length ofpendulum decreases due to the reason that thecentre of gravity rises up. Hence, according to. [rT = = 2 7 1 v i

T will decrease.52. Apparent weight of ballw'=w-RR =: rna (acts upward)

w' = = mg - rna:=m(g - a)Hence, apparent acceleration in the lift isg - a.Now if the man is standing stationary on theground, then the apparent acceleration of thefaHing ball isg.53. In electrochemical cell, the anode produces thecharge q which depends on It.54. The rm s velocity of the molecule of a gas ofmolecular weight M at Kelvin temperature T.isgiven by, C n n s ; m

Let M0 and ,MH are molecular weights ofoxygen and hydrogen and To and Tn th ecorresponding Kelvin temperatures at whichC,ms is same for both gases.That is, Crms(O) =Cnns(H)

To TH,M o ""MHTo = 273+ 47= 320 KMo = 32, MH = 22T 1 '1 =: 32 x 320 =20 K

55. In a circular motion in a uniform magnetic field,the necessary centripetal force to the chargedparticle is provided by the magnetic force, i.e. ,mv2-=qvBr

Hence,Given,

mvr=-qBThus, the time period T is

T o o 2 : 1 ' = 2 v lt(;;) =: 2 ;So, T is independent of its speed.

56. Since, the inclined plane is -frictionless, thenthere will be no rolling and the mass will onlyslide down. .Hence, acceleration is same for all the givenbodies,

or


21/38

57. Given :ip;:4A, NI , '" 140, N, ",280From the formulaip _ N,t.:tr;4 280r;= 140r

So, i = 2A58. The efficiency of Carnot engine is,

'1=1-!'1.Tlwhere, Ii is the temperature of the source and1 '2 that of sink.Since,So,To obtain 100% efficiency (i. e., 11= 1), Q 2 mustbe zero that is, if a sink at absolute zero wouldbe available; all the heat taken from the sourcewould have been converted into work.The temperature of sink means a negativetemperature on the absolute scale at which theefficiency of engine is greater than unity. Thiswould be a violation of the 2nd law ofthermodynamics. Hence, a negative temperatureon the absolute scale is impossible. Hence, vs ecannot reach absolute zero temperature.59. Moment of inertia of circular wire abou t its axisis MR l, Consider two diameters XX and YYmoment of inertia about any of this diameter issame let us say f.From perpendicular axis theorem, I + I "" MR 2

MR2l=~.X

y

x60. The particle is remains stationary under the~~ -taction of three forces FI,Fz, and F3; it meansresultant force is zero.. . . . . .....~}Pl = = - (F2 + Pa)

Since, in second case FI is removed (in terms ofmagnitude we are talking now), the forcesacting are Fl and F] the resultant of which hasthe magnitude as Fl,so acceleration of particleis!i in the direction opposite tothat of 1 \ .m

61, A + B = = 18 ...(1)...(2)212= v I~- 'I' B + lAB cos 0

B sin 8ran ce > A+BcosOtan 900 '" _ _ 1 3 . .~!n0,. =cos 0= -A=> A+BcosO B ...(3)

Solving Eqs, (I), (2) and (3), A = = 5N, B ""13N62. In this question the cars are identical meanscoefficient of friction between the tyre and the

ground is, same for both the cars, as a resultretardation is same for both the cars equal to ug.Let first car travel distance 51' before stoppingwhile second car travel distance s2' then from

v2 = u2 - 2as

an d

S1 1~ $2 =16

63. Using the relation~1 + ll2 =_n~l_ + It2I -1 II - 1 ~2 -11+ 1 1 1---= +----y-1 (5/3-1) (7/5-1). = ;>2 3 5--=-+-=4y -1 2 23 24Y=- =--2 16

64. Let,charge q is placed at mid point of line AB asshown below,Q q QA C BI- 42 xJ2 "I

Also AB '''x (say)x xAC =2,BC =2For the system to be in equilibrium

FQ q + FQQ '" 0


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1 Qq 1 Q Q .--- --- j- --- - '" 04110 (x! 2 i 4TCCo x2

Qq=-- 4

65. Capacitance of spherical conductor = 41tEoawhere, a is radius of conductor.1Therefore, C =----- x I9 x 109

= _ !_ xlO-99'" 0.11 x 10-9 F'" 1.1 X 10-10 F

66.As the string is inextensible, both masses havethe same acceleration a. Also, the pulley ismassless and frictionless, hence, the tension atboth ends of the string is the same. Suppose themass nI2 is greater than mass ml, so, the heaviermass nIz is accelerated downward and thelighter mass ml is accelerated upwards.

Therefore, by Newton's 2nd lawT - m Jg = mla ...(1)

...(2)!zg - T "" 1ll2(1After solving Eqs. (1) and (2)(m2 - ml) ga= (ml + ITI2)-g=s

nl2 (1- m!m2) ._. gmz (1 + ml!m2)

(given)So, g8 ... (3)

~=xrn2Thus Eq, (3) becomes1- x ' 1 7 mz 9-- =._ or x '" - or - = -1+ x '8 9 ml 7

So, the ratio of the masses is 9 : 7.

Let

67. Energy radiated per second by a body which hassurface area A at temperature T is given byStereo's law,

Therefore, ;~ = ( ~ r G ~ r= ( ~ r ( ~ b b 6 r[Since bodies are of same material e] '" e2lE ] _ 16 _ 1E2--16 - I

'" 1: 168. The system of masses is shown below.

