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Phone: 040-66777000 – 03 Fax: 040-66777004 AIEEE–2010 IMPORTANT INSTRUCTIONS                                                                                         !                     ”                               #           !                     $         $           %                 & ‘ (          )        % $   *                  (                       +                                ,-      .% %   /,0      1  %   &    $   $ ‘     2 3  4 .% /        0       1     $    ’          5 6 7  .8 /                           ,  .% %   /,0      1  $ $   % &    * %   9 ‘     2 3  4 .% /       0      1  $   $    * ‘   * $     5 6 7  .8 /                     - ,:   .% %   /,0      1  9   9 & #; $   8    8 *   & ‘     2 3  4  .% /       0      1  ; ‘   ;    8   8 %     5 6 7  .8 /                 9  –                           <           1   *                   (      = .        /                               (      1                                                                   !      ;                                   !   >   !   >       !                      8  1                    )     #        #      #  #           #         < # # )          -         )           &  4                        <                               <                  %   .  ' ,$ /                  '  3               #               <        !          <         4    7 7    < #                                            - 3 ?5          ?:          - 3 ?5       !   >        !                              #                                 <                                   !       ?                         !      AIEEE-2010-2 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004    1. The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N–H bond in NH3 is (1) –964 kJ mol–1 (2) +352 kJ mol–1 (3) + 1056 kJ mol–1 (4) –1102 kJ mol–1 1. (2) Sol : Enthalpy of formation of NH3 = –46 kJ/mole \ N2 + 3H2 ® 2NH3 DHf = – 2 x 46 kJ mol Bond breaking is endothermic and Bond formation is exothermic Assuming ‘x’ is the bond energy of N–H bond (kJ mol–1) \ 712 + (3 x 436)– 6x = –46 x 2 \ x = 352 kJ/mol 2. The time for half life period of a certain reaction A ® products is 1 hour. When the initial concentration of the reactant ‘A’, is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction ? (1) 4 h (2) 0.5 h (3) 0.25 h (4) 1 h 2. (3) Sol : For a zero order reaction x k t = ® (1) Where x = amount decomposed k = zero order rate constant for a zero order reaction [ ]0 1 2 A k 2t = ® (2) Since [A0] = 2M , t1/2 = 1 hr; k = 1 \ from equation (1) 0.25 t 0.25hr 1 = = 3. A solution containing 2.675 g of CoCl3. 6 NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g mol–1). The formula of the complex is (At. Mass of Ag = 108 u) (1) [Co(NH3)6]Cl3 (2) [CoCl2(NH3)4]Cl (3) [CoCl3(NH3)3] (4) [CoCl(NH3)5]Cl2 3. (1) Sol : CoCl3. 6NH3 ® xCl– ¾¾AgN¾O3¾®x AgCl ¯ n(AgCl) = x n(CoCl3. 6NH3) 4.78 2.675 x 143.5 267.5 = \ x = 3 \ The complex is ( ) 3 6 3 Co NH Cl 4. Consider the reaction : Cl2(aq) + H2S(aq) ® S(s) + 2H+(aq) + 2Cl– (aq) The rate equation for this reaction is rate = k [Cl2] [H2S] Which of these mechanisms is/are consistent with this rate equation ? (A) Cl2 + H2 ® H+ + Cl– + Cl+ + HS– (slow) Cl+ + HS– ® H+ + Cl– + S (fast) (B) H2S Û H+ + HS– (fast equilibrium) Cl2 + HS– ® 2Cl– + H+ + S (slow) (1) B only (2) Both A and B (3) Neither A nor B (4) A only 4. (4) AIEEE-2010-3 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 Sol: Rate equation is to be derived wrt slow Step \ from mechanism (A) Rate = k[Cl2] [H2S] 5. If 10–4 dm3 of water is introduced into a 1.0 dm3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ? (Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K–1 mol–1) (1) 5.56 x 10–3 mol (2) 1.53 x 10–2 mol (3) 4.46 x 10–2 mol (4) 1.27 x 10–3 mol 5. (4) Sol : PV n RT = = = 128 x 10–5 moles = 5 1 1 3170 10 atm 1L 0.0821L atm k mol 300K - - - ´ ´ ´ » 1.27 x 10–3 mol 6. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is (1) propene (2) 1–butene (3) 2–butene (4) ethene 6. (3) Sol : 2–butene is symmetrical alkene CH3–CH=CH–CH3 O3 Zn /H2O 3 ¾¾¾¾®2.CH CHO Molar mass of CH3CHO is 44 u. 7. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (DTf), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (Kf = 1.86 K kg mol–1) (1) 0.0372 K (2) 0.0558 K (3) 0.0744 K (4) 0.0186 K 7. (2) Sol : Vant Hoff’s factor (i) for Na2SO4 = 3 \ DTf = (i) kf m = 3 x 1.80 x 0.01 0.0558 K 1 = 8. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous ZnCl2, is (1) 2–Butanol (2) 2–Methylpropan–2–ol (3) 2–Methylpropanol (4) 1–Butanol 8. (2) Sol : 3° alcohols react fastest with ZnCl2/conc.HCl due to formation of 3° carbocation and \ 2–methyl propan–2–ol is the only 3° alcohol 9. In the chemical reactions, NH2 NaNO2 HCl, 278 K A HBF4 B the compounds ‘A’ and ‘B’ respectively are (1) nitrobenzene and fluorobenzene (2) phenol and benzene (3) benzene diazonium chloride and fluorobenzene (4) nitrobenzene and chlorobenzene 9. (3) Sol : AIEEE-2010-4 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 NH2 N2 Cl NaNO2 HCl, 278 K HBF4 F N2 BF3 HCl (A) (B) benzene diazonium chloride fluorobenzene 10. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (1) 59.0 (2) 47.4 (3) 23.7 (4) 29.5 10. (3) Sol : Moles of HCl reacting with ammonia = (moles of HCl absorbed ) – (moles of NaOH solution required) = (20 x 0.1 x 10–3) – (15 x 0.1 x 10–3) = moles of NH3 evolved. = moles of nitrogen in organic compound \ wt. of nitrogen in org. comp = 0.5 x 10–3 x 14 = 7 x 10–3 g % wt = 3 3 7 10 23.7% 29.5 10 - - ´ = ´ 11. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 x 108 ms–1 and NA = 6.02 x 1023 mol–1) (1) 594 nm (2) 640 nm (3) 700 nm (4) 494 nm 11. (4) Sol : Energy required for 1 Cl2 molecule = 3 A 242 10 N ´ Joules. This energy is contained in photon of wavelength ‘l’. 34 8 3 23 hc 6.626 10 3 10 242 10 E 6.022 10 ´ - ´ ´ ´ =  = l l ´ l = 4947 0A » 494 nm 12. Ionisation energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is (1) 4.41 x 10–16 J atom–1 (2) –4.41 x 10–17 J atom–1 (3) –2.2 x 10–15 J atom–1 (4) 8.82 x 10–17 J atom–1 12. (2) Sol : 2 He He 2 2 1 1 IE 13.6 Z 1 + +   =  -   ¥  = 2 ( ) He He 13.6Z where Z 2 + + = Hence 2 18 He 13.6 Z 19.6 10- + ´ = ´ J atom–1 . ( ) 2 1 Li 2 2 2 Li 1 E 13.6 Z 1 + + = - ´ = 2 2 Li 2 He 2 He Z 13.6 Z Z + + +   - ´     = –19.6 x 10–18 x 17 9 4.41 10 J/ atom 4 = - ´ - 13. Consider the following bromides : AIEEE-2010-5 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 Me Br Br Me Me Br Me (A) (B) (C) The correct order of SN1 reactivity is (1) B > C > A (2) B > A > C (3) C > B > A (4) A > B > C 13. (1) Sol : SN1 proceeds via carbocation intermediate, the most stable one forming the product faster. Hence reactivity order for A, B, C depends on stability of carbocation created. Me Me Me > > Me 14. Which one of the following has an optical isomer ? (1) ( )( ) 2 3 2 Zn en NH +   (2) ( ) 3 3 Co en +   (3) ( ) ( ) 3 2 4 Co H O en +   (4) ( ) 2 2 Zn en +   (en = ethylenediamine) 14. (2) Sol : Only option (2) is having non–super imposable mirror image & hence one optical isomer. Zn + NH3 NH3 en no optical isomer. It is Tetrahedral with a plane of symmetry Co+ Co + en en en en en en optical isomer ( 1) ( 2) 2 3 3 H2O H2O Co+ H2O H2O en Horizontal plane is plane of symmtry Zn+ en en no optical isomer, it is tetrahedral with a plane of symmetry 3) 4) 3 2 15. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 an dof octane = 114 g mol–1). (1) 72.0 kPa (2) 36.1 kPa (3) 96.2 kPa (4) 144.5 kPa 15. (1) AIEEE-2010-6 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 Sol : Mole fraction of Heptane = 25 /100 0.25 0.45 25 35 0.557 100 114 = = + XHep tane = 0.45 . \ Mole fraction of octane = 0.55 = Xoctane Total pressure = 0 i i XP = (105 x 0.45) + (45 x 0.55) kPa = 72.0 KPa 16. The main product of the following reaction is C6H5CH2CH(OH)CH(CH3)2 ¾¾conc¾. H2¾SO4¾® ? (1) C C H5C6 H CH(CH3)2 H (2) C C C6H5CH2 H CH3 CH3 (3) C C C6H5 H H CH(CH3)2 (4) C CH2 H5C6CH2CH2 H3C 16. (1) Sol : CH2 CH OH CH CH3 CH3 CH2 CH CH CH3 CH3 CH CH HC CH3 CH3 conc. H2SO4 loss of proton (conjugated system) Trans isomers is more stable & main product here C C H H CH(CH3)2 (trans isomer) 17. Three reactions involving 2 4 H PO- are given below : (i) H3PO4 + H2O ® H3O+ + 2 4 H PO- (ii) 2 4 H PO- + H2O ® 2 4 HPO – + H3O+ (iii) 2 2 4 3 4 H PO- +OH- ®H PO + O – AIEEE-2010-7 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 In which of the above does 2 4 H PO- act as an acid ? (1) (ii) only (2) (i) and (ii) (3) (iii) only (4) (i) only 17. (1) Sol : (i) 3 4 acid H PO + H2 ® 3 H O+ + 2 4 conjugate base H PO- (ii) 2 4 acid H PO- + H2O ® 2 4 conjugate base HPO- + 3 H O+ (iii) 2 4 acid H PO- + acid OH- ® 3 4 conjugate acid H PO + O–2 Only in reaction (ii) 2 4 H PO- acids as ‘acid’. 18. In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10–7 and K2 = 4.8 x 10–11 Select the correct statement for a saturated 0.034 M solution of the carbonic acid. (1) The concentration of 2 3 CO – is 0.034 M. (2) The concentration of 2 3 CO – is greater than that of 3 HCO- . (3) The concentration of H+ and 3 HCO- are approximately equal. (4) The concentration of H+ is double that of 2 3 CO – . 