From the figure.~~F

.;;}f.JjJ t F j :F -1'1 =' rna1 ' 1 - T2 =nta

... (1)

... (2)ndEq. (1) gives

10.2- 1'1"" 2 x 0.6=? 1'1 =10.2-1.2=9NAgain from Eq. (2), we get

9 -1'2'" 2x 0.6; : : ; " > 1'2=9-1.2=7.8N

69. The free body diagram of the person. can bedrawn as

Let the person moves up with an acceleration' a 'then T - 60g = 60a1 'm ax - 60gamax '" ----6()"--"360 - 60g=--60-- '"-ve

Which means it is not possible to climb up onthe rope.Even in this problem it is not possible to remainat rest on rope.No option is right.But if they will ask for the acceleration ofclimbing down then

P " " " 9 ~ '60g

60g-1' =60a


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~ 60g - Tmax = 60ami n~ . = 60g-: :]_60 = 4 m/52rrun 60

70. Angular momentum of particle about 0y

/Ct 1 7r

_ _ _L=m(rxv)-I LI =mrv sin 8

= mv (I " sin 8)=mvl71 . The component dI co s A of element dl is parallel

to the length of the wire 1. Hence, force on thiselemental componentF "" ~ _ Q _ .~i.~.6.dl cos 0)4 rr r

~I0 i i2dl cos fJ2rrr72. Electrons, protons, and helium atoms aredeflected in magnetic field, so the compoundcan emit electrons, protons and He2+.

73. Work function W = h :[Here they are interested in asking thresholdwavelength)where, h = Planck's constant, c ""velocity oflight.Therefore,

74. Formation of covalent bonds due to the wavenature of particles is done in compounds.75. As the side BC is outside the field no emf isinduced across BC . Since, AB and CD are notcutting any flux the emf induced across theserna sides will also be zero.

B~

The side AD is cutting the flux arid emf inducedacross this side is BvL with corner A at higherpotential.

'1 -J +1 -J -i -1 +1-I76. (a) NaCI+ KNOJ --*NaN03 + KCl,2 -2 +1 _ .J +2 -1 +1 -2

(b) CaCP4 + 2 HCl --*CaC12 + H2CP4+2 -I -3 ~I -J +2 -I -3 +1

(c) Ca(OH) 2 + 2NH4Cl -----)0 CaCI2 + 2NH3.,) -2+ 2H20

in all these cases during reaction, there is nochange in oxidation state of ion or moleculeor constituent atom, these are simply ionicreactions.

(d) 2K[Ag(CN)2J + Zn _" 2Ag + K2[Zn(CN)4JAg" -) Ag gaining ofe,reductionZn _" Zn 2+ loss ofe,oxidation

77. According to ideal gas equationPV '" nRT

. PV P xl PNumber ot mole rt = = RT or RT '" R T78.

oI I Number of P-O bonds = 16_/p""""O Actual bonds = 20/1 \o=p~o /.p=o

\ 0 /o 0


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n '" number of unpaired electronsMnzs(Ar) 4s23dSMn2+[Ar]3ds4soI . t l t l t l t l t l

number of unpaired electrons = 5Fe26[Ar] 4s23d 6Fe2+[Ar) 3d64soI t + l t l t l t l t l

number of unpaired electrons = 4Tiz2[Ar] 4s23d 2Ti2+[Arl 3d24so

number of unpaired electrons", 2Cr24[Ar] 3ds 4s1C?+[Ar] 3d4.4s0I tit I tit I

number of unpaired electrons =4Mn 2+ : (Ar) 3ds with five unpaired electronshence maximum magnetic moment.

81.CH 'CH+ Na~ CH-CNa+ + ~H2CH-CH+ ammoniacal AgN03~

CH=CAg+ white ppt.,CH=CH + HCI~ CH2 =CH-CIusually occurs in the presence of Hg2+CH=CH+ NaOH~ no reaction

Because of hard acid hard base soft acid softbase (HASB) theoryCH"""CH+ NaNH2~ CHssCNa+ feasible.

82. o1 1O-C-CH3

6 - COOHo-acetyl salicylic acidAspirin (Analgesic)83. UIS half cell :

H2(g)~ 2H+ (1M) + 2e-1 '1

RHS half cell :2W (1M) + 2e- ~ H2(g)

P2Hz(g) ~ Ii2(g)I'! P :l

P2o e n ",o.aov, K=--. n=21 1 RTE een = E cc lt - _ . log KnF e'" a - ~~ loge ~ :RT PIE < : c l l = 2 F log. P2

84. CH=CH HOCI > Cl2CHCHO~+ +.

CH==:CH + CI ~ CH=CH-C)I~..., OH

tautomerising electrophile is CI+mainly.85. It is carbylamine reaction.aH;zNHZ! CHC J3KOH, C 2HsOHBenzyl amine Benzyl isocyanideCarbylamine reaction (test (or 1 amine)e ~CHCl3 + KOH ~ CCl:l --'" OCCl2Dichloro carbene86. Oxidation takes place at anode. Hence (d)

(c) is not feasible, i . e . , Cr3~ is not oxidised toCr20~- under given condition.