18. (3) Sol : A ® H2CO3  H+ + 3 HCO- K1 = 4.2 x 10–7 B ® 3 HCO-  H+ + 2 3 CO- K2 = 4.8 x 10–11 As K2 << K1 All major total A H+  » H+  and from I equilibrium, A 3 total H+  » HCO- » H+  2 3 CO-  is negligible compared to 3 total HCO- or H+  19. The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is (1) 288 pm (2) 398 pm (3) 618 pm (4) 144 pm 19. (4) Sol : For an ionic substance in FCC arrangement, 2(r+ + r- ) = edge length 2(110 + r- ) = 508 r– = 144 pm 20. The correct order of increasing basicity of the given conjugate bases (R = CH3) is (1) 2 RCOO < HC = C < R < NH (2) 2 R < HC º C < RCOO < NH (3) 2 RCOO < NH < HC º C < R (4) 2 RCOO < HC º C < NH < R 20. (4) Sol : Correct order of increasing basic strength is R–COO(–) < CHºC(–) < ( ) 2 NH - < R(–) 21. The correct sequence which shows decreasing order of the ionic radii of the elements is (1) Al3+ > Mg2+ > Na+ > F– > O2– (2) Na+ > Mg2+ > Al3+ > O2– > F– (3) Na+ > F– > Mg2+ > O2– > Al3+ (4) O2– > F– > Na+ > Mg2+ > Al3+ 21. (4) Sol : For isoelectronic species higher the Z e ratio , smaller the ionic radius AIEEE-2010-8 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 z e for 2 8 O 0.8 10 – = = 9 F 0.9 10 – = = 11 Na 1.1 10 + = = 2 12 Mg 1.2 10 + = = 3 13 Al 1.3 10 + = = 22. Solubility product of silver bromide is 5.0 x 10–13. The quantity of potassium bromide (molar mass taken as 120 g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is (1) 1.2 x 10–10 g (2) 1.2 x 10–9 g (3) 6.2 x 10–5 g (4) 5.0 x 10–8 g 22. (2) Sol : Ag+ + Br–  AgBr Precipitation starts when ionic product just exceeds solubility product sp K = Ag+ Br-  13 sp 11 K 5 10 Br 10 Ag 0.05 – – – + ´  = = =   i.e., precipitation just starts when 10–11 moles of KBr is added to 1L of AgNO3 solution. No. of moles of KBr to be added = 10–11 \ weight of KBr to be added = 10–11 x 120 = 1.2 x 10–9 g 23. The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows : 2 3 Al2O3 ® 4 3 Al + O2, DrG = + 966 kJ mol–1 The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least (1) 4.5 V (2) 3.0 V (3) 2.5 V (4) 5.0 V 23. (3) Sol : DG = – nFE  G E nF = -D 966 103 E 4 96500 = – ´ ´ = –2.5 V \ The potential difference needed for the reduction = 2.5 V 24. At 25°C, the solubility product of Mg(OH)2 is 1.0 x 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ? (1) 9 (2) 10 (3) 11 (4) 8 24. (2) Sol : 2 2 Mg + + 2OH-Mg(OH) 2 2 sp sp 4 2 OH H K Mg OH K OH 10 Mg p 4 and p 10 + – – – + =     = =   \ = = AIEEE-2010-9 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 25. Percentage of free space in cubic close packed structure and in body centred packed structure are respectively (1) 30% and 26% (2) 26% and 32% (3) 32% and 48% (4) 48% and 26% 25. (2) Sol : packing fraction of cubic close packing and body centred packing are 0.74 and 0.68 respectively. 26. Out of the following, the alkene that exhibits optical isomerism is (1) 3–methyl–2–pentene (2) 4–methyl–1–pentene (3) 3–methyl–1–pentene (4) 2–methyl–2–pentene 26. (3) Sol : H2C=HC C2H5 H CH3 only 3–methyl–1–pentene has a chiral carbon 27. Biuret test is not given by (1) carbohydrates (2) polypeptides (3) urea (4) proteins 27. (1) Sol : It is a test characteristic of amide linkage. Urea also has amide linkage like proteins. 28. The correct order of 0 M2 /M E + values with negative sign for the four successive elements Cr, Mn, Fe and Co is (1) Mn > Cr > Fe > Co (2) Cr > Fe > Mn > Co (3) Fe > Mn > Cr > Co (4) Cr > Mn > Fe > Co 28. (1) 29. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is (1) teflon (2) nylon 6,6 (3) polystyrene (4) natural rubber 29. (2) Sol : nylon 6,6 is a polymer of adipic acid and hexamethylene diamine C O (CH2)4 C O NH (CH2)6 NH n 30. For a particular reversible reaction at temperature T, DH and DS were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when (1) Te > T (2) T > Te (3) Te is 5 times T (4) T = Te 30. (2) Sol : DG = DH- TDS at equilibrium, DG = 0 for a reaction to be spontaneous DG should be negative e \T > T    31. A rectangular loop has a sliding connector PQ of length  and resistance R W and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are AIEEE-2010-10 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 (1) 1 2 B v 2B v I I , I R R = – = =   (2) 1 2 B v 2B v I I , I 3R 3R = = =   (3) 1 2 B v I I I R = = =  (4) 1 2 B v B v I I , I 6R 3R = = =   31. 2 Sol. A moving conductor is equivalent to a battery of emf = v B  (motion emf) Equivalent circuit I = I1 + I2 applying Kirchoff’s law 1 IR + IR – vB = 0 ……………(1) 2 I R + IR – vB = 0 ……………(2) adding (1) & (2) R R I1 I2 2IR + IR = 2vB 2vB I 3R =  1 2 vB I I 3R = =  32. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t1/t2 will be (1) 1 (2) 1 2 (3) 1 4 (4) 2 32. 3 Sol. 2 2 t / T 2 0 2t / T 0 1 q 1 q U (q e ) e 2 C 2C 2C = = – = – (where t = CR ) 2t / i U = U e- t 2t1 / i i 1 U Ue 2 = – t 2t1 / 1 e 2 = – t  1 T t ln2 2 = Now t / T 0 q = q et / 2T 0 0 1 q q e 4 = – 2 t = Tln4 = 2Tln2 \ 1 2 t 1 t 4 = Directions: Questions number 33 – 34 contain Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. 33. Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions. AIEEE-2010-11 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 (1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement- 1 (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false. 33. 1 Sol. m1 m2 v1 v2 If it is a completely inelastic collision then 1 1 2 2 1 2 m v +m v = m v +m v 1 1 2 2 1 2 m v m v v m m + = + 2 2 1 2 1 2 p p K.E 2m 2m = + as 1 2 p and p   both simultaneously cannot be zero therefore total KE cannot be lost. 34. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by Xrays, both V0 and Kmax increase. Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light. (1) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1. (2) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement- 1. (3) Statement-1 is false, Statement-2 is true. (4) Statement-1 is true, Statement-2 is false. 34. 4 Sol. Since the frequency of ultraviolet light is less than the frequency of X–rays, the energy of each incident photon will be more for X–rays K.E photoelectron = hn – j Stopping potential is to stop the fastest photoelectron 0 h V e e = n – j so, K.Emax and V0 both increases. But K.E ranges from zero to K.Emax because of loss of energy due to subsequent collisions before getting ejected and not due to range of frequencies in the incident light. 35. A ball is made of a material of density r where roil < r < rwater with roil and rwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position ? (1) (2) AIEEE-2010-12 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 (3) (4) 35. 2 Sol. oil water r < r < r Oil is the least dense of them so it should settle at the top with water at the base. Now the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So it will stay at the oil–water interface. 36. A particle is moving with velocity v = K(y ˆi + x ˆj)  , where K is a constant. The general equation for its path is (1) y = x2 + constant (2) y2 = x + constant (3) xy = constant (4) y2 = x2 + constant 36. 4 Sol. v = Ky ˆi +Kx ˆj  dx dy Ky, Kx dt dt = = dy dy dt Kx dx dt dx Ky = ´ = y dy = x dx y2 = x2 + c. 37. Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line XX' is given by (1) (2) (3) (4) 37. 1 Sol. The magnetic field in between because of each will be in opposite direction Bin between = 0 0 i ˆ i ˆ j ( j) 2 x 2 (2d x) m m - - p p - 0i 1 1 ˆ ( j) 2 x 2d x m   =  -  p  -  at x = d, Bin between = 0 for x < d, Bin between = (ˆj) for x > d, Bin between = (-ˆj) towards x net magnetic field will add up and direction will be (-ˆj) towards x’ net magnetic field will add up and direction will be(ˆj) AIEEE-2010-13 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 38. In the circuit shown below, the key K is closed at t = 0. The current through the battery is (1) 1 2 2 2 1 2 VR R R +R at t = 0 and 2 V R at t = ¥ (2) 2 V R at t = 0 and 1 2 1 2 V(R R ) R R + at t = ¥ (3) 2 V R at t = 0 and 1 2 2 2 1 2 VR R R +R at t = ¥ (4) 1 2 1 2 V(R R ) R R + at t = 0 and 2 V R at t = ¥ 38. 2 Sol. At t = 0, inductor behaves like an infinite resistance So at t = 0, 2 V i R = and at t = ¥ , inductor behaves like a conducting wire 1 2 e q 1 2 V V(R R ) i R R R + = = 39. The figure shows the position – time (x – t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is (1) 0.4 Ns (2) 0.8 Ns (3) 1.6 Ns (4) 0.2 Ns 39. 2 Sol. From the graph, it is a straight line so, uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph. Initial velocity = 2 1m/ s 2 = Final velocity = 2 1m/ s 2 – = – Pi = 0.4  N – s i Pj = -0.4  N – s f i J = P -P    = – 0.4 – 0.4 = – 0.8 N – s (J  = impulse) J  = 0.8 N–s Directions : Questions number 40 – 41 are based on the following paragraph. A nucleus of mass M + Dm is at rest and decays into two daughter nuclei of equal mass M 2 each. Speed of light is c. 40. The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then (1) E2 = 2E1 (2) E1 > E2 (3) E2 > E1 (4) E1 = 2E2 40. 3 Sol. After decay, the daughter nuclei will be more stable hence binding energy per nucleon will be more than that of their parent nucleus. AIEEE-2010-14 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 41. The speed of daughter nuclei is (1) m c M m D + D (2) 2 m c M D (3) m c M D (4) m c M m D + D 41. 2 Sol. Conserving the momentum 1 2 M M 0 V V 2 2 = – V1 = V2 …………….(1) 2 2 2 1 2 1 M 1 M mc . V . .V 2 2 2 2 D = + …………….(2) 2 2 1 M mc V 2 D = 2 2 1 2 mc V M D = 1 2 m V c M = D 42. A radioactive nucleus (initial mass number A and atomic number Z) emits 3 a-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be (1) A Z 8 Z 4 – – – (2) A Z 4 Z 8 – – – (3) A Z 12 Z 4 – – – (4) A Z 4 Z 2 – – – 42. 2 Sol. In positive beta decay a proton is transformed into a neutron and a positron is emitted. p+ ¾¾®n0 + e+ no. of neutrons initially was A – Z no. of neutrons after decay (A – Z) – 3 x 2 (due to alpha particles) + 2 x 1 (due to positive beta decay) The no. of proton will reduce by 8. [as 3 x 2 (due to alpha particles) + 2(due to positive beta decay)] Hence atomic number reduces by 8. 43. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field E  at the centre O is (1) 2 2 0 q ˆj 4p e r (2) – 2 2 0 q ˆj 4p e r (3) – 2 2 0 q ˆj 2p e r (4) 2 2 0 q ˆj 2p e r 43. 3 Sol. Linear charge density q rl = p  2 ˆ K.dq ˆ E dEsin ( j) sin ( j) r =  q – =  q – 2 K qr ˆ E d sin ( j) r r = q q – p  y q q x dq 2 0 K q ˆ sin ( j) r p = q – p  2 2 0 q ˆ ( j) 2 r = – p e AIEEE-2010-15 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 44. The combination of gates shown below yields (1) OR gate (2) NOT gate (3) XOR gate (4) NAND gate 44. 1 Sol. Truth table for given combination is A B X 0 0 0 0 1 1 1 0 1 1 1 1 This comes out to be truth table of OR gate 45. A diatomic ideal gas is used in a Car engine as the working substance. If during the adiabatic expansion part of the cycle, volume of the gas increases from V to 32V the efficiency of the engine is (1) 0.5 (2) 0.75 (3) 0.99 (4) 0.25 45. 2 Sol. The efficiency of cycle is 2 1 T 1 T h = – for adiabatic process TVg–1 = constant For diatomic gas 7 5 g = 1 1 1 1 2 2 T V g- = T V g- 1 2 1 2 1 V T T V g-= 7 1 5 1 2 T T (32) – = 5 2/ 5 2 = T (2 ) = T2 x 4 T1 = 4T2. 1 3 1 0.75 4 4 h = – = = 46. If a source of power 4 kW produces 1020 photons/second, the radiation belong to a part of the spectrum called (1) X–rays (2) ultraviolet rays (3) microwaves (4) g–rays 46. 1 Sol. 4 x 103 = 1020 x hf 3 20 34 4 10 f 10 6.023 10- = ´ ´ ´ f = 6.03 x 1016 Hz The obtained frequency lies in the band of X–rays. 47. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 x 10–3 are (1) 5, 1, 2 (2) 5, 1, 5 (3) 5, 5, 2 (4) 4, 4, 2 47. 1 AIEEE-2010-16 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 48. In a series LCR circuit R = 200 W and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30o. The power dissipated in the LCR circuit is (1) 305 W (2) 210 W (3) Zero W (4) 242 W 48. 4 Sol. The given circuit is under resonance as XL = XC Hence power dissipated in the circuit is V2 P R = = 242 W 49. Let there be a spherically symmetric charge distribution with charge density varying as 0 5 r (r) 4 R r = r – upto r = R, and r(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r(r < R) from the origin is given by (1) 0 0 4 r 5 r 3 3 R pr - e  (2) 0 0 r 5 r 4 3 R r - e  (3) 0 0 4 r 5 r 3 4 R r - e  (4) 0 0 r 5 r 3 4 R r - e  49. 2 Sol. Apply shell theorem the total charge upto distance r can be calculated as followed dq = 4pr2.dr.r 2 0 5 r 4 r .dr. 4 R   = p r  -    3 2 0 5 r 4 r dr dr 4 R   = pr  -    r 3 2 0 0 5 r dq q 4 r dr dr 4 R= = pr -   3 4 0 5 r 1 r 4 4 3 R 4   = pr  -    2 kq E r = 3 4 2 0 0 1 1 5 r r .4 4 r 4 3 4R   = pr  -  pe    0 0 r 5 r E 4 3 R r   =  -  e   50. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by 12 6 a b U(x) x x = - , where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U(x = ¥) – Uat equilibrium], D is (1) b2 2a (2) b2 12a (3) b2 4a (4) b2 6a 50. 3 Sol. 12 6 a b U(x) x x = - U(x = ¥) = 0 as, 13 7 dU 12a 6b F dx x x   = - = - +    at equilibrium, F = 0 AIEEE-2010-17 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 \ 6 2a x b = \ 2 at equilibrium 2 a b b U 2a 2a 4a b b = - = -   \ 2 at equilibrium b D U(x ) U 4a =  = ¥ - = 51. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 16 g cm–3, the dielectric constant of the liquid is (1) 4 (2) 3 (3) 2 (4) 1 51. 3 Sol. From F.B.D of sphere, using Lami’s theorem F tan mg = q ………………(i) when suspended in liquid, as q remains same, \ F' tan mg 1 d = q r - ………………(ii) T q F mg using (i) and (ii) F F' F where, F' mg mg 1 K d = = r - \ F F' mg mg K 1 d = r - or 1 K 2 1 d = = - r 52. Two conductors have the same resistance at 0oC but their temperature coefficients of resistance are a1 and a2. The respective temperature coefficients of their series and parallel combinations are nearly (1) 1 2 1 2 , 2 a + a a + a (2) 1 2 1 2, 2 a + a a + a (3) 1 2 1 2 1 2 , a a a + a a + a (4) 1 2 , 1 2 2 2 a + a a + a 52. 4 Sol. Let R0 be the initial resistance of both conductors \ At temperature q their resistance will be, 1 0 1 2 0 2 R = R (1+ a q) and R = R (1+ a q) for, series combination, Rs = R1 + R2 s0 s 0 1 0 2 R (1+ a q) = R (1+ a q) +R (1+ a q) where s0 0 0 0 R = R + R = 2R \ 0 s 0 0 1 2 2R (1+ a q) = 2R +R q(a + a ) or 1 2 s 2 a + a a = for parallel combination, 1 2 p 1 2 R R R R R = + AIEEE-2010-18 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 0 1 0 2 p0 p 0 1 0 2 R (1 )R (1 ) R (1 ) R (1 ) R (1 ) + a q + a q + a q = + a q + + a q where, 0 0 0 p0 0 0 R R R R R R 2 = = + \ 2 0 0 1 2 1 2 p 0 1 2 R R (1 ) (1 ) 2 R (2 ) + a q + a q + a a q + a q = + a q + a q as a1 and a2 are small quantities \ 1 2 a a is negligible or 1 2 1 2 p 1 2 1 2 [1 ( ) ] 2 ( ) 2 a + a a + a a = = - a + a q + a + a q as 2 1 2 (a + a ) is negligible \ 1 2 p 2 a + a a = 53. A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ‘P’ is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of ‘P’ when t = 2 s is nearly (1) 13 m/s2 (2) 12 m/s2 (3) 7.2 m/s2 (4) 14 m/s2 53. 4 Sol. S = t3 + 5 \ speed, ds 2 v 3t dt = = and rate of change of speed dv 6t dt = = \ tangential acceleration at t = 2s, at = 6 x 2 = 12 m/s2 at t = 2s, v = 3(2)2 = 12 m/s \ centripetal acceleration, 2 2 c v 144 a m/ s R 20 = = \ net acceleration = 2 2 t i a + a » 14 m/ s2 54. Two fixed frictionless inclined plane making an angle 30o and 60o with the vertical are shown in the figure. Two block A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B ? (1) 4.9 ms–2 in horizontal direction (2) 9.8 ms–2 in vertical direction (3) zero (4) 4.9 ms–2 in vertical direction 54. 