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87. Molarity gets affected as number of moles perunit volume (volume increases with inc-rease oftemperature)

88. Cyanide process or Mc. Arthur-Forest cyanideprocess2Ag2S + BNaCN + 0 z + 2HzO -----+

4Na[Ag(CN)J + 4NaOH + 2SSoluble silver complex is filtered and treatedwith zinc dust and silver gets precipitated.2Na[Ag(CNhJ + Zn -----+

Na2[Zn(CN)4] + 2Ag.l-89. (CH3)3 ~CBr + IIp -----+ (CH3)3 ~C ~OH+ HBrBr is substituted by -OH- (nucleophile)SN1 (unimolccular subst itution reaction)

90. On moving down water solubility of alkalineearth metals decreases:Oxides and hydroxides of alkaline earth metalsare basic, except Be, which gives amphoteric.Hence, metal is Be

91.C " , Co ( ~ rtotal time 3Y '" half-life

= 100(~r = 12.5 g92. Stability of complex increases with increase in

charge on the central metal ion increase withbask strength of ligand.In [Fe (CN) 0.512bTJ ==K ; = = 1./36

=.~:~.~~x 0186= 0 051201.86' .


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Cd) CHl-CHz-C EONisp104.109 28', it means pure regular tetrahedral.

(a) NH3 Sp3 pyramidal(b) HP Sp3 bent(c)C~

C d ) N H ! Sp3 tetrahedral107. HPO~- + H20 ~ H2PO;- + OW

HPO!- + H:P ....----'-.O!- + Hp+H2P0'2 is a conjugate base of H3P02 (amonobasic add) and does not give H '" ,.HPO~- isa conjugate base of H2PO:; and does not ionisefurther, s ince H3PO 3 is a dibasic add.

108.C a ) HIH-=Al-H'\H(b) .H20 H301

Sp3 s p 3(2 bp + 2 (3 bp + lip)lp) .

(e) NH3Sp3

(3 bp + 1Ip)N H 4 "s p : !(4 bp)

Change inhybridisatlon -eswell as in shape

Change in shapeonly, bent topyramidal

Change in shapeonly , pyramidalto te t rahed ra I

110.Xe109. Racemic mixture is an equimolar mixture of

enantiomers.

Xe (firstexcited state),j.t 5p5which forms - 2 Three lone pairsXe!'::/; 58 -

5d

Xe (secondexcited state)which formsXe F4

xe (thirdexcited state)w hic h f orm sX e F 6 552 One lone pair

111. NaC] is salt of strong acid and strong base.Its not the case of buffer.

NaC! + Hp ~ NaOH + HClaqueous NaC!, itself exact neutral solution

HCl+ H20 ... H3 0 + +crmakes solution acidic.

112. Unit of rateconstant k '" T .I . __ 1_>-.-1nne (cone. t:For k n '" 0 k '" .!. 1 henceI t (concyO-lconcr" .. M sec"1 .1 1 _ 1k2 n '"1 k =- . .... '"_.'" St (conc)l- l t

113. RNA Rjbonucieic acidSugar present in RNA is D (-) ribose andconsists of cytosine and uracil as pyrimidine andguanine and adenine as purine bases.

114. At HIS (oxidation) 2 x (Ag ~-> Ag ' + e")E~~'" - x

At RHS (reduction) Cu2+ + 2e~ -. -j. CLlE ; . d =- 1 ' - y2Ag + Cu2+ ._> Cll + 2Ag+, E ; o , J = (y - x)

Note: P values remain constant when half-eel!equation is multi pled/divided.115. On the basis of kinetic theory of gases

VI N -2P = " 2 limvand 1 -2 KT2mv =2

1( N ) -2P = " 3 V mv2 l ' N ) 1 -2P =3 V "2mv= ~ ( ~ J ( ~ )KT

PV=NKT

or

or


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...7116. Mn04 -----> Mn2+ change 5

.6___, MnO~- change 1. . . .~-------')< MnO~ change 3

...3~ Mnp3 change 4117. 1 - 1 2 + 1 2 ------7 2HI

Rate of reaction ".,-d[H21 -d[12] 1d[B!]~'"'-"~=2'dt

or118.560 g of Fe 560No. of moles = 56: '" 10mole

70 g orN14g '" 1mole of N70 g =:: 5 moles ofN

20 g H= 20 moles of l-l-atornsci., / ,cH 2 CH2 C H 3119. _J:. =CC 1 BAt least two groups or atoms attached to doublybonded carbon atoms should be different.

120. Na unit cell (body centred cubic). 18 comer atoms ) ( " 8 = 1

1 atom placed inside the unit cell = 1Total awms. '" 2 .Mg unit cell (Face centred cubic)18 corner atoms x " 8 '" 1

6 face atoms x~ = 3Total atoms = 4

121.incorrect 4 :3 2 l - O < ; G ll\lllnbcrill~ CI-I3 CH- CH- CH3~1 2 1 3 1 4'CH on Correct numbering3 Funetional groupCorrect: name 3rnethyl buran-z-ol s hould b? give!!

priority122. CJ - ! ~ lC I l :l COOJ - l - - - > CH : 1CHCOOHred P I

(HVZ reaction) C Iuleohol lc KOH------;0. C H2 =CHCOOl-1(elimination) Acrylic acid

123. Kp '" z, (RT)/!."r,D o I 1 S ' " 1 - 1.5 = - 0 ..5 K - K (RT)-112 _ Kc. P - . c - CRT)I!l&"'(RTF~l ( p

124. Ell = - 13;6 eVnFor second excited state 11 = 3,

E, = _~3.6 =- 1 51 cV._. 9 .125. A square planar geometry is the result of d.spl

hybridisation where inner d sub-shell (Cd. l :I. , .< x orbital) participates.

126. [Cr(NH3)sN02]CI2Nitropentaamine chromium III chloride exhibitslinkage isomerism as -N0:l is ambidenrateligandIsomer is [Cr(NHJJ50NO]CI2.127. Change in volume affects number of mole perunit volume and move in the direction whichundo the change.