4 Sol. mg sin q = ma \ a = g sin q where a is along the inclined plane \ vertical component of acceleration is g sin2 q \ relative vertical acceleration of A with respect to B is AIEEE-2010-19 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 2 2 g 2 g[sin 60 sin 30] 4.9 m/ s 2 - = = in vertical direction. 55. For a particle in uniform circular motion the acceleration a  at a point P(R, q) on the circle of radius R is (here q is measured from the x–axis) (1) v2 v2 cos ˆi sin ˆj R R - q + q (2) v2 v2 sin ˆi cos ˆj R R - q + q (3) v2 v2 cos ˆi sin ˆj R R - q - q (4) v2 v2 ˆi ˆj R R + 55. 3 Sol. For a particle in uniform circular motion, v2 a R =  towards centre of circle \ v2 a ( cos ˆi sin ˆj) R = - q - q  or v2 v2 a cos ˆi sin ˆj R R = - q - q  x y P (R, q) ac ac Directions: Questions number 56 – 58 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index 0 2 m(I) = m + m I , where m0 and m2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. 56. As the beam enters the medium, it will (1) diverge (2) converge (3) diverge near the axis and converge near the periphery (4) travel as a cylindrical beam 56. 2 Sol. As intensity is maximum at axis, \ m will be maximum and speed will be minimum on the axis of the beam. \ beam will converge. 57. The initial shape of the wave front of the beam is (1) convex (2) concave (3) convex near the axis and concave near the periphery (4) planar 57. 4 Sol. For a parallel cylinderical beam, wavefront will be planar. 58. The speed of light in the medium is (1) minimum on the axis of the beam (2) the same everywhere in the beam (3) directly proportional to the intensity I (4) maximum on the axis of the beam 58. 1 AIEEE-2010-20 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 59. A small particle of mass m is projected at an angle q with the x-axis with an initial velocity v0 in the x-y plane as shown in the figure. At a time 0 v sin t g q < , the angular momentum of the particle is (1) 2 0 -mgv t cosqˆj (2) 0 mgv t cosqkˆ (3) 2 0 1 ˆ mgv t cos k 2 - q (4) 2 0 1 ˆ mgv t cos i 2 q where ˆi, ˆj and kˆ are unit vectors along x, y and z–axis respectively. 59. 3 Sol. L = m(r ´ v)    2 0 0 0 0 ˆ 1 ˆ ˆ ˆ L m v cos t i (v sin t gt )j v cos i (v sin gt)j 2 =  q + q -  ´  q + q -         0 1 ˆ mv cos t gt k 2   = q -    2 0 1 ˆ mgv t cos k 2 = - q 60. The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by t x y 0.02(m)sin 2 0.04(s) 0.50(m)   =  p -    . The tension in the string is (1) 4.0 N (2) 12.5 N (3) 0.5 N (4) 6.25 N 60. 4 Sol. 2 2 2 2 2 (2 / 0.004) T v 0.04 6.25 N k (2 / 0.50) = m = m w = p = p      61. Let cos(a + b) = 4 5 and let sin(a – b) = 5 13 , where 0 £ a, b £ 4 p , then tan 2a = (1) 56 33 (2) 19 12 (3) 20 7 (4) 25 16 61. 1 cos (a + b) = 4 5  tan(a + b) = 3 4 sin(a – b) = 5 13  tan(a – b) = 5 12 tan 2a = tan(a + b + a – b) = 3 5 4 12 56 1 3 5 33 4 12 + = - 62. Let S be a non-empty subset of R. Consider the following statement: P: There is a rational number x Î S such that x > 0. Which of the following statements is the negation of the statement P ? (1) There is no rational number x Î S such that x £ 0 (2) Every rational number x Î S satisfies x £ 0 AIEEE-2010-21 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 (3) x Î S and x £ 0  x is not rational (4) There is a rational number x Î S such that x £ 0 62. 2 P: there is a rational number x Î S such that x > 0 ~P: Every rational number x Î S satisfies x £ 0 63. Let a = ˆj – kˆ and c = ˆi – ˆj – kˆ   . Then vector b  satisfying a ´b + c = 0 and a ×b       = 3 is (1) 2ˆi – ˆj + 2kˆ (2) ˆi – ˆj – 2kˆ (3) ˆi + ˆj – 2kˆ (4) -ˆi + ˆj – 2kˆ 63. 4 c = b ´ a     b × c = 0    ( ) ( ) 1 2 3 b ˆi + b ˆj + b kˆ × ˆi – ˆj – kˆ = 0 b1 – b2 – b3 = 0 and a ×b   = 3  b2 – b3 = 3 b1 = b2 + b3 = 3 + 2b3 ( ) ( ) 3 3 3 b = 3 + 2b ˆi + 3 + b ˆj + b kˆ  . 64. The equation of the tangent to the curve y = x + 2 4 x , that is parallel to the x-axis, is (1) y = 1 (2) y = 2 (3) y = 3 (4) y = 0 64. 3 Parallel to x-axis  dy dx = 0  3 8 1 x – = 0  x = 2  y = 3 Equation of tangent is y – 3 = 0(x – 2)  y – 3 = 0 65. Solution of the differential equation cos x dy = y(sin x – y) dx, 0 < x < 2 p is (1) y sec x = tan x + c (2) y tan x = sec x + c (3) tan x = (sec x + c)y (4) sec x = (tan x + c)y 65. 4 cos x dy = y(sin x – y) dx dy 2 y tanx y sec x dx = - 2 1 dy 1 tanx sec x y dx y - = - Let 1 t y = 2 1 dy dt y dx dx - = dy dx - – t tan x = –sec x  dt dx + (tan x) t = sec x. I.F. = e tan x dx = sec x Solution is t(I.F) =  (I.F) sec x dx AIEEE-2010-22 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 1 y sec x = tan x + c 66. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = 3 2 p is (1) 4 2 + 2 (2) 4 2 – 1 (3) 4 2 + 1 (4) 4 2 – 2 66. 4 ( ) ( ) ( ) 5 3 4 4 2 0 5 4 4 cos x sinx dx sinx cos x dx cos x sinx 4 2 2 p p p p p  - +  - +  - = - cos x sin x 4 p 5 4 p 3 2 p 0 2p p 67. If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is (1) 2x + 1 = 0 (2) x = –1 (3) 2x – 1 = 0 (4) x = 1 67. 2 The locus of perpendicular tangents is directrix i.e, x = –a; x = –1 68. If the vectors a = ˆi - ˆj + 2kˆ, b = 2ˆi + 4ˆj + kˆ and c = lˆi + ˆj + mkˆ    are mutually orthogonal, then (l, m) = (1) (2, –3) (2) (–2, 3) (3) (3, –2) (4) (–3, 2) 68. 4 a ×b = 0, b × c = 0, c ×a = 0        2l + 4 + m = 0 l – 1 + 2m = 0 Solving we get: l = –3, m = 2 69. Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S = m p , m, n, p and q are integers such that n, q 0 and qm = pn n q    ¹     . Then (1) neither R nor S is an equivalence relation (2) S is an equivalence relation but R is not an equivalence relation (3) R and S both are equivalence relations (4) R is an equivalence relation but S is not an equivalence relation 69. 2 xRy need not implies yRx S: m p s n q Û qm = pn m m s n n reflexive AIEEE-2010-23 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 m p s n q  p m s q n symmetric m p p r s , s n q q s  qm = pn, ps = rq  ms = rn transitive. S is an equivalence relation. 70. Let f: R ® R be defined by f(x) = k 2x, if x 1 2x 3, if x 1 - £ -   + > – . If f has a local minimum at x = –1, then a possible value of k is (1) 0 (2) 1 2 – (3) –1 (4) 1 70. 3 f(x) = k – 2x if x £ –1 = 2x + 3 if x > –1 2x + 3 k – 2x 1 –1 x 1 lim ®- – f(x) £ –1 This is true where k = –1 71. The number of 3 ´ 3 non-singular matrices, with four entries as 1 and all other entries as 0, is (1) 5 (2) 6 (3) at least 7 (4) less than 4 71. 3 First row with exactly one zero; total number of cases = 6 First row 2 zeros we get more cases Total we get more than 7. Directions: Questions Number 72 to 76 are Assertion – Reason type questions. Each of these questions contains two statements. Statement-1: (Assertion) and Statement-2: (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. 72. Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ….., 20}. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is 1 85 . Statement-2: If the four chosen numbers from an AP, then the set of all possible values of common difference is {±1, ±2, ±3, ±4, ±5}. (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 72. 2 AIEEE-2010-24 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 N(S) = 20C4 Statement-1: common difference is 1; total number of cases = 17 common difference is 2; total number of cases = 14 common difference is 3; total number of cases = 11 common difference is 4; total number of cases = 8 common difference is 5; total number of cases = 5 common difference is 6; total number of cases = 2 Prob. = 20 4 17 14 11 8 5 2 1 C 85 + + + + + = . 73. Statement-1: The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5. Statement-2: The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4). (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 73. 1 A(3, 1, 6); B = (1, 3, 4) Mid-point of AB = (2, 2, 5) lies on the plane. and d.r’s of AB = (2, –2, 2) d.r’s Of normal to plane = (1, –1, 1). AB is perpendicular bisector \ A is image of B Statement-2 is correct but it is not correct explanation. 74. Let S1 = ( ) 10 10 j j 1 j j 1 C =  – , S2 = 10 10 j j 1 j C =  and S3 = 10 2 10 j j 1 j C =  . Statement-1: S3 = 55 ´ 29 Statement-2: S1 = 90 ´ 28 and S2 = 10 ´ 28. (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 74. 2 S1 = ( ) ( )( ) ( ) ( ) ( ( )) 10 10 8 j 1 j 2 10! 8! j j 1 90 90 2 = j j 1 j 2 ! 10 j ! = j 2 ! 8 j 2 ! – = = × – – – – – –   . S2 = ( ) ( ( )) ( ) ( ( )) 10 10 9 j 1 j 1 10! 9! j 10 10 2 = j j 1 ! 9 j 1 ! = j 1 ! 9 j 1 ! = = × – – – – – –   . S3 = ( ) ( ) ( ) 10 10 10 10 10 j j j 1 j 1 j 1 10! j j 1 j j j 1 C j C = j! 10 j ! = =  – +  = – = –    = 90 . 28 + 10 . 29 = 90 . 28 + 20 . 28 = 110 . 28 = 55 . 29. 75. Let A be a 2 ´ 2 matrix with non-zero entries and let A2 = I, where I is 2 ´ 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. Statement-1: Tr(A) = 0 Statement-2: |A| = 1 (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false AIEEE-2010-25 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 75. 2 Let A = a b c d , abcd ¹ 0 A2 = a b a b c d c d×   A2 = 2 2 a bc ab bd ac cd bc d + ++ +   a2 + bc = 1, bc + d2 = 1 ab + bd = ac + cd = 0 c ¹ 0 and b ¹ 0  a + d = 0 Trace A = a + d = 0 |A| = ad – bc = –a2 – bc = –1. 76. Let f: R ® R be a continuous function defined by f(x) = x x 1 e + 2e- . Statement-1: f(c) = 1 3 , for some c Î R. Statement-2: 0 < f(x) £ 1 2 2 , for all x Î R (1) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1 (2) Statement-1 is true, Statement-2 is false (3) Statement-1 is false, Statement-2 is true (4) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1 76. 4 f(x) = x x x 2x 1 e e 2e- e 2 = + + f¢(x) = ( ) ( ) 2x x 2x x 2x 2 2 e 2 e 2e e e + + - × f¢(x) = 0  e2x + 2 = 2e2x e2x = 2  ex = 2 maximum f(x) = 2 1 4 2 2 = 0 < f(x) £ 1 2 2 " x Î R Since 0 < 1 1 3 2 2 <  for some c Î R f(c) = 1 3 77. For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is (1) There is a regular polygon with r 1 R 2 = (2) There is a regular polygon with r 2 R 3 = AIEEE-2010-26 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 (3) There is a regular polygon with r 3 R 2 = (4) There is a regular polygon with r 1 R 2 = 77. 