N2(g)~ 02(g) .. 2NO(g)no. of moles of reactants and products are equal.

128. Heisenberg uncertainty principle6.x.mDov,,-, h4rr

uncertainty in velocity III1 v = ...4x3.14x6xxm6.626 x 10-:14

= 4>


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nCH2=CH2 ---------+(Monomer) -CH2-(CH2-CH2)-CB2-Polym.cr

132. Higher the bond order greater the bondstrengthSpeciesO r

02O20 : 2

Bond order11-52.02.5

133. Cerium Cess [Xe) 4 fl Sdl 6s 2Its most stable oxidation state is +3 but +4 isa ls o e xis ti n.g.... 1134. )~(radius) ecZ

135. 0.005 M calcium acetate (CH3COOhCa(CH3COOhCa---------+Ca2+ + 2CH~OO-

0.005M (2 x 0.005'" 001)

CH:;COO - + Hp ~ CH3COOH+pH '" 7 + pK a + log C

2 2""7 + 2.37 + logg.Ol= = 7+ 2.37- 1"" 8.37

136. Most of the reaction which takes place on thesurface of her rogenous catalyst proceeds beingindependent of concentration.137. Kt '" -log. C ! + log.C o

1r = -K log. C ! + log. Coy ::mx+, straight line, -ve slope

138. Mud is colloidal charged solution which ontreatment with alum (AI3+) gets neutralised orcoagulated.

139. C =C bond is cleaved and oxidised to-eOOH, ..:_c;HQ group is also oxidised to-COOI-L

(CH3)2C =CHCH2CHO(CHa)C =0 + HOOC-CH2-COOH140. Both whereR is attached to vinylic(", arylic)sysremPositive inductive effect occurs in the followingorder:-0- > -:COO- > (CH3~C -> (CH:I-tCH-

> CH3CH2-> CH3-141. Phosphorus (Srd period element) can raisecovalency faCilitatin~ vacantd-orbitals :P (ground) INe] 3s3p3 & toP(ex.cited) (NeJ 3s1 3p3 3dl

(cis-tI'(1!1s)HICH-c:I ~OH

H. . . . . . ) - C l - I 3CH2CH3 CH3CH2

Molecule is optically activeenanriomcrs

Geometrical isomers and enantiorners both arestereo isomers.143. S e c area e m 2)

ee cone, (mollm3)cc _1_._ (m -1)length

S '" k rn2 mol m~lm)k '" S mol' ' m2

144, Joules law suggestsJ", Mechanical work done by the system, IVNet heat given to the systern Qhence J '" _w__!fi+ q2Hence, W = J (ql + q2 )is consant with Joules law of equivalence.

145. A bond is fonned when attractive forcesenl #


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148. SN1 unimolecular nucleophilic substitutionreaction is favoured with stability ofcarbo cation.(a) ~ 3" Carbocaticn(.b) : < > I,' Carbocation(c) @ @ Ar yU c c a rb ~C~ t ;o " n(d) @+C~~mileC~2

- - l' Ca rbocauc n

149. CaOCI2 is written basically as Ca(OCI)CIOCI- ---')ct has +1oxidation stateC)" ---')CI has -1 oxidation state

150. Picric acid, 2, 4, 6 trinitrophenolOHN 0 2 * N ~N02

srrcngty actdic

~------------------1.Key Idea: T h e eq ua ri on h a vi ng 0: and {3asit5roots, i s XZ - (0: + j3)x + O!,~ = 0.Since,a. 2 = = Sa - 3 ~ IX Z - Sa. + 3", 0and~2 = 5~) - 3 => 1 3l - 513+3 = = 0 'These two equations shows that u and'p are theroots of the equation

xl-5x + 3", O.u + j 3 = 5 and a 1 1 = 3

Now ~ + _ ' " ct 2 + 132= (ct + III- 2 C t ! ~1 3 a C t I 3 up25- 6 19~=3andThus the eq ua t ion having ~ and Q _ as its roots is,.., ct .given by

x2 - (~ + Q _ ) x +~. . Q _ = 0j 3 a ~}ao 19x- --x+ 1=03

3x2-19x+3=O>2.',' Y '" (x + ~1 + x2 Y '.: On differentiating with respect to x, we getdy ~ ( 2x 1x '" n (x + ~1 + x r. ,1+ ~2 1+ Xl

.i.;s:,ee . ~1 + x2

2(I+ x2{:) = n2y2

On again differentiating with respect to X, we get2 . '2(1 + x2}2dy. d Y + 2x(dY) = n 2Z Y d .r:d x dx2 dx dx

d~y dy':::;. C l+X 2 )dx2+xdi;=n2y3.',' 1, log3 ~31-" + 2, log3(4 3 " -1) are in A~" 210g3(31 -x + ZF2 ""lO~3 3 + log3(4 3 " - 1)=> 10g3(Y -x + 2) ""log3 3(4 -g - 1)Let 3" '= t

, . :: :; 0

~ + 2",12t - 3t -12t2-51'-3=0

(3t + 1)(4t - 3)= 01 3t " '- '3


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4. Probabilities of solving the problem by A, BandC 11 dl ivelare " 2 ' " 3 an4 respective y.

Probability that the problem is not solved=(1-j J (1 --j ) (1 - ~)123 1=--x-x-oo-2 3 4 4

Hence the probability that the problem is solved1 3=1-4=4'

5. Key Idea: P er io d o f sinu and cos 0 i s 21l.Since sin 2 e =.~_.:::-_( ) ~ _ 2 _ ! : !=!-~os 2 e, 2 2 2

P 'df.20 2rteno a sin . = " " 2 = = 1l6 .. I, In and n are the pth. qth and rth term of anGP whose fi rst term isA and common fa tio isR.

I=ARP-l= log I = log A + (p - L) log RSimilarly, log In= log A + (q -1) log Rand log 11 = log A + (r - 1) log R

Ilog/ p 1 1Now logm q 1logn r 1

Ilog A + (p - 1) log R P l~'" 10gA+(q-l)logR q10gA+ (r-l)logR r -

Io p 11

= 0 q 1 1lOr 1 1[C1 -} Cj - (Cl log A + (C 2 - C 3) log R )]=0

7. Key Idea: Limit of a junc tiOl l e x is ts only, ifLH L = RHL .. ------- r----,--. -.-r J 1 - co s 2x .,..li Ii1- 1 . + 2sm xx l : ~ n. -x~~---J2x

. . J 2 1 sin x I . I sin x I= lim "._,---"-_" = lim ----x-,o.J2x .:x-,o XLet f(x) '" I sinxl, xN If-IJ. - r IS in(O - II) I? W , " " - , , ~ ~ _ h

~ lim sin II ""_ 11,-,0 - 1 1

= lim I sin (0 + 11)IIr-,O 0+ hand RBL

_ sin h= = lim -.--- = 1h-->O II

LHL* RBLI I sin x I d .1m __ . oes not eXIst.x : " " 0 X

S.Let A(4, 0),(-1, -1)andC(3, 5) be thevertices of a I ' l . ABC.

AB=)(:'1- 4} + (-1- 0/r,-._-- ..-.= \125-1-1="\126BC '" ' J r ( ~ i ) : i - : ; :- C 5 ;1 )2 '" , / 4 2- - - ; ( J

f l - - - - - _ - - - , - _ .=\,16 + 36 =--./52_CA - = \~l -I - (0 --~')lnd {1-~-~-"" ,---'c o \,1+ 25 '" ,j 26

and CA2 + A82 = (--J'26i + (- . , /26)2= 26 -t' 26 = 52=BC2

=> (:.4 2 + AB2 "" Be2Thus, the triangle is isosceles and right angledtriangle.

9 .. : Total number of students '" 100and number of boys = 70_._Number of girls = (100. - 70) ""30Now the total marks of 100 students'"I00 x 71'"7200and total marks of 70 boys 70 x 75 -=0 5250.= > Total marks of 30 girls

=nOD - 5250 = 1950.Average marks of 30 girls ~ 1; ~ Q = 65

IO ..We know thatcot' (/eos o ) + tan -1 ( , ( C O S n ) ""~

and given thatcot I " , fr ; ;: ; ; : - tan I /cos --;- '" x

On adding Eqs, (i) and (ii), we get,.'--. 1l2cot" (-,,/os 0:) = - 2 -1 - Xr=:: ( T i x )' " co s o: '" cot .~ + '2

cot . : x : , 1,,/cos u = __ 2_x1+ cot '2

... (i)

. . , O i)

x . xcos '2 - Slll'2,,/COStx = --------x ,xcos - 2 + sin - 2


31/38

1- si n xcoso : '" . . , I+ sm x=

Alternate solutioncot-I(J~os a ) - tan " (""cos (1) '" xIf 1\.1"--:. 11:ln-, -==",=' - tan- C" /cosa)e= xl.j 'cos a, )

( _,I .._..r c o s - X I.11 /Cosa ,r a n l - - - . - " l - - r . - : L j ~ xI+ -.=-="C050:

.,1 cos (XI-coso- . -.=----. !_ . . .. . .. - 3 " '..)

~ lsxS:9 r . lDomain of 5 i 11 . I ! . log~ ( j ) j is IL 9l

16. z 1 / 4. 4i'';.811If...'" 2i/4. i/~iPH".

I ; I .2 ~l=i'~ 'l )iJ~_I_, ._l_l1;1 .. ~" ( 1 \~ ;: 2 I ";- i ,~ 2 ' ,2, J

.1.\2.2J'" 24 '" 2

17. Since fifth term of a GP = = 2a r4 '" 2

Where a and r are first term find common ratioofa GP.Now required product'" a x or x a? x CU,3 x w'~ x ars K 1 . 1 1 . 1 ) . x a / x C 1 r~'" a9 r36 '" (ar4l '"9 .. 512

18. '.' I sin x 1 is a periodic function with period 11 .J .I!I~ . I "[sin x I dx ,-,10 I sin x I dxo II


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"" 10 f: sin x dx=10[-cosx]~= = 10(- COS It-!- cosO]==10 [1 + 1] "" 20

19._' I" = = f J [ / 4 tan"x dx{) >

j.f rrl4 IlT 2 dx1 1,,2= o tan x1 7 1 /4 2Now III -~ I l l ' 2 = (I tan" x (1 + tan x)dxf"!4 2 It dxr: J o sec x tan x

Let tan x = t

1n + 1

1 , I " tlHence imIlt + In _ 2 ] '" 1 1 1 1 --'1,'1_",, . n- )'.1. n +'" lim -]-'-

" ,>~, 1-I- 1n=1

20. f 0 2 [x


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dx sin e e t + y)cos y - sin y cos (a + y)dy sin2(a+y)

sin (a+ y - y)sin2(a + y)

=dx sin a4Y sirha ~ y )dy sin2 (a + y,ldx sin a

25.': xY "" c - . YTaking log on"both sides, we get

y log x "" ( x - y) loge e

On differentiating with respect to X , we get1d (l4o!ogx)-x-y. ..." xdx (1+ logxi

Jo g XC i - : - l o g xi26. The equations of two curves are

x:J - 3zy2 + 2e= 0 (i)and 3x2y - /' - 2"", (ii)On differentiating Eqs, (i) and (ii) with respectto x, we get

( r l Y ) X2 _ y2dx c) 2xyan d .(dY) _. - 2xy" a x -"""2--i'" 2 x-y. : ( : ) C I X ( ; & t 2 ~ ( X 2 2 ~ 2 ) ( ) ~ 2 )

0=-1Hence, the two curves cut at right angle.

27. Key Idea; A junction fe x) is said ro b eincr eas ing junct ion . if f' (x) > o .

f(x) 0= c o t " x + x1 x2f'(x)~ - --2- + 1 ---I+ x-I + x2

Hence, f (x) is increasing function sinceJ' ex) > 0 fur all x.28. We have

f ex}"" (x + IF3 - (x _1)113f' (x) =: ~ [ e x + \ F J - (x _ 11 )2 1: 1 ]

Clear Iy, f' ex) does not exists at x "" 1.Now, f '(x) '" 0, . then e x - 1iJ '" ex + lii3::0;> X 0= 0Clearly, f i (x) "',0 for any other val uc 0 fx E [0, 1]. The value of fCx) at x = 0 is 2.Hence, thegreatest value of f(x)is 2

J ~;2 "[sin; dx29. Lei. J '" 0 - - ' : ; C O S 0 , ' / ; ; \ ; ; - ; . . . (i)J = J O ~ 2 7cot(1i-;4~~~~~~:~~~/i-:~'jxr = = f " 1 2 _ .r _ ._ - j _ ~ C lS ~-= dx ... {ii)- Jo ,isin x + -..Ieos xOn adding Eqs. (i) and (ii)

J ."'2 7 t21=1) Jdx~I"'430. Let J = I -:X(;~:l)

L et x" + ] =t~ nx,,-I dx '" dr

I = = . l I _.dl_" = . ! . _ J ( . _ .1__- ~.)d r. . n ,l (t ~ 1) n - f- L1 . ( t _ . 1 ) 1 ( X " ),-log ._- + c,,-,--Io '" -. -.~ ' + - cn t I'! x" + - 13}. The equations of given curve and a tine arc

y "" 2x - x' .-.0)and .. "(ii)=-xy

On solving Eqs. (i) and (ii), we get [he points ofintersection of curves which are (0. 0) and(3, -3).

3.. Required area = = I n {(2x - x2)_ (--x)} dxJ :I '2'" 0 (3x - x ) dxJ~1~_ x ; - 1 : )l , . .0~ , ? : . ? . - ~ z= _ 2 . 1 . - 92329 .= '2 sq unit


34/38

:i2. The general equation of all non-vertical lines ina plane is ax ~ by =I, where b 7" 0dya+b(JX=O

_ dLvb --"~ '" 0dx2d2v_J =0dx2

Widell is the required differential equation.33. Given lWO vectors lie in xy-plane, Hence a

vector coplanar with them is-ry - ~a70xi"yj~ -; .._); '~a .1 (1'- J) ' 0 , " a (1 - J) = 0

'=> e x i+ y j) ( i - ) ) ' " 0 ~ x - y ""0x=y

-ry .".. a x (i+ j)Required unit vector = - - - - =_-,.~-.1 " 1 1 x.f2e- J " , , , c i + j). . J 2

34. The vector i x j + z i f doubled in magnitude,then ir becomes 4 j + (4x - 2)] + 2 k2[ i+ x ] + 3k I =~I 41 -+ - (4x - 2)1+ .2k/

r+->.._-- (---"--2-"-=:. 2\1 '1 . . . x2 + 9 =\116+ (4x -2) ...4- = : ; ; > 40+-4x2:: 20 + (4x - 2l

.::3

3x2 - 4x - 4 =03x' - 6x + 2x - 4 = 0

3x(x - 2) + 2ex - 2) '" 0(x- 2)(3x + 2) '" 0

2x;= 2. ' - " 335. A parallelepiped is formed by planes drawn

through the points (2. 3, 5) and (5. 9,7).parallel to rhe co-ordinate planes.Let U, b . c be tile lengths of edges, thena '" 5 . - 2 = 3 , b = 9 - 3 -= 6, and c = = 7 - 5 :::2So the length of diagonal of a parallelepiped' " , / , ; ; r - ; - / 7 . , . c2

= 1 9 + 3 6 + 4= - 1 4 ' 9 = 7 unit

36. Key Idea: A line x- Xl, = L::-Jl = : O _ : : : ' . ~ . 1 . lieI III 1 1iT ! th e pla ne a x + by + cz + d '" 0, if(a) al + brn + . en As 1 3 and H t ; ; ; ; Ii~"> p e A ) _ s ; p ( i 3 ) and P C B ) : O : p e A ) . .Note: Two events A and B are said to be muluallyexclusive events if A r.8= < ll '

39. The equation of parabola isyl+4y+4x42",O

=> yl 4y ...4 '= - 4x - 2'" 4z (1"=> (y + 2) = - 4 l x - 2J

Let y + 2:-: Yand x - 1 ' " 'X2Y = = - 4X

Here a= 1 .., Equation of directrix is X c , 11x--=l23'x=2

40.Key Idea: T he n wn be r o j tria ngle s th ose am /lef or me d u siitg 1 1 poin ts " " " c 3

T = ' 'C.II .1'",1 - Til'" 2]I > 1 - 1 e 3 - " e : 1 =21

"C 2 + ' 1 : : : 3 - "C 3 =21(-: "C2 + 't:3 ". ", le:l)

"C2 = 21


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n (n -1) = 21=> n2 _ n - 42= 02(n-7)(n+ 6)=0

n =7 ( ,' n -6)41. We know thatA+B+C"'1!

~ A+C=1!-BA-B+C Jt_.=> . 2=2-B(A - B + C ) . ( 1 t )2casin 2 "'.2casm 2 - B

( a . z + - c2 _ b2)'" 2 ac cos B = 2ac 2ae=a2 + c2_b 2. ( x - 3 ) ' "2. lim --,,__,/ x+ 2

" " - lim [ 1 - _ 2 _ _ ] - xX->OO x+-2.

(-5" ,[ 1 / ( - 5 ) ] ~::1J' " } ~ ~ ( 1 + C - ; 2 ) ) nZ'" e'l~.,(-1 '< ~ ,,)=e-5

Alternative Solution:I , . ( X _ 3 ) X1m--" 4 " ' " X + 2(1-~)"n x

x~ ( 2 ) "+-xe-3 -5=2=e-e

43. Key Idea: If the triangle i s equilateral, thenin ce ntre is coin cid e w ith ce ntro id oj th e tria ngle .Let A (1, -13)B(O, 0), C(2, O)be the vertices of atriangle ABC.

a = BC = )(2 - l + - (0 - Op = 2b= AC '" )(2-- Ii + - (0 - -n} '"2and c = AD =~(O-ll + (o-..J3l = 2

.. The triangle is an equilateral triangle... lncentre is same as centroid of the triangle.~ Co-ordinates of incentre are(1 + - 0 + - 2 13+ 0 + - 0 ) - ( 1 _ ! _ _ )3 ' 3 I, e., '.J3'

44. Key Idea: IJ----:, " ' S andt are th e side s oj atr ia n gl e, th e n " " i t + - - s + -t""

" it +'S+""l =0_,. -rl> --).a+- b =- C

_ , . ~ _ , . _ , . - - -+(a+b)xc",,-Cxc" 1 x " t + - 1 t x " t = O

Similarly,- - ) . - - - * ~ - - - + ----t _,.Hence a x b = D x C = C x a.

45. Key Idea : if (0 is a cub e root o j un ity , th e n1+ (0+ (1)2",' 0 and (oJ", 1 -(1 + - (0 - m 2j '"-(" - til,!

("."1 +r o + (02 '" 0)'" (-2oh7= _ 27 _rol4=-128(m'Jto/= -128 ro" (":(1)3 '" 1)

6i -3i 146.',' 4 3i -1

20 3

16i +-. 4 0 0 1= 4 3i -t,l (R1- - - + R1 + R2)20 3

13i - . . 1 1(6i + 4) 3

=(6i + 4)(3e + 3)",06i -3i 1

: < " . I t 4 3i -1 = x + iY20 3

::) O+Oi= :x+ -iy~ x= 0, y =0

47.': sine~lsin2 0SI4xy $1

(x+- ylO$(x+y)2-4xy

x2 + y2 + 2zy-4xy ~ 0(x _y)2 ;:;0


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Which is true for all real x and y providedx + y ""0, otherwise 4'0'2 will be(x + Y)meaningless.Note: se c20 .;_ 4x y 2 is possible only. if x ;= y.(x + y)

48. The equation of an ellipse isx2 y216+"9 =1

Here a ""4, b '" 3'. e""~1_b2 a2

= ~ 1 - { 6 = ~:. Poc i of an ellipse are (F7 , O }.. Radius of required circle,---c=-------= = ~(.J7 - 0)2 + (0 - 3i

= .fi+9 = , , , " 1 ' 6 ~ 4 unit49. Required probability

=peA lA; A3) + p (A~ A2 A3)"" P(Al) peA2') PCA3) + peAl') P(A2) P(A3)=o r + ( ~ r= ~ + ~ = ~1 l+i+lil o?l-i -1 (()2_1 (Rl~Rl+R3)--i -1+0)-l -150.

l-i -1 ro2-1l-i -1 (1)2_1'-i -1+ro-l -1=0 c ; Two rows are identical).

51. Let q = 1- p. Since, head appears first time inan even throw 2 or 4 or 6....2 3 SS=qp+qp+q p+ ...2 q pS=1_q2

2 (1- p) pS = 1-,(1- pi2 I-pS='2-p

4- 2p=5-5p3p=11P=}

-,

52. Required probability = 7C2 ( i ) 2 ( i ) 5 x-~7C x55_ 2

- 6853. lim x f(2)-Zf(x)

x ... .2 x-Z= lim xf(2) - 2f(2) + 2f(2) - y ex)

x..."z x-2;= lim f(2)(x - 2) - 2{f(x) - f(2)}x .... 2 x-2

=f(2)-2lim fCx)-f(2)x..,2 x-2

=f (2) - 2f' (2)= 4 - 2 x 4 = - 4Alternative Solution:

'lim {If(2) - 2f( X ) }x -,2 x-2

= lim {f(2) - 2f' ex)} (by L' Hospital's Rule).>'-.2;:f(2) - 2f' (2)=4-2)(4=-4

54. Key Idea : E qua tions oj a ngle b ise ctors oj lin esalx + ~y + '1 = 0 and ~x + b2 Y + cl = 0 are

atx + bly + c] _ Q2X + bz Y + C : l~ - ~a? ; + b2' ,Val + 01 2 ~For the two lines 24x + 7y - 20 =0 and4x - 3y - 2;: 0, the angle bisectors are given by

24x + 7y - 20 4x - 3y - 225 . = 5Taking positive sign, we get2x + lly - 5=0Therefore the given ,three lines are concurrentwith one line bisecting the angle between theother two,55. i3x -t;Y = 0 makes an angle of 120 with OXand ..j3x - Y = = 0 makes an angle of 60" with OX.So, the required line is y - 2 = 0, '56. Equation of circle is

x2 + i < l - 4x - 2y _ 20 = O.Whose centre isGe2, 1) and radius is 5 unit.Since SI = = 102 + 72 - 4 x I0 - 2 x 7 - 20 > 0So P lies outside the circle,Now, PC '= ~(2 -loi + (1 -7i

= f 8 2 + 6 2 =M=1 0.. Greatest distance between circle and the pointP ;: 10 + 5;: 15 unit.57. Equation of circle is x2 + y2 + 4x - 4y + 4 = 0whose centre is (-2, 2).


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2 258. Key Idea : T he [oci. o f a n e l lipse ~ \ -I - 1 ..f'"1isa - bgiven by (1a e , 0).Since, e ' ' ' ' 1 , (Ie'" 2

(/:;4b2 0: a~ (1-- e1)

= 0.1 6 (1 -- . ! ) " "12_ 4'2 2Thus, the equation of an ellipse is ~6 + ; 2 = 1.

59. The mid-point of the chord is( X " _ t _ - ;" : :; l . , .i :: L ; h) . The equation of thechord in terms of its mid-point isT =51O f X U ' - c ; Y - ' 1 . ) + Y ( ~ l . . . : < ~ - )

= 2 ( - : 1 ~ ; ~ 2 . )~ ~ ~ )~ x(y! + Y2).j- y(x1 .; XL)

~ (Xl + X2)(Yl + Yz )

--". , , , _) A60. Given that a =xi + y j -I- Z k. and b '" j aresuch that " 1 . - c and b form a right handedsystem.

I~ j f (" t~~x"1 o 1 0

x y z=iz-xk

61. The equation of a line through the centreJ ,! 2k and normal to the given plane is__~ A _ r -'"f- 2kt /. (i+ 2j I- 2k) ... (i)

This meets the plane at a point for which wemust haveld ,. 2l{) -i- ),(i + 2J.r 2k)] c i + 2j ;. 2k)= 15~-.;:, 6 -\- 9},.." 15~On putting A " '] in E g. (i), weget

;'"i+ 3 J - 4 kCentre of the circle is (1, 3, 4).

I - tan:'! 062. Key Idea: cos20'". ----r'1 + tan 0

. . ; ' 3Co. '-2:

63. Key Idea: ian t : J i s negalive ill second andluw-rhqtwdrams .

ran O"".--~ 3Here p :04 and b "'. 3

1 1 =~i;--;-t/i.c, J16'~9" ,/251 1 = = 5

sin H~E "" 4'. h SBut tan 0 is negative which is possible only if 0lie in second and fourth quadrants.

. 0 b 4 4Sill may e ~ i ; ;or - 5'64 .. : sin (a + B)"" I

and

sin ( 1 + fl) '" sin ~ IT{l+~,,"-2 .

1sin (0 " -1\) ,= '2. ( I' . ITsin (X - ,)., SlIlg

n :(X--1J"'6 ... (ii)

... (i)

On solving Eqs. (i) and (ii), we get1 1 : 1tu""j,fl"""6

tan ((1 '1 2P) tan (2a + ~\)~ tan (~;) tan (~~)

( 7t \ ( I T )"" - cot 6 " J - COl : 3G' 1"",,3 x'J3 ",1

65 .. : y '" sin2 0 ' ' I - cosec'l0-= : (sin 0 - cosec 0)2 + 2

y?2Hut at 0 :.,0, y becomes meaningless.

y>2


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. b2 2 266. Key Idea: cosA = + c - a2bc9= 4,b= 3and LA = 600

c2 + 9 16cos 60= -2x 3xc1 ,2_7" 2 = 2 x 3c

,2 -7 = = 3cc2 - 3,-7 = = 0

We have

Thus, c is the root of above equation.67. Key Idea : tan~ '= (s - b)(s - c)2 s(s - a)

A 5 C 2tan 2 " = '6 and tan 2 " = SA C 5 2Now, tan - tan - ,~ - x -2 2 6 5,.-------(s-b)(s-c) (s-a){s-b) 1

s(s - a) s(s - c) = " 3s-b 1'-5- = = " 3

3s-3b==s25'" 3b

a+b+c=3b=;. a+c=2b=;. a , b , c are in AP.

68. Key Idea :~ ~- \10 + b :s ; a sin x + b cosx S "a2 + b .

Since, I c I> ,l a 2 + b 2but - ~a2 + b 2 S a sin x + b cos x $~l + b 2Thus, LHS * RHS. Then no solution exist fora sin x + b cos x = = c .

69.': '(1. is a root of 2S cos2 8 + 5 cos e -12 = 025cos2 (l + Seas a. -12= 0

: : : : : : > 25 eos2 (l + 20 cosCl - 15cos c . : - 12=0: : : : : : > Scos o. (Scos 0. + 4) - 3 (5cosa + 4) = 0

4 3cosu"'-s,sBut 7I2

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