2 r = a cot 2 n p ‘a’ is side of polygon. R = a cosec 2 n p r cot n cos R cosec n n p = = p p 2 cos n 3 p ¹ for any n Î N. 78. If a and b are the roots of the equation x2 – x + 1 = 0, then a2009 + b2009 = (1) –1 (2) 1 (3) 2 (4) –2 78. 2 x2 – x + 1 = 0  x = 1 1 4 2 ± - x = 1 3 i 2 ± a = 1 3 i 2 2 + , b = 1 i 3 2 2 - a = cos isin 3 3 p + p , b = cos isin 3 3 p - p a2009 + b2009 = 2cos2009 3 p = 2 2 2cos 668 2cos 3 3  p p  p + p + = p +    = 2 1 2cos 2 1 3 2 p - = - - = 79. The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals (1) 1 (2) 2 (3) ¥ (4) 0 79. 1 Let z = x + iy |z – 1| = |z + 1|  Re z = 0  x = 0 |z – 1| = |z – i|  x = y |z + 1| = |z – i|  y = –x Only (0, 0) will satisfy all conditions.  Number of complex number z = 1 80. A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle q with the positive z-axis, then q equals (1) 45° (2) 60° (3) 75° (4) 30° 80. 2 AIEEE-2010-27 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004  = cos 45° = 1 2 m = cos 120° = 1 2 - n = cos q where q is the angle which line makes with positive z-axis. Now 2 + m2 + n2 = 1  1 1 2 4 + + cos2q = 1 cos2q = 1 4  cos q = 1 2 (q Being acute)  q = 3 p . 81. The line L given by x y 5 b + = 1 passes through the point (13, 32). The line K is parallel to L and has the equation x y c 3 + = 1. Then the distance between L and K is (1) 17 (2) 17 15 (3) 23 17 (4) 23 15 81. 3 Slope of line L = b 5 - Slope of line K = 3 c - Line L is parallel to line k.  b 3 5 c =  bc = 15 (13, 32) is a point on L.  13 32 32 8 1 5 b b 5 + =  = -  b = –20  c = 3 4 - Equation of K: y – 4x = 3 Distance between L and K = 52 32 3 23 17 17 - + = 82. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ...... = a10 = 150 and a10, a11, ...... are in A.P. with common difference –2, then the time taken by him to count all notes is (1) 34 minutes (2) 125 minutes (3) 135 minutes (4) 24 minutes 82. 1 Till 10th minute number of counted notes = 1500 3000 = n 2 [2 ´ 148 + (n – 1)(–2)] = n[148 – n + 1] AIEEE-2010-28 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 n2 – 149n + 3000 = 0 n = 125, 24 n = 125 is not possible. Total time = 24 + 10 = 34 minutes. 83. Let f: R ® R be a positive increasing function with x f(3x) lim ®¥ f(x) = 1. Then x f(2x) lim ®¥ f(x) = (1) 2 3 (2) 3 2 (3) 3 (4) 1 83. 4 f(x) is a positive increasing function  0 < f(x) < f(2x) < f(3x)  0 < 1 < f(2x) f(3x) f(x) f(x) <  x x x f(2x) f(3x) lim1 lim lim ®¥ ®¥ f(x) ®¥ f(x) £ £ By sandwich theorem.  x f(2x) lim ®¥ f(x) = 1 84. Let p(x) be a function defined on R such that p¢(x) = p¢(1 – x), for all x Î [0, 1], p(0) = 1 and p(1) = 41. Then 1 0 p(x) dx equals (1) 21 (2) 41 (3) 42 (4) 41 84. 1 p¢(x) = p¢(1 – x)  p(x) = –p(1 – x) + c at x = 0 p(0) = –p(1) + c  42 = c now p(x) = –p(1 – x) + 42  p(x) + p(1 – x) = 42 I = 1 1 0 0 p(x) dx = p(1- x) dx 2 I = 1 0 (42) dx  I = 21. 85. Let f: (–1, 1) ® R be a differentiable function with f(0) = –1 and f¢(0) = 1. Let g(x) = [f(2f(x) + 2)]2. Then g¢(0) = (1) –4 (2) 0 (3) –2 (4) 4 85. 1 g¢(x) = 2(f(2f(x) + 2)) ( ( )) d f 2f(x) 2 dx + = 2f(2f(x) + 2) f¢(2f(x) + 2) . (2f¢(x))  g¢(0) = 2f(2f(0) + 2) . f¢(2f(0) + 2) . 2(f¢(0) = 4f(0) f¢(0) = 4(–1) (1) = –4 AIEEE-2010-29 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 86. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is (1) 36 (2) 66 (3) 108 (4) 3 86. 3 Total number of ways = 3C2 ´ 9C2 = 3 ´ 9 8 2 ´ = 3 ´ 36 = 108 87. Consider the system of linear equations: x1 + 2x2 + x3 = 3 2x1 + 3x2 + x3 = 3 3x1 + 5x2 + 2x3 = 1 The system has (1) exactly 3 solutions (2) a unique solution (3) no solution (4) infinite number of solutions 87. 3 D = 1 2 1 2 3 1 3 5 2 = 0 D1 = 3 2 1 3 3 1 1 5 2 ¹ 0  Given system, does not have any solution.  No solution. 88. An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colour is (1) 2 7 (2) 1 21 (3) 2 23 (4) 1 3 88. 1 n(S) = 9C3 n(E) = 3C1 ´ 4C1 ´ 2C1 Probability = 9 3 3 4 2 24 3! 24 6 2 6! C 9! 9 8 7 7 ´ ´ = ´ ´ = ´ = ´ ´ . 89. For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is (1) 11 2 (2) 6 (3) 13 2 (4) 5 2 89. 1 sx 2 = 4 sy 2 = 5 x = 2 y = 4 AIEEE-2010-30 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 i x 5  = 2  xi = 10;  yi = 20 sx 2 = 2 ( )2 ( 2 ) i i 1 1 x x y 16 2 5 - =  -  xi 2 = 40  yi 2 = 105 sz 2 = ( ) ( ) 2 2 2 i i 1 x y 1 145 90 55 11 x y 40 105 9 10 2 10 10 10 2 + -  +  - = + - = = = 90. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if (1) –35 < m < 15 (2) 15 < m < 65 (3) 35 < m < 85 (4) –85 < m < –35 90. 1 Circle x2 + y2 – 4x – 8y – 5 = 0 Centre = (2, 4), Radius = 4 + 16 + 5 = 5 If circle is intersecting line 3x – 4y = m at two distinct points.  length of perpendicular from centre < radius  6 16 m 5 - - < 5  |10 + m| < 25  –25 < m + 10 < 25  –35 < m < 15.  * * * AIEEE-2010-31 (Hyderabad Classes) Limited. 5-9-14/B, Saifabad, (Opp. Secretariat) Hyderabad. 500 063. Phone: 040-66777000 – 03 Fax: 040-66777004 READ THE FOLLOWING INSTRUCTIONS CAREFULLY: